text:
### The Challenge

The broader US stock market has periods of strong correlation. This makes it difficult to implement a portfolio that consistently outperforms the S&P 500 benchmark. The ETFs chosen for the Liquid ETF Competition were selected for their independence from the market, covering very different underlying assets with their own independent variables. Your challenge is to turn this basket into Alpha Streams that are optimized for live Sharpe Ratio.

Whether you are well-seasoned with QuantConnect's API or are new to the platform, we invite you to participate. Your edge will come from using your unique background to create algorithms that no one else has the perspective to make. In addition to the chance of winning a prize, algorithms submitted to Alpha Streams will be available for licensing by other institutional clients after 90 days.

### Requirements

To ensure all Alpha Streams are correctly configured, QuantConnect has provided a starter template. Additionally, all algorithms must:

• Be submitted by January 20, 2020
• Comply with Alpha Streams submission criteria, including using the appropriate transaction models
• Must use only Minute resolution data
• Have a completed five-year backtest attached, using SPY benchmark
• Be long only. Use of Inverse ETFs is allowed
• Be profitable in the backtest
• Have a Probabilistic Sharpe Ratio greater than 80%
• Average 5 trades per week.
• Trade only using the ETFs provided

72 ETFs were chosen for their liquidity and unique underlying assets. Each has over $200 million daily volume, ensuring sufficient order book depth for institutional capital. The universe includes inverse ETFs, which allow you to express short sentiment. You can view a full implementation, along with a list of the constituents, on GitHub. ### Applications To apply, simply create an Alpha Stream using the eligible universe module and submit it for licensing to Alpha Streams. From the tag-wizard, select the universe of your alpha to enter it into a category. The submission process is demonstrated in this video. ### Support There are two support channels that we encourage you to go through early and often, depending on what you need: • QC Support: For API, algorithm mechanics, and data questions. • QC Forum: For ideas, finding a team, or trading concepts. Both the QC team and the QC community remain committed to helping you submit algorithms that best reflect the strength of your ideas. However, in the interest of fairness, we cannot assist with idea generation. Don't be shy about asking questions! ### Evaluation Algorithms will be displayed on the leaderboard according to the live Sharpe ratio they accrue during the evaluation period from January 21, 2020 through March 20, 2020. The leaderboard will be updated daily according to the live results. The ranking criteria can be expressed as: Where is portfolio returns, is the risk-free rate, and is the standard deviation of excess returns. Final judgment of winners is reserved for our sponsors, who will view the performances of all submitted algorithms and compare them with their internal benchmarks. For this reason, an algorithm with a lower Sharpe ratio may ultimately be chosen over one with a higher Sharpe ratio because it has certain characteristics that make it more amenable to the sponsor. High daily turnover, lower return over maximum drawdown (annualized return divided by maximum drawdown), low correlation with common benchmarks like SPY, and low correlation with other submitted algorithms will all contribute to final judgment. Sharpe ratios will be calculated in real time and update the leaderboard automatically. Algorithms that encounter runtime errors or fail to produce any trades during a trading day will be subject to disqualification. ### Competition Winners ### Competition Prizes The Competition winners will be decided on March 20, 2020 at 8 pm EST. The authors of the top three (3) algorithms per category will be notified via email. Winners will be required to respond to the notification email no later than April 20, 2020 at 11:59 pm EST. Failure to respond by the deadline, or failure to meet the participant qualifications, may result in the prize being forfeited and awarded to the author of the next highest algorithm. Out of the algorithms submitted which are accepted into the competition, the top three algorithms will receive prizes according to the following distribution: Place Prize 1st$8,000
2nd $1,500 3rd$500

In the event of a tie, the Competition prize will be split evenly between the winners. For example, in the event of a tie for second place, the prize for second and the prize for third will be combined, halved, and given equally to the tied algorithms.

Odds of winning will depend on the total number of entrants.

Participants may win a maximum of one (1) prize. In the event a Participant has more than one algorithm that qualifies for a prize, they will be entitled only to the prize for their top placing submission. The Participant’s remaining algorithm(s) would then cede its place(s) to the next highest scoring algorithm.

### 1. Overview

This Themed Algorithm Development Competition (the “Competition”) is hosted by QuantConnect Corporation, a Delaware Corporation (“QuantConnect”).

All algorithms accepted into the Alpha Marketplace that meet the criteria in the "Eligibility" and "Submissions" sections will be automatically entered into the Competition. All eligible algorithms are therefore required to follow the standard guidelines that have been set for listed Alphas which include using a reliable data source, calling the History and WarmUp methods, and being grounded in reality. Learn more about creating and submitting and Alpha in the documentation.

### 2. Eligibility

To be eligible to take part in the Competition, Participants must enter their algorithm on or before 11:59 pm EST on January 20, 2020.

Algorithms must meet the mechanical and performance criteria set forth in the “Judgement” section.

Participants must have a registered QuantConnect account and be registered Alpha Stream developers. Registration is available here. QuantConnect registration and Alpha Streams registration are both free. Additional identity verification may be required before prize distribution.

Participants must be at least 18 years old or older at time of entry.

Participants must be from a country or region where their entry is not prohibited by law.

There is no fee for entry and no purchase is necessary for this contest of skill.

### 3. Submissions

Submitted algorithms will be evaluated by their performance with live market data from January 21, 2020 through March 20, 2020. They will not be able to be changed or updated during this period.

To be eligible for participation, algorithms must have been submitted as an Alpha Stream and include a project name, description, subscription price, and exclusive subscription price, with a full backtest attached that can be reproduced perfectly by running an unedited version of the submitted algorithm at the time of submission. They must trade within the pre-defined universe, accessible from the "Create New Algorithm" section of terminal:

• “Liquid ETF Competition Universe”

Submissions must possess a Probabilistic Sharpe Ratio of greater than 80%, the formula for which is viewable here.

Submissions must also pass a standard manual approval process during which QuantConnect reserves the right to reject it for any reason. These include but are not limited to: not adhering to Competition criteria, suspected overfitting, making use of information that would not have been knowable at time of trade (lookahead bias), improper use of the API, algorithm lacking resiliency by not using History() or WarmUp() methods, using unreliable data sources, uncertainty about the author’s identity, technical failures during the approval process, or any violations of our Terms and Conditions.

A Participant can decide to remove their algorithm from consideration in the Competition at any time. This will cause them to forgo any potential prize or features associated with winning the Competition. The algorithm itself will remain in the marketplace for one (1) year after removal according to Section 2.4 of the Alpha Stream Author Agreement.

Limit of one algorithm submission per day. QuantConnect reserves the right to limit or refuse the submissions of Participants who we deem, in our sole discretion, to be abusing the submission process by, for example, repeatedly attempting to submit more than one algorithm per day or submitting algorithms from multiple accounts..

Past winners and all currently performing algorithms will be available at quantconnect.com/competition.

Algorithms must use data provided by QuantConnect and cannot make use of external or other non-permitted data. They must not incorporate or include any material, non-public information.

Participant agrees the Alpha Stream Author Agreement governs the Participant’s relationship with the Competition sponsor except as set forth in these Competition rules. Participant agrees that all algorithms submitted for the Competition will be available to license exclusively to the Competition sponsor until June 30, 2020, after which they become publicly available on the Alpha Streams marketplace. Participants retain intellectual property rights over their submissions in accordance with Section 6 of the Alpha Stream Author Agreement.

### 4. Judgement

Algorithms will be displayed on the leaderboard according to the live Sharpe ratio they accrue during the evaluation period from January 21, 2020 through March 20, 2020. The leaderboard will be updated daily according to the live results. The ranking criteria can be expressed as:

Where is portfolio returns, is the risk-free rate, and is the standard deviation of excess returns.

Final judgment of winners is reserved for our sponsors, who will view the performances of all submitted algorithms and compare them with their internal benchmarks. For this reason, an algorithm with a lower Sharpe ratio may ultimately be chosen over one with a higher Sharpe ratio because it has certain characteristics that make it more amenable to the sponsor. High daily turnover, lower return over maximum drawdown (annualized return divided by maximum drawdown), low correlation with common benchmarks like SPY, and low correlation with other submitted algorithms will all contribute to final judgment.

Sharpe ratios will be calculated in real time and update the leaderboard automatically. Algorithms that encounter runtime errors or fail to produce any trades during a trading day will be subject to disqualification.

All algorithms will be run with a default of a simulated $1,000,000 starting balance. Slippage models, fee models, settlement models, brokerage models, fill models, and buying power models will all be set to the applicable Alpha Stream Model. Any models that the Participant has applied will be overridden at the approval phase. Backtests must run over a period of 5 years from the date of submission, without runtime errors, and restart in a stateful way which resets any indicators by using the History API. Algorithms must be long only. Algorithms must be profitable and not have a drawdown recovery of greater than 6 months in their backtest Algorithms must use Minute resolution data only Algorithms must use data that has been approved by QuantConnect and cannot use the download() function. Each participant may have only one QuantConnect account. If the Participant submits entries from more than one account, all entries may be disqualified. Collaboration on algorithms is allowed, but QuantConnect will not be involved with prize division. Submissions are the sole responsibility of the registered account holder of the account submitting the algorithm. By submitting an algorithm, the participant represents that he/she holds all rights necessary to submit the algorithm, including all copyright and other intellectual property rights. Any algorithm that takes longer than 6 hours on QuantConnect’s servers to run a backtest will be disqualified. ### 6. Awards, Payout The Competition winners will be decided on March 20, 2020 at 8 pm EST. The authors of the top three (3) algorithms will be notified via email. Winners will be required to respond to the notification email no later than April 20, 2020 at 11:59 pm EST. Failure to respond by the deadline, or failure to meet the participant qualifications, may result in the prize being forfeited and awarded to the author of the next highest algorithm. Out of the algorithms submitted which are accepted into the competition, the top three algorithms per category will receive prizes according to the following distribution: Place Prize 1st$8,000
2nd $1,500 3rd$500

In the event of a tie, the Competition prize will be split evenly between the winners. For example, in the event of a tie for second place, the prize for second and the prize for third will be combined, halved, and given equally to the tied algorithms.

Odds of winning will depend on the total number of entrants.

