Posted by Jason Polak on 21. September 2016 · Write a comment · Categories: commutative-algebra · Tags:

Suppose $R$ is a Noetherian local ring with unique maximal ideal $m\subset R$. We say that $R$ is regular if the dimension of $R$ is equal to the dimension of $m/m^2$ as an $R/m$-vector space. Regular local rings arise as the local rings of varieties over a field corresponding to smooth points, and this gives an abundant supply of them: for a field $k$, the rings $k[x,y]_{(x,y)}$ and $k[x,y]_{(x-a,y-a^2)}/(y – x^2)$ for instance are regular local rings.
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Posted by Jason Polak on 10. September 2016 · Write a comment · Categories: math

Let $R$ be an integral domain and let $f:R\to R$ be an automorphism of $R$. Is it always true that $\mathrm{Frac}(R^f) = [\mathrm{Frac}(R)]^f$ where $\mathrm{Frac}$ denotes the fraction field, and $(-)^f$ denotes the ring of fixed elements under $f$?
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Posted by Jason Polak on 23. August 2016 · Write a comment · Categories: group-theory · Tags: ,

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Let $k$ be a commutative ring. Let $\G_a$ be group functor $\G_a(R) = R$ and $\G_m$ be the group functor $\G_m(R) = R^\times$, both over the base ring $k$. What are the homomorphisms $\G_a\to \G_m$? In other words, what are the characters of $\G_a$? This depends on the ring, of course!

The representing Hopf algebra for $\G_a$ is $k[x]$. And, the representing Hopf algebra for $\G_m$ is $k[x,x^{-1}]$. Homomorphisms $\G_a\to \G_m$ correspond to Hopf algebra maps $k[x,x^{-1}]\to k[x]$. Such a map is a $k$-algebra homomorphism that satisfies the additional conditions for being a Hopf algebra homomorphism.
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Posted by Jason Polak on 14. August 2016 · Write a comment · Categories: math

Let $V$ be a vector space over a field $k$ and $S = \oplus S_n$ the symmetric algebra on $V$. If $f$ is a $k$-endomorphism of $V$, then $f$ extends to a linear operator $f_n$ on $S_n$ for each $n$. What is the trace of $f$ on $S_n$? There’s a surprisingly elegant way to compute this, which encapsulates the combinatorics of computing $f_n$ directly into a formal power series, which I learnt from Bourbaki’s “Groupes et algebres de Lie”:
$$
\sum_{m=0}^\infty {\rm Tr}(f_n)T^n = \det(1 – fT)^{-1}
$$
The proof of this identity is not difficult. First, we may as well assume $k$ is algebraically closed since computing the trace does not depend on the conjugacy class of a matrix chosen for $f$. In this case we can choose a basis $v_1,\dots, v_k$ of $V$ such that $f$ is in lower-triangular form, with diagonal entries $\lambda_1,\dots,\lambda_k$.

Choose a basis of the symmetric algebra $S_n$ to be the vectors $v_1^{i(1)}\otimes\cdots\otimes v_k^{i(k)}$ with $\sum_j i(j) = n$, ordered lexicographically. Then the diagonal entries of $f_n$ will be the products $\lambda_1^{i(1)}\cdots\lambda_k^{i(k)}$ (this part requires a little thought), so
$$
{\rm Tr}(f_n) = \sum_{i(1) + \cdots + i(k) = n} \lambda_1^{i(1)}\cdots\lambda_k^{i(k)}
$$
Hence,
$$
\sum_{n=0}^\infty {\rm Tr}(f_n)T^n \\
= \sum_{n=0}^\infty\sum_{i(1) + \cdots + i(k)=n}\lambda_1^{i(1)}\cdots\lambda_k^{i(k)}T^n\\
=\sum_{n=0}^\infty \lambda_1^nT^n \cdots \sum_{n=0}^\infty \lambda_nT^n\\
=(1-\lambda_1T)^{-1}\cdots (1-\lambda_kT)^{-1}
=\det(1 – fT)^{-1}
$$
The key observation was that term $\lambda_1^{i(1)}\cdots\lambda_k^{i(k)}T^n$ for $i(1) + \cdots + i(k) = n$ corresponds to the way power series are multiplied. To give an example, consider the matrix
$$
\begin{pmatrix}1 & 2\\-1 & 4\end{pmatrix}
$$
on a $2$-dimensional vector space. Then $\det(1 – fT)$ is $(1- 2T)(1 – 3T)$. To compute its inverse we must find the following product of power series:
$$
(1 + 2T + 4T^2 + 8T^3 + \cdots)(1 + 3T + 9T^2 + 27T^3 + \cdots)
$$
For example, ${\rm Tr}(f_3) = 8 + 4\cdot 3 + 2\cdot 9 + 27 = 65$. In general,
$$
{\rm Tr}(f_k) = \sum_{j=0}^k 2^{k-j}3^j.
$$
Getting power series to keep track of this calculation really makes the calculation straightforward – for example,
$$
{\rm Tr}(f_{100}) = 1546132562196033990574082188840405015112916155251
$$

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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