Let $R$ be a ring and $M$ and $R$-module. If every finitely generated submodule of $M$ is flat, then so is $M$, because direct limits commute with the $\mathrm{Tor}$-functor. What about the converse? If $M$ is flat, are all its finitely generated submodules flat too?

Not necessarily! In fact, here’s a roundabout argument without an actual counterexample: we’ve already seen that the weak dimension of a ring is less than or equal to one iff every ideal is flat. And, for Noetherian rings, the weak dimension is the same as the global dimension. For a field, the global dimension of $k[X]:=k[x_1,\dots,x_n]$ is $n$ and so if $n\geq 2$ then $k[X]$ must have ideals that are not flat, and yet each ideal is finitely generated. Hence $k[X]$ as a $k[X]$-module is flat (as it’s free) but has finitely generated $k[X]$-submodules that cannot be flat.

Amusingly, this counterexample is also a counterexample to the statement that to any conjecture one should give either a proof or an explicit counterexample!

Hint: for an actual counterexample, $(x,y)$ in $k[x,y]$ works!

Given an idempotent $e$ in a ring $R$, the right ideal $eR$ is projective as a right $R$-module. In fact, $eR + (1-e)R$ is actually a direct sum decomposition of $R$ as a right $R$-module. An easy nontrivial example is $\Z\oplus\Z$ with $e = (1,0)$.

Fix an $a\in R$. If $aR$ is a projective right $R$-module, however, that doesn’t mean that $a$ is an idempotent. In fact $aR$ is projective whenever $a$ is a nonzerodivisor, and in this case $aR$ is just isomorphic to $R$ itself as a right $R$-module.

So how do idempotents come into play in general? It turns out we have to look at annihilators! The right annihilator of $e$ is the right ideal $(1-e)R$. Indeed, $e(1-e) = 0$. And, if $er = 0$, then $(1 – e)r = r$, so anything that annihilates $e$ is a multiple of $(1-e)$. So we see that the annihilator of $eR$ is $(1-e)R$.

What about in general? It turns out that if $aR$ is projective, the right annihilator of $a$ must be of the form $eR$ for an idempotent $e$. Indeed, if $aR$ is projective, then the map $R\to aR$ given by $r\mapsto ar$ has a splitting $\varphi:aR\to R$. I’ll leave it as an exercise to show that the right annihilator of $a$ is $(1 – \varphi(a))R$, and that $1 – \varphi(a)$ is in fact an idempotent.

Conversely, if the right annihilator of an $a\in R$ is of the form $eR$ for some idempotent, then multiplication by $1-e$ gives the splitting of the natural map $R\to aR$, so $aR$ must be projective.

Posted by Jason Polak on 16. June 2016 · Write a comment · Categories: commutative-algebra, homological-algebra, modules

Projective modules are the algebraic analogues of vector bundles, and they satisfy some strong properties. To state one we will first introduce the notation $P^* := {\rm Hom}_R(P,R)$ for any right $R$-module $P$. (Working with right $R$-modules is just a convention)

Here’s one property that projective modules satisfy: if $P$ is a right projective module over a ring $R$ then the natural map
$$
e:P\to P^{**}$$
given by $e(p)(f) = f(p)$ is a monomorphism—which, in the category of $R$-modules, just means that $e$ is injective. The first question should be: is it ever not an isomorphism? The lack of surjectivity for $e$ can already be found when $R = k$ is a field.

Here, if $P = \oplus_I k$ then ${\rm Hom}_k(\oplus_I k,k) = \prod_I k$ so the dual has strictly greater cardinality as soon as $I$ is an infinite set. In fact, this same argument shows that the $P$ cannot be isomorphic to $P^{*}$, let alone $P^{**}$ whenever $P$ is not finitely generated.

But $e$ is always a monomorphism whenever $P$ is projective. If $P$ is arbitrary, then $e$ may not be a monomorphism. For example if $R = \Z$ then $P=\Z/2$ is a counterexample. ${\rm Hom}_\Z(\Z/2,\Z) = 0$. Another more striking example is $P = \Q$, the rational numbers. So, $e$ may fail to be a monomorphism even when $P$ is flat.