Participants may win a maximum of one (1) prize. In the event a Participant has more than one algorithm that qualifies for a prize, they will be entitled only to the prize for their top placing submission. The Participant’s remaining algorithm(s) would then cede its place(s) to the next highest scoring algorithm.

### 7. Other Conditions

The Competition begins at 12:00 am EST on November 1, 2019. All algorithms that are approved by QuantConnect before 11:59 pm January 20, 2019, will be eligible for consideration.

QuantConnect reserves the right to extend the length of the competition’s submission period if fewer than 100 algorithms are accepted by the submission deadline.

Prizes are non-transferrable.

QuantConnect reserves the right to alter, amend, modify, suspend, or terminate this Competition or any aspect of it at any time with no advance notice. These include but are not limited to prize value, prize distribution, and scoring criteria. In the event of a change, QuantConnect will make reasonable efforts to notify all Participants by email.

Algorithms that are not executed or processed successfully, for whatever reason, will be disqualified. These include but are not limited to any malfunction or other problem with the Competition website or any error in the collection, processing, or retention of any submitted algorithm.

QuantConnect may remove algorithms from the Competition for any reason, including underperformance, a Participant misrepresenting their identity, or intellectual property concerns.

The Participant gives his/her express consent for QuantConnect to use or publish, without additional compensation and in any medium, information for publicity purposes including Participant name, written and spoken statements, and algorithm performance for all or part of the Competition.

All current employees, interns, and contractors of QuantConnect; and their immediate family members or household members are prohibited from participating in the Competition.

All federal, state, provincial, and local laws and regulations apply.

Whenever possible, each provision of these rules will be interpreted in such a manner as to be effective and valid under applicable law. In the event that any provision is not enforceable or invalid under applicable law, all remaining provisions will remain in full force and effect.

All taxes imposed on prizes are the sole responsibility of the winners. Payments to potential winners are subject to the express requirement that they submit all documentation requested by QuantConnect for compliance with applicable state, federal, local and foreign (including provincial) tax reporting and withholding requirements. If a potential winner fails to provide any required documentation or comply with applicable laws, the Prize may be forfeited and Competition Sponsor may select an alternative potential winner. Any winners who are U.S. residents will receive an IRS Form-1099 in the amount of their Prize if required.

QuantConnect is not responsible for the actions of Participants in this Competition, and any attempts to circumvent the rules or otherwise interfere with the Competition will result in disqualification. QuantConnect reserves sole discretion in deeming when a Participant is acting in violation of the rules and administering the appropriate consequence.

These rules shall be governed by and subject to the QuantConnect Terms of Use including the jurisdictional and dispute processes specified therein.

### 8. Liability

To the maximum extent permitted by law, you indemnify and agree to keep indemnified QuantConnect at all times from and against any liability, claims, demands, losses, damages, costs and expenses resulting from any of your acts, defaults or omissions and/or a breach of any warranty set forth herein. To the maximum extent permitted by law, you agree to defend, indemnify and hold harmless QuantConnect from and against any and all claims, actions, suits or proceedings, as well as any and all losses, liabilities, damages, costs and expenses (including reasonable attorneys fees) arising out of or accruing from: (a) your algorithm or other material uploaded or otherwise provided by you that infringes any third party proprietary rights, intellectual property rights, industrial property rights, personal or moral rights or any other rights, including without limitation, copyright, trademark, patent, trade secret, privacy, publicity or confidentiality obligations, or defames any person; (b) any misrepresentation made by you in connection with the Competition; (c) any non-compliance by you with these rules or any applicable U.S. or foreign state or federal law; (d) claims brought by persons or entities other than the parties to these rules arising from or related to your involvement with the Competition; and (e) your acceptance, possession, misuse or use of any Prize, or your participation in the Competition and any Competition-related activity.


text:
# Combinatorics: Gal Kronenberg (Oxford)

Speaker: Gal Kronenberg (Oxford)

Title: The chromatic index of random multigraphs

Abstract:
For a (multi)graph G=(V,E), we denote by χ'(G) the minimum number of colors needed to color the edges of G properly. Clearly, χ'(G)≥Δ. Let ρ(G)=max{ e(S)/ \floor{|S|/2} | S⊆V }. By the fact that every color class forms a matching, we have that χ'(G)≥ρ(G). We say that a multigraph is first class if the upper bound for the chromatic index matches the trivial lower bound of max{Δ,ρ(G)}, that is, if χ'(G) = max{Δ,ρ(G)}. In the 70s, Goldberg, and independently Seymour, conjectured that for any multigraph G, χ'(G) ≤  max{Δ+1,ρ(G)}. We show that their conjecture (in a stronger form) is true for random multigraphs, and that a typical multigraph is, in fact, first class.

Joint work with Penny Haxell and Michael Krivelevich.

## Date:

Mon, 30/12/2019 - 10:00 to 12:00

## Location:

C-400, CS building

format: plain


text:
0
Research Papers

# Solar Thermal Electrolytic Process for the Production of Zn From ZnO: An Ionic Conductivity Study

[+] Author and Article Information
L. Venstrom, K. Krueger, N. Leonard, B. Tomlinson, S. Duncan

Department of Mechanical Engineering, Valparaiso University, Valparaiso, IN 46383

R. D. Palumbo1

Department of Mechanical Engineering, Valparaiso University, Valparaiso, IN 46383robert.palumbo@valpo.edu

1

Corresponding author.

J. Sol. Energy Eng 131(3), 031005 (Jun 10, 2009) (9 pages) doi:10.1115/1.3142802 History: Received June 05, 2008; Revised December 17, 2008; Published June 10, 2009

## Abstract

The ionic conductivities of mixtures of ZnO in $Na3AlF6$ and in $xCaF2–yNa3AlF6$ mixtures were established with a swept-sine measurement technique. A millivolt sinusoidal voltage at frequencies from 1000 Hz to 25,000 Hz was impressed on a system containing the electrolytes. The system’s frequency response was used to establish the conductivities. The influence of these conductivities on the potential of a solar thermal electrolytic process was evaluated using two process performance parameters: the back-work ratio and the fraction of minimum solar thermal energy required to drive the metal production reaction. We found the conductivity of mixtures of $ZnO–Na3AlF6$ to be independent of the concentration of ZnO for weight percentages of ZnO from 0.5% to 5%. For temperatures 1240–1325 K the conductivity is close to that of pure $Na3AlF6$, $3±0.5 Ω−1 cm−1$. At temperatures from 1350 K to 1425 K it jumps to $6±0.5 Ω−1 cm−1$ When $CaF2$ is added to the mixture, the electrolyte’s conductivity drops. We thus expect that calcium cations are not present to any important extent in the electrolyte. When $CaF2$ is part of the chemical system, the concentration of ZnO can have a measurable impact on the electrolyte’s conductivity. Combining the conductivity results with the two solar process performance parameters illustrates the importance of operating the solar process at low current densities when the temperature range is 1200–1500 K. The results further suggest that one should consider studying the electrolytic process at 1800 K.

<>
Copyright © 2009 by American Society of Mechanical Engineers
Your Session has timed out. Please sign back in to continue.

## Figures

Figure 1

The minimum required energy input for the decomposition of ZnO into its elements each at 1 bar. ΔH is the minimum amount of energy required for the reaction and ΔGo is the minimum amount of energy that must be supplied as electric work.

Figure 2

Schematic of the model of the electrolytic cell and the measurement system used to establish its ionic resistance

Figure 3

Isometric view of our reaction chamber and the cell used to establish the ionic conductivity of various electrolytes. The crucible is made from hot pressed boron nitride. The electrodes are 1 mm diameter molybdenum wire.

Figure 4

Conductivity of 2 wt % alumina in Na3AlF6 from 1260 K to 1300 K. The thin-line values were reported in Ref. 24. The rectangular values were determined by our measurements.

Figure 5

Conductivity of 5 wt % alumina in Na3AlF6 from 1280 K to 1300 K. The thin-line values with their uncertainty intervals were reported in Ref. 24. The rectangular values and their uncertainty intervals were determined by our measurements.

Figure 6

Conductivity versus temperature for various Na3AlF6, CaF2, and ZnO solutions. Uncertainty intervals are at a 95% confidence level.

Figure 7

Conductivity versus temperature for 35–65 wt %Na3AlF6–CaF2 with ZnO solutions

Figure 8

Conductivity versus temperature for 50–50 wt %Na3AlF6–CaF2 with ZnO solutions

Figure 9

Back-work ratio versus temperature, the ideal values, and those based on this study’s conductivity results

Figure 10

Fraction of energy required that is thermal versus temperature, the ideal values, and those based on this study’s conductivity results

Figure 12

Typical example of electrolytic resistance versus frequency obtained from the swept-sine method

Figure 11

Schematic of the uncertainty analysis for the conductivity measurements

## Discussions

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### Related Content

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text:
Tagged: Berkeley

Problem 457

Let $A$ be a real symmetric $n\times n$ matrix with $0$ as a simple eigenvalue (that is, the algebraic multiplicity of the eigenvalue $0$ is $1$), and let us fix a vector $\mathbf{v}\in \R^n$.

(a) Prove that for sufficiently small positive real $\epsilon$, the equation
$A\mathbf{x}+\epsilon\mathbf{x}=\mathbf{v}$ has a unique solution $\mathbf{x}=\mathbf{x}(\epsilon) \in \R^n$.

(b) Evaluate
$\lim_{\epsilon \to 0^+} \epsilon \mathbf{x}(\epsilon)$ in terms of $\mathbf{v}$, the eigenvectors of $A$, and the inner product $\langle\, ,\,\rangle$ on $\R^n$.

(University of California, Berkeley, Linear Algebra Qualifying Exam)

Problem 381

Consider the matrix
$A=\begin{bmatrix} 3/2 & 2\\ -1& -3/2 \end{bmatrix} \in M_{2\times 2}(\R).$

(a) Find the eigenvalues and corresponding eigenvectors of $A$.

(b) Show that for $\mathbf{v}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}\in \R^2$, we can choose $n$ large enough so that the length $\|A^n\mathbf{v}\|$ is as small as we like.

(University of California, Berkeley, Linear Algebra Final Exam Problem)

Problem 317

Suppose that $A$ is a real $n\times n$ matrix.

