Posted by Jason Polak on 21. September 2017 · Write a comment · Categories: modules · Tags:

Here is one characterisation of commutative rings of Krull dimension zero:

Theorem. A commutative ring $R$ has Krull dimension zero if and only if every element of the Jacobson radical ${\rm Jac}(R)$ of $R$ is nilpotent and the quotient ring $R/{\rm Jac}(R)$ is von Neumann regular.

Recall that a ring $R$ is von Neumann regular if for every $x\in R$ there exists a $y\in R$ such that $xyx = x$. This odd property is equivalent to saying that every $R$-module is flat.

Here are two examples of what happens when we drop various assumptions in the “if” direction of the theorem:

  1. The ring $\Z_{(p)}$ of integers localised away from the prime $(p)$ is an example of a ring such that $R/{\rm Jac}(R)$ is von Neumann regular but ${\rm Jac}(R)$ has no nontrivial nilpotent elements. The ring $\Z_{(p)}$ has Krull dimension one.
  2. Another type of example is given by $\Z[[t]]/t^n$ where $\Z[[t]]$ denotes the power series ring with integer coefficients. Unlike our first example, the Jacobson radical of this ring is the ideal $(t)$, which is also the nilradical (=set of nilpotent elements), but $R/{\rm Jac}(R) = \Z$, which is not von Neumann regular and has Krull dimension one.

Note that we were forced look for counterexamples to dropped assumptions in the class of infinite rings. That’s because every finite commutative ring has Krull dimension zero.

Posted by Jason Polak on 19. September 2017 · Write a comment · Categories: homological-algebra, modules

Consider a field $k$. Define an action of $k[x,y]$ on $k[x]$ by $f*g = f(x,x)g(x)$ for all $f\in k[x,y]$ and $g\in k[x]$. In other words, the action is: multiply $f$ and $g$ and then replace every occurrence of $y$ by $x$.

Is $k[x]$ a projective $k[x,y]$-module? Consider first the map $k[x,y]\to k[x]$ given by $f\mapsto f(x,x)$. It’s easy to check that this map is in fact a $k[x,y]$-module homomorphism. It would be tempting to try and split this map with the inclusion map $k[x]\to k[x,y]$. But this doesn’t work: this inclusion is not a $k[x,y]$-module homomorphism.

In fact, the $k[x,y]$-module homomorphism $k[x,y]\to k[x]$ given by $f\mapsto f(x,x)$ cannot split simply because there are no nonzero $k[x,y]$-module homomorphisms $k[x]\to k[x,y]$. Therefore, $k[x]$ is not projective as a $k[x,y]$-module, using the module structure we gave it.

Here are two more ways to see this:

  1. Through the notion of separability: by definition, $k[x]$ being a projective $k[x,y]\cong k[x]\otimes_k k[x]$-module under the structure that we have defined means that $k[x]$ is a separable $k$-algebra. However, all separable $k$-algebras are finite-dimensional as vector spaces over $k$, whereas $k[x]$ is infinite-dimensional.
  2. Through Seshradi’s theorem: this theorem says that every finitely-generated projective module over $k[x,y]$ is actually free. Therefore, we just have to show that $k[x]$ is not free because $k[x]$ is certainly finitely-generated as a $k[x,y]$-module. But $(x^2y – xy^2)$ annihilates all elements of $k[x]$, which cannot happen in a free module.
Posted by Jason Polak on 27. August 2017 · Write a comment · Categories: math, modules

Let $R$ be an associative ring with identity. The Jacobson radical ${\rm Jac}(R)$ of $R$ is the intersection of all the left maximal ideals of $R$. So, ${\rm Jac}(R)$ is a left ideal of $R$. It turns out that the Jacobson radical of $R$ is also the intersection of all the right maximal ideals of $R$, and so ${\rm Jac}(R)$ is also an ideal!