Can you give any examples of $e$ being a monomorphism even when $P$ is not projective?

Posted by Jason Polak on 21. April 2016 · 1 comment · Categories: math

I was looking through the past episodes of Wild Spectral Sequences, where we did the snake lemma, five lemma, cohomological dimension formula, Schanuel’s lemma (my favourite), and an application of the LHS sequence to calculating H^1 of tori. However, I’m surprised we never did the 3×3 lemma:

Theorem.If we have a 3×3 diagram:

3x3diag

where all the rows are exact, and two out of the three columns are exact, then all three columns are exact.

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Posted by Jason Polak on 29. March 2016 · Write a comment · Categories: math

Let $V$ be a variety over a field $k$ with separable closure $\bar k$. A $\bar k/k$-form of $V$, or simply form, is a variety $W$ such that $W_{\bar k} = V_{\bar k}$, where $\bar k$ is the separable closure of $k$. If $V$ is quasiprojective then then forms of $V$ are in bijection with the Galois cohomology set $H^1(k,\mathrm{Aut}(V))$. Here, the Galois action on $\mathrm{Aut}(V)$ is given by $\sigma\circ f = (1\otimes \sigma^{-1})f(1\otimes\sigma)$.

Let’s illustrate this with an example: the multiplicative group $\G_m$. Here, we are looking for forms that are also algebraic groups. Are there any nontrivial ones? Yes indeed! Let $E/k$ be a quadratic extension with $\langle\sigma\rangle = \mathrm{Gal}(E/k)$. Then the algebraic $k$-group $F$ defined for each $k$-algebra $R$ by

$$F(R) = (\mathrm{Res}^1_{E/k}\G_m)(R) = \{ x \in E\otimes_k R : x\sigma(x) = 1\}$$

is a form of $\G_m$, because $F_E\cong \G_{m,E}$. It’s also not isomorphic to $\G_m$ over $k$ (why? though we’ll see one reason why later). In any case, we’ve found two kinds of one-dimensional tori: $\G_m$, and $F(R)$ for a quadratic extension $E/k$. Are there any other ones? In this post we’ll classify all one-dimensional tori.
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Posted by Jason Polak on 26. March 2016 · Write a comment · Categories: math

It’s time for another episode of Wild Spectral Sequences! I haven’t written about this in a long time! Truthfully, I didn’t want to prove any more diagram lemmas, even though I still think the spectral sequence approach is much more fun than the diagram chasing approach. Today however, I’m going to talk a little about the Lyndon-Hochschild-Serre (LHS) spectral sequence for group cohomology applied to the first cohomology of tori.

A torus $T$ over a field $F$ is an algebraic group such that $T_{\overline{F}}$ is isomorphic to $\G_{m,\overline{F}}^n$. In other words, it’s a form of $\G_m^n$ for some $n$. The cohomology I’m talking about is the Galois cohomology group $H^1(F,T)$. Where does this group show up? If $G$ acts on an $F$-variety $V$, and $T\subset G$ is an $F$-torus that is the stabiliser of some $v\in V(F)$, then the “stable class” $G(\overline{F})v\cap V(F)$ of $v$ in $V(F)$ may be different than the usual orbit $G(F)v$.
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Posted by Jason Polak on 29. February 2016 · 1 comment · Categories: math

In the last post we proved that $F_1\Ccl$ is a category. Consider now the category $F_1^+\Ccl$ whose objects are pairs $(A\to A’, A’/A)$ where $A\to A’$ is an object of $F_1\Ccl$ and $A’/A$ is a choice of pushout of $*\leftarrow A\to A’$. In other words, we can think of $F_1^+\Ccl$ to be the category of sequences $A\to A’\to A’/A$ where $A’/A$ is the pushout. Perhaps even better, we can think of $F_1^+\Ccl$ as the category whose objects are all commutative squares

pushoutAst

that are pushout squares. Here I used the symbol $B$ to denote an arbitrary pushout as there may be more than one choice of pushout. In general I’ll just write $A’/A$ as I’ll just be talking about one choice at a time.
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Posted by Jason Polak on 29. February 2016 · Write a comment · Categories: math

TikZ is a drawing language for LaTeX that can produce all sorts of diagrams, including commutative diagrams with the tikz-cd package, which is possibly the best package for commutative diagrams, at least with regards to typesetting quality and usability. Sometimes when you draw a diagram, though, you might want a little more programming language thrown in to make drawing easier. A classic example is a timeline. Though probably not useful for research math papers, a timeline could be helpful for a survey paper or history text.