(a) Is it true that $A$ must commute with its transpose?

(b) Suppose that the columns of $A$ (considered as vectors) form an orthonormal set.
Is it true that the rows of $A$ must also form an orthonormal set?

(University of California, Berkeley, Linear Algebra Qualifying Exam)

Problem 190

Prove that the matrix
$A=\begin{bmatrix} 1 & 1.00001 & 1 \\ 1.00001 &1 &1.00001 \\ 1 & 1.00001 & 1 \end{bmatrix}$ has one positive eigenvalue and one negative eigenvalue.

(University of California, Berkeley Qualifying Exam Problem)

Problem 133

Find a square root of the matrix
$A=\begin{bmatrix} 1 & 3 & -3 \\ 0 &4 &5 \\ 0 & 0 & 9 \end{bmatrix}.$

How many square roots does this matrix have?

(University of California, Berkeley Qualifying Exam)

Problem 58

Let $A$ be an $n \times n$ matrix over a field $K$. Prove that
$\rk(A^2)-\rk(A^3)\leq \rk(A)-\rk(A^2),$ where $\rk(B)$ denotes the rank of a matrix $B$.

(University of California, Berkeley, Qualifying Exam)

Problem 55

Let $A$ and $B$ are $n \times n$ matrices with real entries.
Assume that $A+B$ is invertible. Then show that
$A(A+B)^{-1}B=B(A+B)^{-1}A.$

(University of California, Berkeley Qualifying Exam)

Problem 35

Let $A$ be an $n$ by $n$ matrix with entries in complex numbers $\C$. Its only eigenvalues are $1,2,3,4,5$, possibly with multiplicities. What is the rank of the matrix $A+I_n$, where $I_n$ is the identity $n$ by $n$ matrix.

(UCB-University of California, Berkeley, Exam)

Problem 15

Let $p_1(x), p_2(x), p_3(x), p_4(x)$ be (real) polynomials of degree at most $3$. Which (if any) of the following two conditions is sufficient for the conclusion that these polynomials are linearly dependent?

(a) At $1$ each of the polynomials has the value $0$. Namely $p_i(1)=0$ for $i=1,2,3,4$.

(b) At $0$ each of the polynomials has the value $1$. Namely $p_i(0)=1$ for $i=1,2,3,4$.

(University of California, Berkeley)


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# The Sixth Annual Large Hadron Collider Physics conference LHCP 2018

4-9 June 2018
Europe/Zurich timezone

## f$_0$(980) resonance production in pp collisions with the ALICE detector at LHC

5 Jun 2018, 16:00
1h 30m
Library, Centro San Domenico

LHC experiments

### Speaker

Alessandra Lorenzo (Universita e INFN, Bologna (IT))

### Description

We report on a preliminary study of the production of f$_{0}$(980)$\rightarrow \pi^{+}\pi^{-}$ at mid-rapidity ($\vert y \vert$~$<$~0.5) performed with the ALICE detector at the LHC in minimum bias pp collisions at the centre-of-mass energy $\sqrt{\mathit{s}}$ = 5.02 TeV. The f$_{0}$(980) signal extraction is challenging due to the large background from correlated $\pi^{+}\pi^{-}$ pairs from other resonance decays in the invariant mass window under study, as well as due to the combinatorics from uncorrelated pairs. We present in detail the strategy followed for the signal extraction and first results in terms of \textit{p}$_{\mathrm{T}}$-dependent production yields. Results are discussed and compared with production yields of other resonances and stable hadrons.

Short-lived hadronic resonances are useful probes for the investigation of the late hadronic phase of ultra-relativistic heavy-ion collisions since their lifetimes are of the same order of magnitude as the time span between the chemical and kinetic freeze-out, typically estimated to be about 10 fm/\textit{c} for central collisions. Our study in pp collisions provides a feasibility check and constitutes a reference for the measurement in high-multiplicity events (p-Pb, Pb-Pb).

The nature of the f$_0$(980) remains elusive: different interpretation of this resonance including $q\bar{q}$ states, bound states of hadrons such as $K\bar{K}$, and as tetraquark candidate are available. Studies in different collision systems are particularly interesting because they can provide information about the nature of this particle.

### Primary author

Alessandra Lorenzo (Universita e INFN, Bologna (IT))


text:
# Thread: i do need some help....

1. ## i do need some help....

here's the question and has to do with order and string-series:
let A be the set of all bit string-series of the size of 11. Let also R be a relation on A where 2 string-series are related iff the 3 first bits are the same.sHOW that R is reflexive,symmetric, and transitive....i hope MY translation from greek to english is not TOO bad...
ps:actually the symbol series will be 8...
we are interested in only about the 3 first bits
000.....
001.....
010....
011....
100....
101....
110...
111...
we intuitively understand that is reflexive...how we prove...and what aboutsymmetric & transitive....?

2. Originally Posted by aegean
we intuitively understand that is reflexive...how we prove...and what aboutsymmetric & transitive....?
As I said before, anytime you has the relation defined as “same as” it is easily shown to be an equivalence relation.
If a has the same first three bits as b then b has the first three bits as a: symmetric.

If a has the same first three bits as b and b has the same first three bits as c, then clearly a has the same first three bits as c: transitive.


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# What is the Quotient Rule?

0

What is the Quotient Rule?

Quotient rule is an important topic in Mathematics. Let’s learn about it in this article. Quotient Rule is a method of finding out the derivative of a function which is given in the form of the ratio of two differentiable functions. The quotient rule is a rule in calculus for taking the derivative of a quotient of two functions. It is the most important topic of differentiation.

Let g(x) and h(x) are two differentiable functions and h(x) is not equal to 0.

Then, we can apply the Quotient rule in the form of f(x). i.e;

f(x)= g(x)/h(x), d(f(x)/g(x))= (g(x) x ff(x) – f(x) x dg(x)) / (g(x))2

Here, d denotes a derivative.

df(x) is derivative of function f.

dg(x) is derivative of function g.

For finding the derivative of f(x) divided by g(x), you must:

Take g(x) times the derivative of f(x). Then from that product, you must subtract the product of f(x) and the derivative of g(x). And, finally divide those terms by g(x) squared. For more detailed knowledge you can join Cuemath classes.

## Steps to Find The Derivatives Using The Quotient Rule

1. First check the given function, it should be in the form of division.
2. Differentiate both sides of the function.
3. Let the function on the Left Hand Side be y and equal to some other function of x.  Then, derivatives of the Left Hand Side will be dy/dx.
4. The derivatives on the Right Hand Side will be denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. At the end, the answer you will get after simplification will be the derivative of the function y which is given to you. To learn more about the quotient rule, join classes.

## Quotient Rule Examples

Quotient rule is quite similar to the product rule in calculus. The only difference between the product rule and the quotient rule is that in the product rule we have the function in the product form of type f(x)*g(x) and in the quotient rule, we have the function in the division form of type f(x)/g(x).

Examples are 3x/x^4, e^3x/2x, 4x^2/x^5, etc.

Example :

Consider a function f(x) which is given by,

f(x) = g(x) /  h(x)

Then the derivative of the function f(x) is as follows:-

f'(x)= g(x)/h(x)’

The quotient rule gives us derivatives of a function in a unique way, It is the combination of the original function and derivatives of that given function. The integral quotient rule is the method of integrating two functions given in the form of numerator and denominator(p/q). This rule is also called the division rule or Antiderivative quotient.

## Summary

If a function is a product, sum, or quotient of a simpler function, then we can use the product, sum, or quotient rule to differentiate the overall function in terms of simpler functions and their derivatives.

• The product rule states that if P is a product of two functions g and h which are differentiable according to the rule  P(x)=g(x)h(x), then

P′(x)=g(x)h′(x)+h(x)g'(x).

• If  Q  is a quotient of differentiable functions  g  and  h  then according to the quotient rule Q(x) = g(x) h(x) , then

Q′(x)=h(x)g′(x)−g(x)h′(x)

g(x)^2

Remember that the quotient rule begins with the bottom function or the denominator and ends with the bottom function squared.

Kim recently joined the team, and she writes for the Headline column of the website. She has done major in English, and a having a diploma in Journalism.


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# Re: [tlaplus] How to choose a record field when a record is an element of a set?

On 11.11.2015 18:50, Y2i wrote:
> Looking for a practical advice...
>
> Let's assume I'm reading a record by its ID.  There is at most one record
> with such an ID.
>
> TypeInvariant == \A d \in data : Cardinality({x \in data : x.id = d.id}) = 1
>
>
> If such record exists, I need to return its *d* field.  I currently use
> CHOOSE to select such a record:
>
>
> R(r) == CHOOSE x \in r : TRUE
>
>
>
>     /\ r.op = "R"
>
>     /\ LET d == {x \in data : x.id = r.id}
>
>        IN  IF d = {} THEN
>
>                qout' = Append(qout, [code|->"ERR"])
>
>            ELSE
>
>                qout' = Append(qout, [code|->"OK", d|->R(r).d])
>
>     /\ UNCHANGED <<data>>
>
>
> Is there a  better way to achieve this?

I recently had similar questions and found no better way to do it than
what you just demonstrated. In the end, I found that converting the sets
to records with the ID as the domain was a better overall solution for
me. To check whether an element with the ID in is in the record, one
just has to check (id \in DOMAIN theSet), and use theSet[id] to retrieve
it. This looks a lot nicer when written in TLA+, but I'm not sure how it
affects TLC/TLAPS performance.

Best regards,
Jaak

• Follow-Ups:
• References:


text:
# glmertree

0th

Percentile

##### (Generalized) Linear Mixed Model Trees

Model-based recursive partitioning based on (generalized) linear mixed models.

