Posted by Jason Polak on 29. March 2018 · 2 comments · Categories: modules · Tags:

I already mentioned the idea of stably isomorphic for a ring $R$: two $R$-modules $A$ and $B$ are stably isomorphic if there exists a natural number $n$ such that $A\oplus R^n\cong B\oplus R^n$.

Let's examine a specific case: if $A$ is stably isomorphic to a free module, then let's call it stably free.

So, to reiterate: a module $A$ is called stably free if there exists a natural number $n$ such that $A\oplus R^n$ is free. We already saw an example of a stably free module, where $R = \mathbb{H}[x,y]$, the two variable polynomial ring over the quaternions.

One might wonder: why don't we allow infinite $n$ in this definition? It's because of the Eilenberg swindle, named after mathematician Samuel Eilenberg.

The Eilenberg swindle goes like this: suppose $P$ is a projective $R$-module. Then, there exists a module $Q$ such that $P\oplus Q \cong F$ where $F$ is a free module. Now, let $E = \oplus_{i=1}^\infty F$.

Then:
$$P\oplus E \cong P\oplus (P\oplus Q)\oplus (P\oplus Q)\cdots\\
P\oplus (Q\oplus P)\oplus (Q\oplus P)\oplus\\
F\oplus F\oplus F\oplus\cdots$$
Therefore, $P\oplus F$ is free. Hence, if we allowed infinite $n$ in the definition of stably free, every projective module would be stably free and there wouldn't be much point in the definition 'stably free'.

Here is an exercise for you:

Show that if $A$ is a stably free $R$ module that is not finitely generated, then $A$ is free.
Posted by Jason Polak on 22. March 2018 · Write a comment · Categories: modules · Tags: ,

If $a$ and $b$ are two real numbers and $ax = bx$, then we can't conclude that $a = b$ because $x$ may be zero. The same is true for tensor products of modules: if $A$ and $B$ are two left $R$-modules and $X$ is a right $R$-module, then an isomorphism $X\otimes_R A\cong A\otimes_R B$ does not necessarily mean that $A\cong B$. Of course, $X$ not even need be zero for this to happen.

Addition for real numbers is a little different. If $a$ and $b$ are two real numbers then $x + a = x + b$ is equivalent to $a = b$. What about for direct sums? If $A$ and $B$ are two $R$-modules, and $X$ is a third $R$ module, what if $X\oplus A\cong A\oplus B$? Is it true that $A\cong B$?

The answer is no. Perhaps this is surprising from the way direct sums work. After all, in a direct sum $X\oplus A$, it "feels like" what happens in $X$ is independent from what happens in $A$. And for vector spaces, this is true: for $k$-modules where $k$ is a field, if $X\oplus A\cong X\oplus B$, then certainly $A\cong B$, because $A$ and $B$ have the same dimension.
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Posted by Jason Polak on 14. February 2018 · Write a comment · Categories: homological-algebra, modules · Tags: , ,

Over a field $k$, an arbitrary product of copies of $k$ is a free module. In other words, every vector space has a basis. In particular, this means that arbitrary products of projective $k$-modules are projective.

Over the ring of integers, an arbitrary product of projective modules is not necessarily projective. In fact, a product of countably infinitely many copies of $\Z$ is not projective!

So while arbitrary direct sums of projective modules are projective, the same is not true of arbitrary products for some rings.

Besides fields, which other rings have the property that arbitrary direct products of their projective modules are projective?
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Posted by Jason Polak on 28. September 2017 · 3 comments · Categories: commutative-algebra, homological-algebra, modules

Here is an interesting question involving free, projective, and flat modules that I will leave to the readers of this blog for now.

First, consider free modules. If $R$ is a ring, then every $R$-module is free if and only if $R$ is a division ring. The property of $R$ being a division ring can be expressed in terms of first-order logic in the language of rings: $\forall x[x\not=0 \rightarrow \exists y(xy = 1)]$.

The meat of this first-order statement is the equation $xy = 1$. Now, multiply by $x$ on the right to get the equation $xyx = x$. Now we can put this in a first-order sentence: $\forall x\exists y[xyx = x]$. Notice how we removed the condition $x\not=0$ from this one. That's because $x=0$ satisfies $xyx = x$ for any $y$ in all rings. Rings that model $\forall x\exists y[xyx = x]$ are called von Neumann regular. More importantly, these are exactly the rings for which every $R$-module is flat.

By weakening the statement that $R$ is a division ring, we got a statement equivalent to the statement that every $R$-module is flat. One might wonder: where did the projective modules go? Is there a first-order sentence (or set of sentences perhaps) in the language of rings whose models are exactly those rings $R$ for which every $R$-module is projective? Diagrammatically:

Can we replace the question mark with a first-order sentence, or a set of them?

My initial thoughts are no because of ultraproducts, but I have not yet come up with a rigorous argument.

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 · Tags: , ,

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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