Suppose you have a tab-delimited text file of dates of birthdays:

1809-02-12	Charles Darwin
1822-12-27	Louis Pasteur
1826-09-17	Bernhard Riemann
1877-02-07	Godfrey Harold Hardy
1845-03-03	Georg Cantor

You could draw them on a timeline in TikZ directly, but the problem is you’d have to worry about manually calculating how far down the line these dates are. This is a good case where writing a program in Python 3 that outputs the TikZ picture drawing commands is much easier and painless, especially if you need several timelines:
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Posted by Jason Polak on 19. February 2016 · 1 comment · Categories: math

MathJobs.org by the AMS has been the standard for most North American jobs for years. Two PhD students Daniel Luetgehetmann and Sebastian Meinert have launched a new platform called MathHire.

So, how does it compare to MathJobs? Obviously one comparison is the price point. Both sites have two types of listings: advertising only, and a application management system. For MathJobs, the current pricing is, quoted from their page:

Regular Account (for up to seven ads, full functioning, 12 months from date of sign-up), US610
Regular Account (for one ad only, full functioning, 12 months from date of sign-up), US415
Advertising-only (for up to seven ads, no online applications, 12 months from date of sign-up), US495
Advertising-only (for one ad only, no online applications, 12 months from date of sign-up), US305
Upgrade from existing single-ad account to seven ad account, US295

For MathHire (for academic jobs), the ‘advertising only option is free, and the advertising and application management system is supposed to be 200 euros (currently about 220USD), but all jobs posted before April 1, 2016 will be free as well. This makes sense — the system is new and it needs users to be sustainable.

MathJobs, despite its antiquated 90’s look, is definitely the more polished looking of the two in terms of usability. For example, the UI of MathJobs is much better at presenting information in a succint way. The MathHire website does not take up the full browser screen width and requires lots of scrolling to see a small amount of information. Also, it was difficult to tell from the job listings page which ones had the feature for applying on MathHire, whereas on MathJobs this is evident. The top bar on MathHire takes up too much screen space, especially on small-screened laptops.

However, I admire MathHire and I think it’s an interesting creation that deserves attention. For one, as we’ve seen, posting only is free. Given the extremely low technical requirements for managing a posting-only system, it seems a bit exorbitant to charge 305 dollars for one ad, as MathJobs does. MathHire also promises more features for reviewing applications from their website. I can’t comment too much on this, since I’ve not reviewed jobs on either sites.

If MathHire would work on making an intuitve, polished UI (which is one of the most important factors for the success of an online community, cf. stackexchange), I believe it would have a chance to overtake MathJobs eventually.

Update: since this post, the MathHire website user interface has been updated in response to some of the comments in this post and looks good. I encourage you to check it out!

Posted by Jason Polak on 13. February 2016 · Write a comment · Categories: math

Let $\Ccl$ be a category with cofibrations. Recall that we have introduced a subcategory of the arrow category of $\Ccl$ as follows:

Definition. Define a new category $F_1\Ccl$ to be the full subcategory of ${\rm Ar}\Ccl$ whose objects are the cofibrations of $\Ccl$, and whose cofibrations $(A\to A’)\to (B\to B’)$ are those pairs $(\varphi:A\to B,\psi:A’\to B’)$ that satisfy the following two properties:

  1. $A\to B$ is a cofibration.
  2. $A’\cup_A B\to B’$ is a cofibration.

We have already discussed this definition and hopefully motivated it. Now comes the hard part: we need to prove the next theorem! The tricky part of course will be the third axiom (isn’t it always the third axiom?) because it requires more three-dimensional-ish diagram reasoning.

Theorem. The category $F_1\Ccl$ is a category with cofibrations.

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