Keywords
tree
##### Usage
lmertree(formula, data, weights = NULL, cluster = NULL,
ranefstart = NULL, offset = NULL, joint = TRUE,
abstol = 0.001, maxit = 100, dfsplit = TRUE, verbose = FALSE,
plot = FALSE, REML = TRUE, lmer.control = lmerControl(), …)glmertree(formula, data, family = "binomial", weights = NULL,
cluster = NULL, ranefstart = NULL, offset = NULL, joint = TRUE,
abstol = 0.001, maxit = 100, dfsplit = TRUE, verbose = FALSE,
plot = FALSE, nAGQ = 1L, glmer.control = glmerControl(), …)
##### Arguments
formula

formula specifying the response variable and a three-part right-hand-side describing the regressors, random effects, and partitioning variables, respectively. For details see below.

data

data.frame to be used for estimating the model tree.

family

family specification for glmtree and glmer. See glm documentation for families.

weights

numeric. An optional numeric vector of weights. Can be a name of a column in data or a vector of length nrow(data).

cluster

optional vector of cluster IDs to be employed for clustered covariances in the parameter stability tests. Can be a name of a column in data or a vector of length nrow(data). If cluster = NULL (the default), observation-level covariances are employed in the parameter stability tests. If partitioning variables are measured on the cluster level, this will likely yield spurious splits, which can be mitigated by specification of the cluster argument, which results in cluster-level covariances being employed in the parameter stability tests.

ranefstart

NULL (the default), TRUE, or a numeric vector of length nrow(data). Specifies the offset to be used in estimation of the first tree. NULL by default, yielding a zero offset initialization. If ranefstart = TRUE is specified, the random effects will be estimated first and the first tree will be grown using the random-effects predictions as an offset.

offset

optional numeric vector to be included in the linear predictor with a coeffcient of one. Note that offset can be a name of a column in data or a a numeric vector of length nrow(data).

joint

logical. Should the fixed effects from the tree be (re-)estimated jointly along with the random effects?

abstol

numeric. The convergence criterion used for estimation of the model. When the difference in log-likelihoods of the random-effects model from two consecutive iterations is smaller than abstol, estimation of the model tree has converged.

maxit

numeric. The maximum number of iterations to be performed in estimation of the model tree.

dfsplit

logical or numeric. as.integer(dfsplit) is the degrees of freedom per selected split employed when extracting the log-likelihood.

verbose

Should the log-likelihood value of the estimated random-effects model be printed for every iteration of the estimation?

plot

Should the tree be plotted at every iteration of the estimation? Note that selecting this option slows down execution of the function.

REML

logical scalar. Should the fixef-effects estimates be chosen to optimize the REML criterion (as opposed to the log-likelihood)? Will be passed to funtion lmer(). See lmer for details.

nAGQ

integer scalar. Specifies the number of points per axis for evaluating the adaptive Gauss-Hermite approximation to the log-likelihood, to be passed to function glmer(). See glmer for details.

lmer.control, glmer.control

list. An optional list with control parameters to be passed to lmer() and glmer(), respectively. See lmerControl for details.

Additional arguments to be passed to lmtree() or glmtree(). See mob_control documentation for details.

##### Details

(G)LMM trees learn a tree where each terminal node is associated with different fixed-effects regression coefficients while adjusting for global random effects (such as a random intercept). This allows for detection of subgroups with different fixed-effects parameter estimates, keeping the random effects constant throughout the tree (i.e., random effects are estimated globally). The estimation algorithm iterates between (1) estimation of the tree given an offset of random effects, and (2) estimation of the random effects given the tree structure. See Fokkema et al. (2018) for a detailed introduction.

To specify all variables in the model a formula such as y ~ x1 + x2 | random | z1 + z2 + z3 is used, where y is the response, x1 and x2 are the regressors in every node of the tree, random is the random effects, and z1 to z3 are the partitioning variables considered for growing the tree. If random is only a single variable such as id a random intercept with respect to id is used. Alternatively, it may be an explicit random-effects formula such as (1 | id) or a more complicated formula such as ((1+time) | id). (Note that in the latter two formulas, the brackets are necessary to protect the pipes in the random-effects formulation.)

In the random-effects model from step (2), two strategies are available: Either the fitted values from the tree can be supplied as an offset (joint = FALSE) so that only the random effects are estimated. Or the fixed effects are (re-)estimated along with the random effects using a nesting factor with nodes from the tree (joint = TRUE). In the former case, the estimation of each random-effects model is typically faster, but more iterations are required.

The code is still under development and might change in future versions.

##### Value

The function returns a list with the following objects:

tree

The final lmtree/glmtree.

lmer

The final lmer random-effects model.

ranef

The corresponding random effects of lmer.

varcorr

The corresponding VarCorr(lmer).

variance

The corresponding attr(VarCorr(lmer), "sc")^2.

data

The dataset specified with the data argument including added auxiliary variables .ranef and .tree from the last iteration.

loglik

The log-likelihood value of the last iteration.

iterations

The number of iterations used to estimate the lmertree.

maxit

The maximum number of iterations specified with the maxit argument.

ranefstart

The random effects used as an offset, as specified with the ranefstart argument.

formula

The formula as specified with the formula argument.

randomformula

The formula as specified with the randomformula argument.

abstol

The prespecified value for the change in log-likelihood to evaluate convergence, as specified with the abstol argument.

mob.control

A list containing control parameters passed to lmtree(), as specified with ….

lmer.control

A list containing control parameters passed to lmer(), as specified in the lmer.control argument.

joint

Whether the fixed effects from the tree were (re-)estimated jointly along with the random effects, specified with the joint argument.

##### References

Fokkema M, Smits N, Zeileis A, Hothorn T, Kelderman H (2018). “Detecting Treatment-Subgroup Interactions in Clustered Data with Generalized Linear Mixed-Effects Model Trees”. Behavior Research Methods, 50(5), 2016-2034. https://doi.org/10.3758/s13428-017-0971-x

lmer, glmer, lmtree, glmtree, plot.lmertree, plot.glmertree

##### Aliases
• lmertree
• glmertree
• print.lmertree
• ranef.lmertree
• logLik.lmertree
• model.frame.lmertree
• predict.lmertree
• terms.lmertree
• as.party.lmertree
• residuals.lmertree
• resid.lmertree
• VarCorr.lmertree
• print.glmertree
• ranef.glmertree
• logLik.glmertree
• model.frame.glmertree
• predict.glmertree
• terms.glmertree
• as.party.glmertree
• residuals.glmertree
• resid.glmertree
• VarCorr.glmertree
##### Examples
# NOT RUN {
## artificial example data
data("DepressionDemo", package = "glmertree")

## fit normal linear regression LMM tree for continuous outcome
lt <- lmertree(depression ~ treatment | cluster | age + anxiety + duration,
data = DepressionDemo)
print(lt)
plot(lt, which = "all") # default behavior, may also be "tree" or "ranef"
coef(lt)
ranef(lt)
predict(lt, type = "response") # default behavior, may also be "node"
predict(lt, re.form = NA) # excludes random effects, see ?lme4::predict.merMod
residuals(lt)
VarCorr(lt) # see lme4::VarCorr

## fit logistic regression GLMM tree for binary outcome
gt <- glmertree(depression_bin ~ treatment | cluster | age + anxiety + duration,
data = DepressionDemo)
print(gt)
plot(gt, which = "all") # default behavior, may also be "tree" or "ranef"
coef(gt)
ranef(gt)
predict(gt, type = "response") # default behavior, may also be "node" or "link"
predict(gt, re.form = NA) # excludes random effects, see ?lme4::predict.merMod
residuals(gt)
VarCorr(gt) # see lme4::VarCorr
# }

Documentation reproduced from package glmertree, version 0.2-0, License: GPL-2 | GPL-3

### Community examples

Looks like there are no examples yet.


text:
# Differential Equation- Separation of Variables

1. May 19, 2008

1. The problem statement, all variables and given/known data
y'=xsec^2(x^2)

2. The attempt at a solution
dy/dx=xsec^2(x^2)
dy=xsec^2(x^2)dx
$$\int$$dy=$$\int$$xsec^2(x^2)dx
lny= (here i'll do a u substitution)
----
u=x^2 du=1/3x^3dx

... and here's my problem. It seems like that creates a very difficult u-sub to try and manage. Any suggestions/help?

2. May 19, 2008

### Defennder

This is incorrect. Your method is correct, though.

3. May 19, 2008

### soul

I think you should write sec function in terms of cos. Function will be dy = x / (cos^2(x^2)). Then if you say x^2 = u and du = 2x.dx, you can solve the problem easily.

4. May 19, 2008

u=x^2 du=2xdx
2$$\int$$du/cos^2(u)

5. May 19, 2008

### Defennder

You're off by a factor of 1/4. It should be $$\frac{1}{2} \int sec^2u \ du$$.


text:
# Thread: Birthday Coincidence Probability (w/ leap years)

1. ## Birthday Coincidence Probability (w/ leap years)

What is the probability of k people having a coincidence of birthdays, taking leap-years into account? Hint: Consider all the days in a 4-year period, in which the leap-day 2/29 occurs once.

Help tonight/before tomorrow afternoon would be lovely.

Thanks!!

2. Originally Posted by cassie00
What is the probability of k people having a coincidence of birthdays, taking leap-years into account? Hint: Consider all the days in a 4-year period, in which the leap-day 2/29 occurs once.

Help tonight/before tomorrow afternoon would be lovely.

Thanks!!
It's easier to find the probability that among k people there are no coincidences and subtract this probability from 1.

3. I originally tried splitting up the years (3 "normal years" for every 1 leap-year) and weighting them as 3/4 and 1/4.

Would you recommend using a larger sample of days? 3(365)+366?

Thanks!

4. Originally Posted by cassie00
I originally tried splitting up the years (3 "normal years" for every 1 leap-year) and weighting them as 3/4 and 1/4.

Would you recommend using a larger sample of days? 3(365)+366?

Thanks!
Yes, this is what the hint suggests. The probability of having a birthday other than Feb. 29 is going to be $\frac{4}{3(365) + 366}$, whereas the probability of having Feb. 29 will be $\frac{1}{3(365) + 366}$.

5. Originally Posted by icemanfan
Yes, this is what the hint suggests. The probability of having a birthday other than Feb. 29 is going to be $\frac{4}{3(365) + 366}$, whereas the probability of having Feb. 29 will be $\frac{1}{3(365) + 366}$.

Why isn't the numerator of the first probability 1460?

6. ## I'm taking the hint :)

In the course of 4 years (i.e. 3 regular years and 1 leap year), each regular day occurs 4 times while February 29 occurs once. There are 365 regular days. The total number of days in the course of 4 years is:

365 x 4 + 1 = 1461

During that period, a person celebrates his/her birthday 4 times if it falls on a regular day and 1 time if it falls on February 29 (of course, he/she can choose to celebrate it on February 28 or March 1, but that's not the point of the problem).

