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

2016-10-01-fdomain

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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Posted by Jason Polak on 05. October 2016 · Write a comment · Categories: commutative-algebra

Let $R$ be a ring and $M$ be a nonzero left $R$-module. If we take a nonzero $m\in M$, then the map $R\to M$ given by $r\mapsto rm$ has some kernel $I$, which is a left ideal of $R$ and thus $M$ admits a left $R$-submodule isomorphic to $R/I$. So, arbitrary modules contain submodules isomorphic to quotients of $R$ by left ideals.

In the commutative world, a remarkable fact is that sometimes you can ensure that an $R$-module contains a submodule isomorphic to $R/P$ for some prime $P$! This happens when $R$ is Noetherian and $M$ is finitely generated. Then the zero divisors of $R$ on $M$ (thos elements $r\in R$ such that $rm = 0$ for some nonzero $m\in M$) by the theory of associated primes is a set that is the union of primes, each of which is the annihilator of a nonzero element of $M$.

Therefore, if $R$ is Noetherian and $M$ is finitely generated, you can always find an $m\in M$ such that $R\to M$ given by $r\mapsto rm$ has kernel $P$ where $P$ is a prime ideal! Comes in handy on occasion.

Can you find a counterexample of a Noetherian ring $R$ and an infinitely generated module $M$ where $M$ does not contain any submodule isomorphic to one of the form $R/P$ where $P$ ranges over the primes?

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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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 22. December 2015 · Write a comment · Categories: algebraic-geometry, commutative-algebra · Tags: ,

Let $F$ be a field and $E/F$ be a nontrivial Galois extensions with Galois group $\Gamma$. If $V$ is an $F$-scheme then the points $V(E)$ carry a natural action of $\Gamma$ via the action on $\mathrm{Spec}(E)$. Sometimes, however, $V$ might have two Galois actions. How does this arise?

Perhaps the most natural setting is when $V$ is defined using the restriction of scalars functor. For example, if $X$ is an $E$-scheme, then $V = \mathrm{Res}_{E/F}(X)$ is by definition the scheme whose points in an $F$-algebra $R$ are given by
$$V(R) = X(R\otimes_F E).$$
Then, for any such $R$, the set $V(R)$ has a natural action of $\Gamma$ acting on $E$ in the tensor product $R\otimes_F E$. On the other hand, if $R$ is an $E$-algebra, then $V(R)$ will also have a $\Gamma$-action, via the action of $\Gamma$ on $R$. And, these two actions won’t be the same!
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Let $R$ be any commutative ring. The content of a polynomial $f\in R[x]$ is by definition the two-sided ideal in $R$ generated by the coefficients of $f$. If $f,g\in R[x]$, then $c(fg)\subseteq c(f)c(g)$, because each coefficient of $fg$ is a linear combination of elements of $c(f)c(g)$. Sometimes, however, this inclusion is strict. For example, if $k$ is a field of characteristic two, and $R = k[u,v]$ then $f = u + vX$ satisfies $c(f^2)\subset c(f)^2$, where the inclusion is strict. Indeed, $f^2 = u^2 + v^2X^2$ so $c(f^2) = (u^2,v^2)$, whereas $c(f)c(f) = (u^2,v^2,uv)$. A ring $R$ in which $c(fg) = c(f)c(g)$ for all $f,g\in R[x]$ is called Gaussian. We have just seen that $k[u,v]$ is not Gaussian, and in fact, we didn’t even have to specify that $k$ is characteristic two. What about a polynomial ring $k[u]$ over a field $k$ in one variable? Since $k$ is a field, $k[u]$ is a principal ideal domain (PID), and PIDs are always Gaussian. These observations can be clarified by looking at the concept of weak dimension.
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Let $R$ be a ring and $M$ be an $R$-module. The flat dimension of $M$ is the infimum over all lengths of flat resolutions of $M$. Usually, the flat dimension of $M$ is denoted by $\mathrm{fd}_R(M)$. For example, $\mathrm{fd}_{\mathbb{Z}}(\mathbb{Q}) = 0$. Since $\mathbb{Q}$ has projective dimension $1$, the flat dimension and projective dimension of a module can be different. Sometimes they can be the same: $\mathbb{Z}/n$ for $n$ a positive integer has the same flat and projective dimension as $\mathbb{Z}$-modules.

The weak dimension of a ring $R$ is defined to be $\mathrm{w.dim}(R) = \sup_{M} \{ \mathrm{fd}_R(M) \}$ where $M$ runs over all left $R$-modules. Due to the symmetric nature of the tensor product, we can also take the supremum over all right $R$-modules, in contrast to the asymmetric nature of global dimension.
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In the post Examples: Projective Modules that are Not Free, we saw nine examples of projective modules that are not free. On in particular was ‘the’ submodule $M = \oplus_{i=1}^\infty \mathbb{Z}$ of $\prod_{i=1}^\infty\mathbb{Z}$. Now, that’s a cool example to be sure, but the way we showed that $M$ was not free was to cite that $\prod_{i=1}^\infty\mathbb{Z}$ is uncountable. Actually, I like the argument a lot, but it’s possible to use the idea of that example and choose $M$ instead to be finite and different from all the examples in the aforementioned post. In fact, we’ll see a large class of examples that can be constructed from the ideas here.

The idea is to take an abelian group $A$ an consider $A$ as a module over its endomorphism ring $E = \mathrm{Hom}(A,A)$, where the endomorphisms are just homomorphisms $A\to A$ of abelian groups. Sometimes, $A$ can be projective over $E$. Actually, for a while it was believed that the projective dimension of $A$ over $E$ could only be $0$ or $1$, but eventually I.V. Bobylev showed in [1] that $A$ could have any projective dimension over $E$, including infinity!
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