The idea behind the Jacobson radical is that one might be able to explore the properties of a ring $R$ by first looking at the less complicated ring $R/{\rm Jac}(R)$. Since the ideals of $R$ containing ${\rm Jac}(R)$ correspond to the ideals of $R/{\rm Jac}(R)$, the ring $R/{\rm Jac}(R)$ has zero Jacobson radical. Often the rings $R$ for which ${\rm Jac}(R) = 0$ are called Jacobson semisimple.

This terminology might be a tad bit confusing because typically, a ring $R$ is called semisimple if every left $R$-module is projective, or equivalently, if every left $R$-module is injective. How does the notion of semisimple differ from Jacobson semisimple? The Wedderburn-Artin theorem gives a classic characterisation of semisimple rings: they are exactly the rings that are finite direct products of full matrix rings over division rings. Since a full matrix ring over a division ring has no nontrivial ideals, the product of such rings must have trivial Jacobson radical. Thus:

A semisimple ring is Jacobson semisimple.

The converse is false: there exists a ring that is Jacobson semisimple but not semisimple. For example, let $R$ be an infinite product of fields. Then ${\rm Jac}(R) = 0$. However, $R$ is not semisimple. Why not? If it were, by Wedderburn-Artin it could also be written as a finite product of full matrix rings over division rings, which must be a finite product of fields because $R$ is commutative. But a finite product of fields only has finitely many pairwise orthogonal idempotents, whereas $R$ has infinitely many.

Incidentally, because $R$ is not semisimple, there must exist $R$-modules that are not projective. However, $R$ does have the property that every $R$-module is flat!

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?

The flat dimension of an $R$-module $M$ is the infimum over lengths of flat resolutions of $M$, and the weak dimension (or $\mathrm{Tor}$-dimension) of $R$ is the supremum over all possible flat dimensions of modules. Let’s use $\mathrm{w.dim}(R)$ to denote the weak dimension of $R$. As with the global dimension, the weak dimension of $R$ can be computed as the supremum over the set of flat dimensions of the modules $R/I$ for $I$ running over the set of all left-ideals or right-ideals, either is fine!

So, if every ideal is flat, then $\mathrm{w.dim}(R) \leq 1$. What about the converse? If $\mathrm{w.dim}(R) \leq 1$, is it true that every ideal is flat? Let’s make a side remark in that if we replace weak dimension with global dimension, and flat with projective, then the answer follows from Schanuel’s lemma. However, as far as I know there is no Schanuel’s lemma when ‘projective’ is replaced by ‘flat’.

However, we can get away with using part of the proof of Schanuel’s lemma. Before continuing, the reader may wish to check out the statement and proof of Schanuel’s lemma using a double complex spectral sequence.
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Here are nine examples of projective modules that are not free, some of which are finitely generated.

Direct Products

Consider the ring $R= \Z/2\times\Z/2$ and the submodule $\Z/2\times \{0\}$. It is by construction a direct summand of $R$ but certainly not free. And it’s finitely generated! Another example is the submodule $\Z/2\subset \Z/6$, though this is the same kind of thing because $\Z/6\cong\Z/2\times\Z/3$. This was the first example I ever saw of a nonfree projective module.

Infinite Direct Products

One can modify the above construction for infinite direct products of rings, too. For instance, $R = \prod_{i=1}^\infty \Z$ contains $\Z$ as a direct summand. Hence $\oplus_{i=1}^\infty\Z$ is a projective $R$ module, yet cannot be free since nonzero free modules are uncountable.

Ideals in Dedekind Domains

In a Dedekind domain $R$, take an ideal representing a nontrivial element in the class group. It will then be projective. As an example, the class number of $\Z[\sqrt{5}]$ is two, and the ideal $(2,1+\sqrt{5})$ represents the nontrivial element in the class group. It is not free since it is not principal, and it is finitely generated projective since it is invertible.

More generally, for any ring extension of commutative rings $R\subseteq S$, one may define invertible $R$-submodules of $S$ as it is done for Dedekind domains. Then any invertible $R$-submodule of $S$ will be finitely-generated and projective. For more details and a further example, see Lam’s ‘Lectures on Modules and Rings’, Sections 2B-2C.