The probability that his/her birthday falls on January 1 is 4/1461. The probability is the same for each regular day. Meanwhile, the probability that his/her birthday falls on February 29 is 1/1461.

The probability that k people have their birthdays on January 1 is:

$
( \frac {4}{1461} )^k
$

The probability is the same for the other 365 regular days. Meanwhile, the probability that k people have their birthdays on February 29 is:

$
( \frac {1}{1461} )^k
$

Hence, the total probability that k people have the same birthdays is:

$
365 ( \frac {4}{1461} )^k + ( \frac {1}{1461} )^k
$

7. Originally Posted by cassie00
What is the probability of k people having a coincidence of birthdays, taking leap-years into account? Hint: Consider all the days in a 4-year period, in which the leap-day 2/29 occurs once.

Thanks!!
Of related interest: The Birthday Paradox


text:
LaTeX/Linguistics

LaTeX

Getting Started

Common Elements

Mechanics

Technical Texts

Special Pages

Special Documents

Creating Graphics

Programming

Miscellaneous

Help and Recommendations

Appendices

edit this boxedit the TOC

There are a number of LaTeX packages available for writing linguistics papers. Various packages have been created for enumerated examples, syntactic trees, OT tableaux, feature matrices, IPA fonts, and many other applications. Some packages such as the tipa package are effectively standard within the field, while others will vary by author preference.

Some recommended packages[1]:

• Glosses: gb4e;
• IPA symbols: tipa;
• OT Tableaux: OTtablx;
• Syntactic trees: qtree + tree-dvips (for drawing arrows);
• Alternatively, xyling is very powerful but not as user friendly as qtree;
• The xy package itself has a steep learning curve, but allows a lot of control; for simplest trees use the xymatrix feature and arrows;
• tikz-qtree has the same syntax as qtree, but uses PGF/TikZ, which allows more options for drawing arrows, etc.
• Dependency trees and bubble parses:
• The TikZ-dependency package provides a high-level, convenient interface to draw dependency graphs. It is based on PGF/TikZ but does not require prior knowledge of TikZ in order to be used productively.
• Attribute-Value Matrices (AVMs): avm
• John Frampton's expex: expex

Enumerated examplesEdit

There are several commonly used packages for creating the kinds of numbered examples that are used in linguistics publications.

gb4eEdit

The gb4e package[2] is called with:

\usepackage{gb4e}

IMPORTANT: If you use gb4e package, let it be the last \usepackage call in the document's preamble. Otherwise you may get exceeded parameter stack size error.

Examples for this package are placed within the exe environment, and each example is introduced with the \ex command.

\begin{exe}
\ex This is an example.
\end{exe}

produces:

Multiple examples can be included within the environment, and each will have its own number.

\begin{exe}
\ex This is the first example.
\ex This is the second example.
\ex This is the third.
\end{exe}

produces:

To create nested lists of examples, the xlist enviroment is used.

\begin{exe}
\ex \begin{xlist}
\ex This is a sub-example.
\ex This is a second sub-example.
\ex \begin{xlist}
\ex This is a sub-sub-example.
\ex This is a second sub-sub-example.
\end{xlist}
\end{xlist}
\end{exe}

produces:

For notating acceptability judgments, the \ex command can take an optional argument. When including a judgment marker, the corresponding sentence must be surrounded by braces.

\begin{exe}
\ex This sentence is grammatical English.
\ex[*] {This sentence English in ungrammatical is.}
\end{exe}

produces:

Referencing examples in text works as it does in normal LaTeX documents. See the labeling and cross-referencing section for more details.

\begin{exe}
\ex\label{ex1} Godzilla destroyed the city.
\ex\label{ex2} Godzilla roared.
\end{exe}
Sentence (\ref{ex1}) contains two arguments, but (\ref{ex2}) contains only one.

Further details can be found in the full documentation available here.

lingmacrosEdit

The lingmacros package[3] created by Emma Pease is an alternate method for example numbering. This package uses two main commands, \enumsentence and \eenumsentence. The former is used for singleton examples, while the latter command is used for nested examples.

\enumsentence{This is an example.}

\enumsentence{This is the first example.}
\enumsentence{This is the second example.}
\enumsentence{This is the third.}

Multiply nested examples make use of the normal LaTeX list environments.

\eenumsentence{\item This is a sub-example.
\item This is a second sub-example.
\item \begin{enumerate}
\item This is sub-sub-example.
\item This is a second sub-sub-example.
\end{enumerate}
}

produces:

Full documentation can be found here.

Syntactic treesEdit

Often, linguists will have to illustrate the syntactic structure of a sentence. One device for doing this are syntactic trees. Unfortunately, trees look very different in different grammar formalisms, and different LaTeX packages are suited for different formalisms.

Constituent treesEdit

While there are several packages for drawing syntactic trees available for LaTeX, this article focuses on the qtree and xyling packages.

qtreeEdit

Drawing trees with qtree is relatively straightforward. First, the qtree package has to be included in the document's preamble:

\usepackage{qtree}

A new tree is started using the \Tree command, each (sub-)tree is indicated by brackets [ ]. The root of a (sub-)tree is always preceded by a ., leaf nodes are simply expressed by their labels.

For example, the following code

\Tree [.S [.NP LaTeX ] [.VP [.V is ] [.NP fun ] ] ]

produces this syntactic tree as output:

Note that the spaces before the closing brackets are mandatory.

By default, qtree centers syntactic trees on the page. This behaviour can be turned off by either specifying the behaviour when loading the package

\usepackage[nocenter]{qtree} % do not center trees

or via the command

\qtreecenterfalse % do not center trees from here on

anywhere in the document. The effect of the latter can be undone by using the command

\qtreecentertrue % center trees from here on

IMPORTANT: If you use gb4e package, let it be the last \usepackage call in the document's preamble. Otherwise you may get exceeded parameter stack size error.

tikz-qtreeEdit

Using the same syntax as qtree, tikz-qtree is another easy-to-use alternative for drawing syntactic trees.

For simple trees, tikz-qtree is completely interchangable with qtree. However, some of qtree's advanced features are implemented in a different way, or not at all. On the other hand, tikz-qtree provides other features such as controlling the direction of the tree's growth (top to bottom, left to right etc.) or different styles for edges.

To use the tikz-qtree package for drawing trees, put the following into the document's preamble:

\usepackage{tikz}
\usepackage{tikz-qtree}

The syntax of tikz-qtree and result when drawing a simple tree is the same as for qtree.

\Tree [.S [.NP LaTeX ] [.VP [.V is ] [.NP fun ] ] ]

Note that, other than for qtree, trees are not centered by default. To center them, put them into a centered environment:

\begin{center}
\Tree [.S [.NP LaTeX ] [.VP [.V is ] [.NP fun ] ] ]
\end{center}

For setting the style of trees, tikz-qtree provides the \tikzset command. For example, to make a tree grow from left to right instead of from top to bottom, use the following code:

\tikzset{grow'=right} % make trees grow from left to right
\tikzset{every tree node/.style={anchor=base west}} % allign nodes of the tree to the left (west)
\Tree [.S [.NP LaTeX ] [.VP [.V is ] [.NP fun ] ] ]

The above code changes the default orientation for all trees that are defined after \tikzset commands. To only change the direction of a single tree, it has to be put into a \tikzpicture environment:

\begin{tikzpicture} % all changes only affect trees within this environment
\tikzset{grow'=right} % make trees grow from left to right
\tikzset{every tree node/.style={anchor=base west}} % allign nodes of the tree to the left (west)
\Tree [.S [.NP LaTeX ] [.VP [.V is ] [.NP fun ] ] ]
\end{tikzpicture}

Dependency TreesEdit

Dependency trees can take multiple visual forms. Commonly, they quite resemble phrase structure trees. Alternatively, they can be captured by brackets drawn above running text.

Two-dimensional Dependency TreesEdit

These can be either achieved using the fairly universal drawing package TikZ, like so:

% In the preamble:
\usepackage{tikz}

% In the document:
\begin{tikzpicture}
\node (is-root) {is}
[sibling distance=3cm]
child { node {this} }
child {
node {tree}
[sibling distance=1.5cm]
child { node {an} }
child { node {example} }
child { node {.} }
child[missing]
};
\path (is-root) +(0,-2.5\tikzleveldistance)
node {\textit{This is an example tree.}};
\end{tikzpicture}

which gives you the following drawing:

A dependency tree created using TikZ

TikZ has the advantage that it allows for generating PDF directly from the LaTeX source, without need for any detour of compiling to DVI using latex, and then converting to PDF probably via PS using tools such as dvips and ps2pdf. Latter is the case of another package based on the package xy, namely xyling.

The code for a similar tree using xyling might look like:

% In the preamble:
\usepackage{xytree}

% In the document:
\Tree{	& \K{is}\B{dl}\B{drr} \\
\K{this} &&& \K{tree}\B{dll}\B{dl}\B{dr} \\
& \K{an} & \K{example} && \K{.} }

\medskip
\textit{This is an example tree.}

which gives you a drawing like this:

A dependency tree created using xyling

Dependency Trees as Brackets above TextEdit

One way to typeset dependency brackets above running text is using the package xytree. It gives you fairly good control of how the brackets are typeset but requires compiling the LaTeX code to DVI (and perhaps converting to PDF using the tools dvips and ps2pdf later).

