Posted by Jason Polak on 28. September 2017 · 2 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 20. September 2017 · Write a comment · Categories: homological-algebra, model-theory

There are all sorts of notions of dimension that can be applied to rings. Whatever notion you use though, the ones with dimension zero are usually fairly simple compared with the rings of higher dimension. Here we’ll look at three types of dimension and state what the rings of zero dimension look like with respect to each type. Of course, several examples are included.

All rings are associative with identity but not necessarily commutative. Some basic homological algebra is necessary to understand all the definitions.

Global Dimension

The left global dimension of a ring $R$ is the supremum over the projective dimensions of all left $R$-modules. The right global dimension is the same with “left” replaced by “right”. And yes, there are rings where the left and right global dimensions differ.

However, $R$ has left global dimension zero if and only if it has right global dimension zero. So, it makes sense to say that such rings have global dimension zero. Here is their characterisation:

A ring $R$ has global dimension zero if and only if it is semisimple; that is, if and only if it is a finite direct product of full matrix rings over division rings.

Examples of such rings are easy to generate by this characterisation:

  1. Fields and finite products of fields
  2. $M_2(k)$, the ring of $2\times 2$ matrices over a division ring $k$
  3. etc.

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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 22. February 2017 · Write a comment · Categories: commutative-algebra, homological-algebra · Tags:

Let $R$ be a commutative ring and $A$ and $R$-module. We say that $x_1,\dots,x_n\in R$ is a regular sequence on $A$ if $(x_1,\dots,x_n)A\not = A$ and $x_i$ is not a zero divisor on $A/(x_1,\dots,x_{i-1})A$ for all $i$. Regular sequences are a central theme in commutative algebra. Here’s a particularly interesting theorem about them that allows you to figure out a whole bunch of Ext-groups:

Theorem. Let $A$ and $B$ be $R$-modules and $x_1,\dots,x_n$ a regular sequence on $A$. If $(x_1,\dots,x_n)B = 0$ then
{\rm Ext}_R^n(B,A) \cong {\rm Hom}_R(B,A/(x_1,\dots,x_n)A)$$

This theorem tells us we can calculate the Ext-group ${\rm Ext}_R^n(B,A)$ simply by finding a regular sequence of length $n$, and calculating a group of homomorphisms. We get two cool things out of this theorem: first, a corollary of this theorem is that any two maximal regular sequences on $A$ have the same length if they are both contained in some ideal $I$ such that $IA\not= A$, and second, it enapsulates a whole range of Ext-calculations in an easy package.

For example, let’s say we wanted to calculate ${\rm Ext}_\Z^1(\Z/2,\Z)$. Well, $2\in\Z$ is a regular sequence, and so the above theorem tells us that this Ext-group is just ${\rm Hom}_\Z(\Z/2,\Z/2) \cong\Z/2$.

Another example: is ${\rm Ext}_{\Z[x]}^1(\Z,\Z[x])\cong\Z$.

Of course, the above theorem is really just a special case of a Koszul complex calculation. However, it can be derived without constructing the Koszul complex in general, and so offers an instructive and minimalist way of seeing that for Noetherian rings and finitely generated modules, the notion of length of a maximal regular sequence is well-defined.

Posted by Jason Polak on 25. January 2017 · Write a comment · Categories: commutative-algebra, homological-algebra

We already saw that an abelian group with a $\Z$-direct summand is projective over its endomorphism ring. Finitely generated abelian groups are also projective over their endomorphism rings by essentially the same argument. What’s an example of an abelian group that is not projective over its endomorphism ring?

Here’s one: the multiplicative group $Z(p^\infty)$ of all $p$-power roots of unity. Another way to define this group is $\Z[p^{-1}]/\Z$. What is the endomorphism ring of this group? In fact it is the $p$-adic integers $\Z_p$. Indeed, an endomorphism $Z(p^\infty)\to\Z(p^\infty)$ has to send $1/p$ to an element $a_1$ such that $pa_1 = 0$. So we have the choice of the elements $0/p, 1/p,\cdots, (p-1)/p$, which form the cyclic subgroup $\Z/p$.

Similarly, $1/p^2$ has to be sent to an element $a_2$ such that $p^2a_2 = 0$, but also $pa_2 = a_1$. So $a_2$ has to be of the form $n/p^2$ where $n\in \Z$; in other words, $a_2$ can be in the cyclic subgroup $\Z/p^2$ generated by $1/p^2$. Hence, an endomorphism of $Z(p^\infty)$ is specified by an element of the inverse system $\cdots\to \Z/p^3\to \Z/p^2\to \Z/p$ where the transition maps are multiplication by $p$: in other words the $p$-adic integers $\Z_p$.

Now, we see that $Z(p^\infty)$ cannot be a projective $\Z_p$-module. Indeed, $\Z_p$ is a local ring and hence any projective $\Z_p$-module is in fact free (Kaplansky’s theorem) and in particular torsionfree. However, $Z(p^\infty)$ has nothing but torsion! In fact we can say more: since $\Z_p$ is a principal ideal domain, it has global dimension one, so the projective dimension of $Z(p^\infty)$ as a $\Z_p$-module is one.

A ring of left global dimension zero is a ring $R$ for which every left $R$-module is projective. These are also known as semisimple rings of the Wedderburn-Artin theory fame, which says that these rings are precisely the finite direct products of full matrix rings over division rings. Note the subtle detail that “semisimple” is used here instead of “left semisimple” because left semisimple is the same thing as right semsimple.

In the commutative world, the story for Krull dimension zero is not so simple. For example, every finite commutative ring has Krull dimension zero. Indeed, if $R$ is a ring with Krull dimension greater than zero, then there would exist two distinct primes $P\subset Q$ so that $R/P$ is an integral domain that is not a field. Thus, $R$ is infinite, as every finite integral domain is a field.

The story becomes simpler if we require $R$ to have no nilpotent elements: i.e., that $R$ is reduced. In this case, a commutative ring is reduced and of Krull dimension zero if and only if every principal ideal is idempotent. Every principal ideal being idempotent means that for every $x\in R$ there is an $a\in R$ such that $xax = x$. Rings, commutative or not, satisfying this latter condition are called von Neumann regular. So:

Theorem. A commutative ring has Krull dimension zero and is reduced if and only if it is von Neumann regular.

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Posted by Jason Polak on 05. October 2016 · 2 comments · Categories: commutative-algebra, homological-algebra · Tags:

For finite commutative rings, integral domains are the same as fields. This isn’t too surprising, because an integral domain $R$ is a ring such that for every nonzero $a\in R$ the $R$-module homomorphism $R\to R$ given by $r\mapsto ra$ is injective. Fields are those rings for which all these maps are surjective. But injective and surjective coincide for endofunctions of finite sets. Therefore, domains are the same thing as fields for finite rings.

But did you know that there is another class of commutative rings for which fields are the same as integral domains? Indeed, for self-injective rings, fields are the same as domains. By definition, a commutative ring $R$ is self-injective if $R$ is injective as an $R$-module. Note: for noncommutative rings, which we don’t consider here, there is a difference between left and right self-injective; that is, an arbitrary ring may be injective as a left module over itself, but not right self-injective, and vice-versa.

In other words, self-injective integral domains are fields. And, the proof is sort of along the lines of the one for finite rings:

Proof. Let $a\in R$ be nonzero. Then the multiplication map $R\xrightarrow{a} R$ is injective, and fits into a diagram


Where the dotted arrow exists because $R$ is injective as an $R$-module; since it is a map $R\to R$ it is given by multiplication by some $b\in R$. Therefore $1 = ab$. QED.

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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?