Rings of Continuous Functions

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Posted by Jason Polak on 19. December 2014 · Write a comment · Categories: homological-algebra, modules · Tags:

There are many ways to define the propery of semisimple for a ring $R$. My favourite is the “left global dimension zero approach”: a ring $R$ is left semisimple if every left $R$-module is projective, which is just the same thing as saying that every left $R$-module is injective. In particular, ideals are direct summands, and an easy application of Zorn’s lemma shows that $R$ can be written as a direct product of minimal left ideals, which is actually a finite sum because $R$ contains $1$.

An attack of Schur’s lemma yields the famous Wedderburn-Artin theorem: a ring $R$ is semisimple if and only if it is the finite direct product of matrix rings over division rings.

Since $R$ can be written as a finite direct product of minimal left ideals, we see that $R$ must be Noetherian and Artinian. Is the converse true?

Of course not! Here is a minimal counterexample: $\Z/4$. This ring cannot be semisimple. Indeed if it were, by the Wedderburn-Artin theorem, it would be a direct product of fields since it is commutative. It is not a field so it is not $\F_4$, and the only other possibility is $\Z/2\times\Z/2$, which it is also not isomorphic to since $\Z/4$ is cyclic.

We don’t have to appeal to the Wedderburn-Artin theorem however: the reduction map $\Z/4\to\Z/2$ makes $\Z/2$ into a $\Z/4$-module. If $\Z/4$ were semisimple, then $\Z/2$ would be a projective $\Z/4$-module, and hence at the very least as abelian groups, $\Z/2$ would be a direct summand of $\Z/4$, which is also nonsensical.

Can you think of a noncommutative example?

Let $R$ be a ring and $M$ an $R$-module with a finite free resolution (an “FFR module”). That is, there exists an exact sequence $0\to F_n\to F_{n-1}\to \cdots\to F_0\to M\to 0$ with each $F_i$ a finitely generated free $R$-module. If we denote by $r_i$ the rank of $F_i$, then the Euler characteristic of $M$ is defined to be $\chi(M) = r_0 – r_1 + \cdots + (-1)^nr_n$. One can easily prove that this is independent of the finite free resolution chosen and hence is a well-defined integer. In the post The Alternating Binomial Sum Vanishes we saw that it’s possible to prove that the alternating sum $\sum (-1)^i\binom{n}{i}$ vanishes by using some facts about Euler characteristics together with the Koszul complex.

Today’s problem shows that the Euler characteristic ins’t that interesting for ideals or homomorphic images of rings. First, here are some facts about the Euler characteristic, some of which might be useful for solving the problem:

  1. Any FFR module over a commutative ring has nonnegative Euler characteristic
  2. If an FFR module over a commutative ring has nonzero Euler characteristic, then the annihilator of $A$ is nil
  3. (Stallings) If $A$ is an FFR module with annihilator $I$ then $\chi(A)I = 0$.

We refer to these as Fact 1, Fact 2, and Fact 3.

Problem. If $I$ is a nonzero (!) ideal of a ring $R$ with a finite free resolution, show that $R/I$ also has a finite free resolution and that $\chi(I) = 1$ and $\chi(R/I) = 0$.

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Posted by Jason Polak on 03. May 2014 · Write a comment · Categories: modules · Tags:

Let \(R\) be a finite ring. The example we’ll have in mind at the end is the ring of \(2\times 2\) matrices over a finite field, and subrings. A. Kuku proved that \(K_i(R)\) for \(i\geq 1\) are finite abelian groups. Here, \(K_i(R)\) denotes Quillen’s \(i\)th \(K\)-group of the ring \(R\). In this post we will look at an example, slightly less simple than \(K_1\) of finite fields, showing that these groups can be arbitrarily large. Before we do this, let us briefly go over why this is true

But even before this, can you think of an example showing why this is false for \(i=0\)?
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