An example code:

% In the preamble:
\usepackage{xytree}

% In the document:
\xytext{
\xybarnode{Peter} &~~~&
\xybarnode{and}
\xybarconnect(UL,U){-2}"_{\small conj}"
\xybarconnect(UR,U){2}"^{\small conj}"
&~~~&
\xybarnode{Mary} &~~~&
\xybarnode{bought}
\xybarconnect[8](UL,U){-4}"_{\small subj}"
\xybarconnect[13]{6}"^{\small punct}"
\xybarconnect[8](UR,U){4}"^{\small obj}"
&~~~&
\xybarnode{a} &~~~&
\xybarnode{car}
\xybarconnect(UL,U){-2}"_{\small det}"
&~~~&
\xybarnode{.}
}

results in:

A dependency tree above running text created using xytree

Dependency Trees using TikZ-dependencyEdit

The package provides high level commands to design and style dependency graphs. To draw a graph, you only need to create a dependency environment, write the text of the sentence within the deptext environment and use depedge commands to draw the edges. Global and local optional parameters can be used to style and fine tune the looks of the graph, as shown in the following example:

% In the preamble:
\usepackage{tikz-dependency}

% In the document:
\begin{dependency}[theme = simple]
\begin{deptext}[column sep=1em]
A \& hearing \& is \& scheduled \& on \& the \& issue \& today \& . \\
\end{deptext}
\deproot{3}{ROOT}
\depedge{2}{1}{ATT}
\depedge[edge start x offset=-6pt]{2}{5}{ATT}
\depedge{3}{2}{SBJ}
\depedge{3}{9}{PU}
\depedge{3}{4}{VC}
\depedge{4}{8}{TMP}
\depedge{5}{7}{PC}
\depedge[arc angle=50]{7}{6}{ATT}
\end{dependency}

This code snippet would produce the following result:

GlossesEdit

Below, it is explained how to make glossed examples with different packages.

With gb4eEdit

To create a glossed example, use the normal exe environment. But after the \ex tag, introduce the example and its gloss using \gll and the translation after it with \trans tag.

\begin{exe}
\ex
\gll Кот ест сметану\\
cat.NOM eat.3.SG.PRS sour-cream.ACC\\
\trans The cat eats sour cream'
\end{exe}

The code will produce the following output:

Vertically aligned glosses are separated by spaces, so if it's necessary to include a space in part the gloss, simply enclose the connected parts inside braces.

\begin{exe}
\ex
\gll Pekka pel\"astyi karhusta.\\
Pekka {became afraid} bear.ELA\\
\trans Pekka became afraid because of the/a bear.'
\end{exe}

With lingmacrosEdit

The lingmacros package uses the \shortex command to introduce glossed examples inside the \enumsentence and \eenumsentence commands. This command takes four arguments and builds off the normal tabular environment. Its first argument specifies the number of columns in the gloss. The second and third arguments give the text and its gloss respectively, and items within each column are divided by the usual & tabular separator. The fourth argument is the translation.

\enumsentence{\shortex{3}
{Pekka & pel\"astyi & karhu-sta.}
{Pekka & became afraid & bear.ELA}
{Pekka became afraid because of the/a bear.'}
}

IPA charactersEdit

The tipa package is the standard LaTeX package for International Phonetic Alphabet symbols.

\usepackage{tipa}

There are two methods for getting IPA symbols into a document. The first way is to use the IPA environment.

\begin{IPA}
text in IPA format here
\end{IPA}

This method is useful for long stretches of text that need to be in IPA. Alternatively, there is the \textipa command that will format the text in its argument into IPA. This command is similar to other font typesetting commands.

\textipa{text in IPA format here}

Basic symbolsEdit

The IPA format works by translating ASCII characters into corresponding IPA symbols. Lower case letters are rendered as usual,

\textipa{abcdefghijklmnopqrstuvwxyz}

however capital letters are rendered differently.

\textipa{ABCDEFGHIJKLMNOPQRSTUVWXYZ}

produces:

Punctuation marks that are normally used in LaTeX are also rendered faithfully in the IPA environment.

\textipa{! * + = ? . , / [ ] ( )  ' | ||}

produces:

Numerals and @ also have variants in the tipa environment.

\textipa{1234567890 @}

produces:

In addition, there are a number of special macros for representing symbols that don't have other associations, some of which are listed here. For a complete list see the official TIPA Manual[4].

The \; macro preceding a capital letter produces a small caps version of the letter.

\textipa{\;A \;B \;E \;G \;H \;I \;L \;R \;Y}

produces:

The \: macro produces retroflex symbols.

\textipa{\:d \:l \:n \:r \:s \:t \:z}

gets you:

The \! macro produces implosive symbols and the bilabial click.

\textipa{\!b \!d \!g \!j \!G \!o}

gets you:

ReferencesEdit

1. [1] LaTeX for Linguists presentation
2. [2] The gb4e package on CTAN
3. [3] The lingmacros package on CTAN
4. TIPA manual


text:
## 15.11 Henselian pairs

Some of the results of Section 15.9 may be viewed as results about henselian pairs. In this section a pair is a pair $(A, I)$ where $A$ is a ring and $I \subset A$ is an ideal. A morphism of pairs $(A, I) \to (B, J)$ is a ring map $\varphi : A \to B$ with $\varphi (I) \subset J$. As in Section 15.9 given an object $\xi$ over $A$ we denote $\overline{\xi }$ the “base change” of $\xi$ to an object over $A/I$ (provided this makes sense).

Definition 15.11.1. A henselian pair is a pair $(A, I)$ satisfying

1. $I$ is contained in the Jacobson radical of $A$, and

2. for any monic polynomial $f \in A[T]$ and factorization $\overline{f} = g_0h_0$ with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$, there exists a factorization $f = gh$ in $A[T]$ with $g, h$ monic and $g_0 = \overline{g}$ and $h_0 = \overline{h}$.

Observe that if $A$ is a local ring and $I = \mathfrak m$ is the maximal ideal, then $(A, I)$ is a henselian pair if and only if $A$ is a henselian local ring, see Algebra, Lemma 10.152.3. In Lemma 15.11.6 we give a number of equivalent characterizations of henselian pairs (and we will add more as time goes on).

Lemma 15.11.2. Let $(A, I)$ be a pair with $I$ locally nilpotent. Then the functor $B \mapsto B/IB$ induces an equivalence between the category of étale algebras over $A$ and the category of étale algebras over $A/I$. Moreover, the pair is henselian.

Proof. Essential surjectivity holds by Algebra, Lemma 10.142.10. If $B$, $B'$ are étale over $A$ and $B/IB \to B'/IB'$ is a morphism of $A/I$-algebras, then we can lift this by Algebra, Lemma 10.137.17. Finally, suppose that $f, g : B \to B'$ are two $A$-algebra maps with $f \mod I = g \mod I$. Choose an idempotent $e \in B \otimes _ A B$ generating the kernel of the multiplication map $B \otimes _ A B \to B$, see Algebra, Lemmas 10.150.4 and 10.150.3 (to see that étale is unramified). Then $(f \otimes g)(e) \in IB$. Since $IB$ is locally nilpotent (Algebra, Lemma 10.31.3) this implies $(f \otimes g)(e) = 0$ by Algebra, Lemma 10.31.6. Thus $f = g$.

It is clear that $I$ is contained in the Jacobson radical of $A$. Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0h_0$ be a factorization of $\overline{f} = f \bmod I$ with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$. By Lemma 15.9.5 there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ such that the factorization lifts to a factorization into monic polynomials over $A'$. By the above we have $A = A'$ and the factorization is over $A$. $\square$

Lemma 15.11.3. Let $A = \mathop{\mathrm{lim}}\nolimits A_ n$ where $(A_ n)$ is an inverse system of rings whose transition maps are surjective and have locally nilpotent kernels. Then $(A, I_ n)$ is a henselian pair, where $I_ n = \mathop{\mathrm{Ker}}(A \to A_ n)$.

Proof. Fix $n$. Let $a \in A$ be an element which maps to $1$ in $A_ n$. By Algebra, Lemma 10.31.4 we see that $a$ maps to a unit in $A_ m$ for all $m \geq n$. Hence $a$ is a unit in $A$. Thus by Algebra, Lemma 10.18.1 the ideal $I_ n$ is contained in the Jacobson radical of $A$. Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_ nh_ n$ be a factorization of $\overline{f} = f \bmod I_ n$ with $g_ n, h_ n \in A_ n[T]$ monic generating the unit ideal in $A_ n[T]$. By Lemma 15.11.2 we can successively lift this factorization to $f \bmod I_ m = g_ m h_ m$ with $g_ m, h_ m$ monic in $A_ m[T]$ for all $m \geq n$. At each step we have to verify that our lifts $g_ m, h_ m$ generate the unit ideal in $A_ n[T]$; this follows from the corresponding fact for $g_ n, h_ n$ and the fact that $\mathop{\mathrm{Spec}}(A_ n[T]) = \mathop{\mathrm{Spec}}(A_ m[T])$ because the kernel of $A_ m \to A_ n$ is locally nilpotent. As $A = \mathop{\mathrm{lim}}\nolimits A_ m$ this finishes the proof. $\square$

Lemma 15.11.4. Let $(A, I)$ be a pair. If $A$ is $I$-adically complete, then the pair is henselian.

Proof. By Algebra, Lemma 10.95.6 the ideal $I$ is contained in the Jacobson radical of $A$. Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0h_0$ be a factorization of $\overline{f} = f \bmod I$ with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$. By Lemma 15.11.2 we can successively lift this factorization to $f \bmod I^ n = g_ n h_ n$ with $g_ n, h_ n$ monic in $A/I^ n[T]$ for all $n \geq 1$. As $A = \mathop{\mathrm{lim}}\nolimits A/I^ n$ this finishes the proof. $\square$

Lemma 15.11.5. Let $(A, I)$ be a pair. Let $A \to B$ be a finite type ring map such that $B/IB = C_1 \times C_2$ with $A/I \to C_1$ finite. Let $B'$ be the integral closure of $A$ in $B$. Then we can write $B'/IB' = C_1 \times C'_2$ such that the map $B'/IB' \to B/IB$ preserves product decompositions and there exists a $g \in B'$ mapping to $(1, 0)$ in $C_1 \times C'_2$ with $B'_ g \to B_ g$ an isomorphism.

Proof. Observe that $A \to B$ is quasi-finite at every prime of the closed subset $T = \mathop{\mathrm{Spec}}(C_1) \subset \mathop{\mathrm{Spec}}(B)$ (this follows by looking at fibre rings, see Algebra, Definition 10.121.3). Consider the diagram of topological spaces

$\xymatrix{ \mathop{\mathrm{Spec}}(B) \ar[rr]_\phi \ar[rd]_\psi & & \mathop{\mathrm{Spec}}(B') \ar[ld]^{\psi '} \\ & \mathop{\mathrm{Spec}}(A) }$

By Algebra, Theorem 10.122.12 for every $\mathfrak p \in T$ there is a $h_\mathfrak p \in B'$, $h_\mathfrak p \not\in \mathfrak p$ such that $B'_ h \to B_ h$ is an isomorphism. The union $U = \bigcup D(h_\mathfrak p)$ gives an open $U \subset \mathop{\mathrm{Spec}}(B')$ such that $\phi ^{-1}(U) \to U$ is a homeomorphism and $T \subset \phi ^{-1}(U)$. Since $T$ is open in $\psi ^{-1}(V(I))$ we conclude that $\phi (T)$ is open in $U \cap (\psi ')^{-1}(V(I))$. Thus $\phi (T)$ is open in $(\psi ')^{-1}(V(I))$. On the other hand, since $C_1$ is finite over $A/I$ it is finite over $B'$. Hence $\phi (T)$ is a closed subset of $\mathop{\mathrm{Spec}}(B')$ by Algebra, Lemmas 10.40.6 and 10.35.22. We conclude that $\mathop{\mathrm{Spec}}(B'/IB') \supset \phi (T)$ is open and closed. By Algebra, Lemma 10.23.3 we get a corresponding product decomposition $B'/IB' = C'_1 \times C'_2$. The map $B'/IB' \to B/IB$ maps $C'_1$ into $C_1$ and $C'_2$ into $C_2$ as one sees by looking at what happens on spectra (hint: the inverse image of $\phi (T)$ is exactly $T$; some details omitted). Pick a $g \in B'$ mapping to $(1, 0)$ in $C'_1 \times C'_2$ such that $D(g) \subset U$; this is possible because $\mathop{\mathrm{Spec}}(C'_1)$ and $\mathop{\mathrm{Spec}}(C'_2)$ are disjoint and closed in $\mathop{\mathrm{Spec}}(B')$ and $\mathop{\mathrm{Spec}}(C'_1)$ is contained in $U$. Then $B'_ g \to B_ g$ defines a homeomorphism on spectra and an isomorphism on local rings (by our choice of $U$ above). Hence it is an isomorphism, as follows for example from Algebra, Lemma 10.22.1. Finally, it follows that $C'_1 = C_1$ and the proof is complete. $\square$

Lemma 15.11.6. Let $(A, I)$ be a pair. The following are equivalent

1. $(A, I)$ is a henselian pair,

2. given an étale ring map $A \to A'$ and an $A$-algebra map $\sigma : A' \to A/I$, there exists an $A$-algebra map $A' \to A$ lifting $\sigma$,

3. for any finite $A$-algebra $B$ the map $B \to B/IB$ induces a bijection on idempotents,

4. for any integral $A$-algebra $B$ the map $B \to B/IB$ induces a bijection on idempotents, and

5. (Gabber) $I$ is contained in the Jacobson radical of $A$ and every monic polynomial $f(T) \in A[T]$ of the form

$f(T) = T^ n(T - 1) + a_ n T^ n + \ldots + a_1 T + a_0$

with $a_ n, \ldots , a_0 \in I$ and $n \ge 1$ has a root $\alpha \in 1 + I$.

Moreover, in part (5) the root is unique.

Proof. Assume (2) holds. Then $I$ is contained in the Jacobson radical of $A$, since otherwise there would be a nonunit $f \in A$ congruent to $1$ modulo $I$ and the map $A \to A_ f$ would contradict (2). Hence $IB \subset B$ is contained in the Jacobson radical of $B$ for $B$ integral over $A$ because $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is closed by Algebra, Lemmas 10.40.6 and 10.35.22. Thus the map from idempotents of $B$ to idempotents of $B/IB$ is injective by Lemma 15.10.2. On the other hand, since (2) holds, every idempotent of $B/IB$ lifts to an idempotent of $B$ by Lemma 15.9.10. In this way we see that (2) implies (4).

The implication (4) $\Rightarrow$ (3) is trivial.

Assume (3). Let $\mathfrak m$ be a maximal ideal and consider the finite map $A \to B = A/(I \cap \mathfrak m)$. The condition that $B \to B/IB$ induces a bijection on idempotents implies that $I \subset \mathfrak m$ (if not, then $B = A/I \times A/\mathfrak m$ and $B/IB = A/I$). Thus we see that $I$ is contained in the Jacobson radical of $A$. Let $f \in A[T]$ be monic and suppose given a factorization $\overline{f} = g_0h_0$ with $g_0, h_0 \in A/I[T]$ monic. Set $B = A[T]/(f)$. Let $\overline{e}$ be the idempotent of $B/IB$ corresponding to the decomposition

$B/IB = A/I[T]/(g_0) \times A[T]/(h_0)$

of $A$-algebras. Let $e \in B$ be an idempotent lifting $\overline{e}$ which exists as we assumed (3). This gives a product decomposition

$B = eB \times (1 - e)B$

Note that $B$ is free of rank $\deg (f)$ as an $A$-module. Hence $eB$ and $(1 - e)B$ are finite locally free $A$-modules. However, since $eB$ and $(1 - e)B$ have constant rank $\deg (g_0)$ and $\deg (h_0)$ over $A/I$ we find that the same is true over $\mathop{\mathrm{Spec}}(A)$. We conclude that

\begin{align*} f & = \text{CharPol}_ A(T : B \to B) \\ & = \text{CharPol}_ A(T : eB \to eB) \text{CharPol}_ A(T : (1 - e)B \to (1 - e)B) \end{align*}

is a factorization into monic polynomials reducing to the given factorization modulo $I$. Here $\text{CharPol}_ A$ denotes the characteristic polynomial of an endomorphism of a finite locally free module over $A$. If the module is free the $\text{CharPol}_ A$ is defined as the characteristic polynomial of the corresponding matrix and in general one uses Algebra, Lemma 10.23.2 to glue. Details omitted. Thus (3) implies (1).

Assume (1). Let $f$ be as in (5). The factorization of $f \bmod I$ as $T^ n$ times $T - 1$ lifts to a factorization $f = gh$ with $g$ and $h$ monic by Definition 15.11.1. Then $h$ has to have degree $1$ and we see that $f$ has a root reducing to $1$ modulo $1$. Finally, $I$ is contained in the Jacobson radical by the definition of a henselian pair. Thus (1) implies (5).

Before we give the proof of the last step, let us show that the root $\alpha$ in (5), if it exists, is unique. Namely, Due to the explicit shape of $f(T)$, we have $f'(\alpha ) \in 1 + I$ where $f'$ is the derivative of $f$ with respect to $T$. An elementary argument shows that

$f(T) = f(\alpha + T - \alpha ) = f(\alpha ) + f'(\alpha ) \cdot (T - \alpha ) \bmod (T - \alpha )^2 A[T]$

This shows that any other root $\alpha ' \in 1 + I$ of $f(T)$ satisfies $0 = f(\alpha ') - f(\alpha ) = (\alpha ' - \alpha )(1 + i)$ for some $i \in I$, so that, since $1 + i$ is a unit in $A$, we have $\alpha = \alpha '$.

Assume (5). We will show that (2) holds, in other words, that for every étale map $A \to A'$, every section $\sigma : A' \to A/I$ modulo $I$ lifts to a section $A' \to A$. Since $A \to A'$ is étale, the section $\sigma$ determines a decomposition

15.11.6.1
$$\label{more-algebra-equation-GCHP} A'/IA' \cong A/I \times C$$

of $A/I$-algebras. Namely, the surjective ring map $A'/IA' \to A/I$ is étale by Algebra, Lemma 10.142.8 and then we get the desired idempotent by Algebra, Lemma 10.142.9. We will show that this decomposition lifts to a decomposition

15.11.6.2
$$\label{more-algebra-equation-GCHP-want} A' \cong A'_1 \times A'_2$$

of $A$-algebras with $A'_1$ integral over $A$. Then $A \to A'_1$ is integral and étale and $A/I \to A'_1/IA'_1$ is an isomorphism, thus $A \to A'_1$ is an isomorphism by Lemma 15.10.3 (here we also use that an étale ring map is flat and of finite presentation, see Algebra, Lemma 10.142.3).

Let $B'$ be the integral closure of $A$ in $A'$. By Lemma 15.11.5 we may decompose

15.11.6.3
$$\label{more-algebra-equation-dec-mod-I} B'/IB' \cong A/I \times C'$$

as $A/I$-algebras compatibly with (15.11.6.1) and we may find $b \in B'$ that lifts $(1, 0)$ such that $B'_ b \to A'_ b$ is an isomorphism. If the decomposition (15.11.6.3) lifts to a decomposition

15.11.6.4
$$\label{more-algebra-equation-want-2} B' \cong B'_1 \times B'_2$$

of $A$-algebras, then the induced decomposition $A' = A'_1 \times A'_2$ will give the desired (15.11.6.2): indeed, since $b$ is a unit in $B'_1$ (details omitted), we will have $B'_1 \cong A'_1$, so that $A'_1$ will be integral over $A$.

Choose a finite $A$-subalgebra $B'' \subset B'$ containing $b$ (observe that any finitely generated $A$-subalgebra of $B'$ is finite over $A$). After enlarging $B''$ we may assume $b$ maps to an idempotent in $B''/IB''$ producing

15.11.6.5
$$\label{more-algebra-equation-again-dec-mod-I} B''/IB'' \cong C''_1 \times C''_2$$

Since $B'_ b \cong A'_ b$ we see that $B'_ b$ is of finite type over $A$. Say $B'_ b$ is generated by $b_1/b^ n, \ldots , b_ t/b^ n$ over $A$ and enlarge $B''$ so that $b_1, \ldots , b_ t \in B''$. Then $B''_ b \to B'_ b$ is surjective as well as injective, hence an isomorphism. In particular, we see that $C''_1 = A/I$! Therefore $A/I \to C''_1$ is an isomorphism, in particular surjective. By Lemma 15.10.4 we can find an $f(T) \in A[T]$ of the form

$f(T) = T^ n(T - 1) + a_ n T^ n + \ldots + a_1 T + a_0$

with $a_ n, \ldots , a_0 \in I$ and $n \ge 1$ such that $f(b) = 0$. In particular, we find that $B'$ is a $A[T]/(f)$-algebra. By (5) we deduce there is a root $a \in 1 + I$ of $f$. This produces a product decomposition $A[T]/(f) = A[T]/(T - a) \times D$ compatible with the splitting (15.11.6.3) of $B'/IB'$. The induced splitting of $B'$ is then a desired (15.11.6.4). $\square$

Lemma 15.11.7. Let $A$ be a ring. Let $I, J \subset A$ be ideals with $V(I) = V(J)$. Then $(A, I)$ is henselian if and only if $(A, J)$ is henselian.

Proof. For any integral ring map $A \to B$ we see that $V(IB) = V(JB)$. Hence idempotents of $B/IB$ and $B/JB$ are in bijective correspondence (Algebra, Lemma 10.20.3). It follows that $B \to B/IB$ induces a bijection on sets of idempotents if and only if $B \to B/JB$ induces a bijection on sets of idempotents. Thus we conclude by Lemma 15.11.6. $\square$

Lemma 15.11.8. Let $(A, I)$ be a henselian pair and let $A \to B$ be an integral ring map. Then $(B, IB)$ is a henselian pair.

Proof. Immediate from the fourth characterization of henselian pairs in Lemma 15.11.6 and the fact that the composition of integral ring maps is integral. $\square$

Lemma 15.11.9. Let $I \subset J \subset A$ be ideals of a ring $A$. The following are equivalent

1. $(A, I)$ and $(A/I, J/I)$ are henselian pairs, and

2. $(A, J)$ is an henselian pair.

Proof. Assume (1). Let $B$ be an integral $A$-algebra. Consider the ring maps

$B \to B/IB \to B/JB$

By Lemma 15.11.6 we find that both arrows induce bijections on idempotents. Hence so does the composition. Whence $(A, J)$ is a henselian pair by Lemma 15.11.6.

Conversely, assume (2) holds. Then $(A/I, J/I)$ is a henselian pair by Lemma 15.11.8. Let $B$ be an integral $A$-algebra. Consider the ring maps

$B \to B/IB \to B/JB$

By Lemma 15.11.6 we find that the composition and the second arrow induce bijections on idempotents. Hence so does the first arrow. It follows that $(A, I)$ is a henselian pair (by the lemma again). $\square$

Lemma 15.11.10. Let $A$ be a ring and let $(A, I)$ and $(A, I')$ be henselian pairs. Then $(A, I + I')$ is an henselian pair.

Proof. By Lemma 15.11.8 the pair $(A/I, (I' + I)/I)$ is henselian. Thus we get the conclusion from Lemma 15.11.9. $\square$

Lemma 15.11.11. Let $J$ be a set and let $\{ (A_ j, I_ j)\} _{j \in J}$ be a collection of pairs. Then $(\prod _{j \in J} A_ j, \prod _{j\in J} I_ j)$ is Henselian if and only if so is each $(A_ j, I_ j)$.

Proof. For every $j \in J$, the projection $\prod _{j \in J} A_ j \rightarrow A_ j$ is an integral ring map, so Lemma 15.11.8 proves that each $(A_ j, I_ j)$ is Henselian if $(\prod _{j \in J} A_ j, \prod _{j\in J} I_ j)$ is Henselian.

Conversely, suppose that each $(A_ j, I_ j)$ is a Henselian pair. Then every $1 + x$ with $x \in \prod _{j \in J} I_ j$ is a unit in $\prod _{j \in J} A_ j$ because it is so componentwise by Algebra, Lemma 10.18.1 and Definition 15.11.1. Thus, by Algebra, Lemma 10.18.1 again, $\prod _{j \in J} I_ j$ is contained in the Jacobson radical of $\prod _{j \in J} A_ j$. Continuing to work componentwise, it likewise follows that for every monic $f \in (\prod _{j \in J} A_ j)[T]$ and every factorization $\overline{f} = g_0h_0$ with monic $g_0, h_0 \in (\prod _{j \in J} A_ j / \prod _{j \in J} I_ j)[T] = (\prod _{j \in J} A_ j/I_ j)[T]$ that generate the unit ideal in $(\prod _{j \in J} A_ j / \prod _{j \in J} I_ j)[T]$, there exists a factorization $f = gh$ in $(\prod _{j \in J} A_ j)[T]$ with $g$, $h$ monic and reducing to $g_0$, $h_0$. In conclusion, according to Definition 15.11.1 $(\prod _{j \in J} A_ j, \prod _{j\in J} I_ j)$ is a Henselian pair. $\square$

Lemma 15.11.12. The property of being Henselian is preserved under limits of pairs. More precisely, let $J$ be a preordered set and let $(A_ j, I_ j)$ be an inverse system of henselian pairs over $J$. Then $A = \mathop{\mathrm{lim}}\nolimits A_ j$ equipped with the ideal $I = \mathop{\mathrm{lim}}\nolimits I_ j$ is a henselian pair $(A, I)$.

Proof. By Categories, Lemma 4.14.10, we only need to consider products and equalizers. For products, the claim follows from Lemma 15.11.11. Thus, consider an equalizer diagram

$\xymatrix{ (A, I) \ar[r] & (A', I') \ar@<1ex>[r]^{\varphi } \ar@<-1ex>[r]_{\psi } & (A'', I'') }$

in which the pairs $(A', I')$ and $(A'', I'')$ are henselian. To check that the pair $(A, I)$ is also henselian, we will use the Gabber's criterion in Lemma 15.11.6. Every element of $1 + I$ is a unit in $A$ because, due to the uniqueness of the inverses of units, this may be checked in $(A', I')$. Thus $I$ is contained in the Jacobson radical of $A$, see Algebra, Lemma 10.18.1. Thus, let

$f(T) = T^{N - 1}(T - 1) + a_{N - 1} T^{N - 1} + \dotsb + a_1 T + a_0$

be a polynomial in $A[T]$ with $a_{N - 1}, \dotsc , a_0 \in I$ and $N \ge 1$. The image of $f(T)$ in $A'[T]$ has a unique root $\alpha ' \in 1 + I'$ and likewise for the further image in $A''[T]$. Thus, due to the uniqueness, $\varphi (\alpha ') = \psi (\alpha ')$, to the effect that $\alpha '$ defines a root of $f(T)$ in $1 + I$, as desired. $\square$

Lemma 15.11.13. The property of being Henselian is preserved under filtered colimits of pairs. More precisely, let $J$ be a directed set and let $(A_ j, I_ j)$ be a system of henselian pairs over $J$. Then $A = \mathop{\mathrm{colim}}\nolimits A_ j$ equipped with the ideal $I = \mathop{\mathrm{colim}}\nolimits I_ j$ is a henselian pair $(A, I)$.

Proof. If $u \in 1 + I$ then for some $j \in J$ we see that $u$ is the image of some $u_ j \in 1 + I_ j$. Then $u_ j$ is invertible in $A_ j$ by Algebra, Lemma 10.18.1 and the assumption that $I_ j$ is contained in the Jacobson radical of $A_ j$. Hence $u$ is invertible in $A$. Thus $I$ is contained in the Jacobson radical of $A$ (by the lemma).

Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0 h_0$ be a factorization with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$. Write $1 = g_0 g'_0 + h_0 h'_0$ for some $g'_0, h'_0 \in A/I[T]$. Since $A = \mathop{\mathrm{colim}}\nolimits A_ j$ and $A/I = \mathop{\mathrm{colim}}\nolimits A_ j/I_ j$ are filterd colimits we can find a $j \in J$ and $f_ j \in A_ j$ and a factorization $\overline{f}_ j = g_{j, 0} h_{j, 0}$ with $g_{j, 0}, h_{j, 0} \in A_ j/I_ j[T]$ monic and $1 = g_{j, 0} g'_{j, 0} + h_{j, 0} h'_{j, 0}$ for some $g'_{j, 0}, h'_{j, 0} \in A_ j/I_ j[T]$ with $f_ j, g_{j, 0}, h_{j, 0}, g'_{j, 0}, h'_{j, 0}$ mapping to $f, g_0, h_0, g'_0, h'_0$. Since $(A_ j, I_ j)$ is a henselian pair, we can lift $\overline{f}_ j = g_{j, 0} h_{j, 0}$ to a factorization over $A_ j$ and taking the image in $A$ we obtain a corresponding factorization in $A$. Hence $(A, I)$ is henselian. $\square$

Example 15.11.14 (Moret-Bailly). Lemma 15.11.13 is wrong if the colimit isn't filtered. For example, if we take the coproduct of the henselian pairs $(\mathbf{Z}_ p, (p))$ and $(\mathbf{Z}_ p, (p))$, then we obtain $(A, pA)$ with $A = \mathbf{Z}_ p \otimes _\mathbf {Z} \mathbf{Z}_ p$. This isn't a henselian pair: $A/pA = \mathbf{F}_ p$ hence if $(A, pA)$ where henselian, then $A$ would have to be local. However, $\mathop{\mathrm{Spec}}(A)$ is disconnected; for example for odd primes $p$ we have the nontrivial idempotent

$(1/2 \otimes 1) \left(1 \otimes 1 - (1 + p)^{-1}u \otimes u\right)$

where $u \in \mathbf{Z}_ p$ is a square root of $1 + p$. Some details omitted.

Lemma 15.11.15. Let $A$ be a ring. There exists a largest ideal $I \subset A$ such that $(A, I)$ is a henselian pair.

Lemma 15.11.16. Let $(A, I)$ be a henselian pair. Let $\mathfrak p \subset A$ be a prime ideal. Then $V(\mathfrak p + I)$ is connected.

Proof. By Lemma 15.11.8 we see that $(A/\mathfrak p, I + \mathfrak p/\mathfrak p)$ is a henselian pair. Thus it suffices to prove: If $(A, I)$ is a henselian pair and $A$ is a domain, then $\mathop{\mathrm{Spec}}(A/I) = V(I)$ is connected. If not, then $A/I$ has a nontrivial idempotent by Algebra, Lemma 10.20.4. By Lemma 15.11.6 this would imply $A$ has a nontrivial idempotent. This is a contradiction. $\square$

Comment #2997 by Peng Du on

In Lemma 15.10.11 (1), the pair (J/I, A/I) should be written as (A/I, J/I).

Comment #3120 by on

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