# Two Elements Mutually Divisible, Unit Multiple?

Posted by Jason Polak on 29. December 2016 · Write a comment · Categories: commutative-algebra

Let $R$ be an integral domain and $a,b\in R$. Suppose that $a | b$ and $b | a$. By definition, this means that $ax = b$ and $by = a$ for some $x,y\in R$, and so $b(1 – yx) = 0$. If $b\not = 0$ then $yx = 1$ and so $x$ and $y$ are units. Of course, if $b$ is zero, then $a = 0$, and in either case $b = ax$ where $x$ is a unit.

Concluding, we see that in an integral domain, if $a | b$ and $b | a$ then $b = ax$ where $x$ is a unit of $R$. Did you know that this implication can fail when $R$ is not an integral domain?

Here’s a counterexample that I learned in a paper of Anderson and Valdez-Leon. Let $R = k[X,Y,Z]/(X – XYZ)$ where $k$ is a field, and let $x,y,z$ denote respectively the images of $X,Y,Z$ in $R$. Then $x | xy$ and $xy | x$, the latter because $xyz = x$. However, it is not possible to write $xy$ as a unit multiple of $x$.

Why is this? Well, first of all, $y$ is definitely not a unit in $R$ because in the quotient ring obtained by setting $x=z=0$, this would mean that $y$ would be a unit in the polynomial ring $k[y]$. On the other hand, this argument does not preclude the existence of an $f\in R$ such that $fx = xy$.

So suppose such an $f$ did exist. Then in $k[X,Y,Z]$, we would have the relation $fX – YX = gX(1 – YZ)$ for some $g\in k[X,Y,Z]$. Since $k[X,Y,Z]$, is an integral domain, we have $f = Y + g(1 – YZ)$. If $f$ were a unit in $R$, then by setting $X = 0$ we see that $Y + g(1 – YZ)$ would be a unit in $k[Y,Z]$. Setting $Y = Z$ shows that $Z + g(1 – Z^2) = Z + g – gZ^2$ would be a unit in $k[Z]$, which is impossible since the term $gZ^2$ shows $Z + g – gZ^2$ is not a constant polynomial.

I like this example not only because it’s neat, but because it illustrates the principle that a good place to look for counterexamples in commutative ring theory is in the land of quotients of polynomial rings.

# Example: Separability Idempotent for a Field Extension

Posted by Jason Polak on 26. November 2016 · Write a comment · Categories: commutative-algebra · Tags:

Let $R$ be a commutative ring and let $A$ be an $R$-algebra. We say that $A$ is separable if $A$ is projective as an $A\otimes_RA^{\rm op}$-module. There is a multiplication map $\mu:A\otimes_RA^{\rm op}\to A$ given by $a\otimes a’\mapsto aa’$, whose kernel we’ll call $J$.

It’s a fact that $A$ is separable if and only if there exists an idempotent $e\in A\otimes_RA^{\rm op}$ such that $\mu(e) = 1$ and $Je = 0$. The most familiar examples of separable algebras to many readers are probably the finite separable field extensions of a given field $R$. So let’s see an example of the separability idempotent!

For example, $\Q(\sqrt{2})$ is a separable $\Q$-algebra. One can then write down explicitly the above definitions. For example, $\Q(\sqrt{2})\otimes_\Q\Q(\sqrt{2})\cong \Q(\sqrt{2})\oplus \Q(\sqrt{2})$ via the map given on pure tensors by $(a,b)\mapsto (ab,a\sigma(b))$ where $\sigma:\Q(\sqrt{2})\to\Q(\sqrt{2})$ is the $\Q$-algebra isomorphism given by $\sqrt{2}\mapsto -\sqrt{2}$. The inverse of this map is given by
$$(u,v)\mapsto \frac{u+v}{2}\otimes 1 + \frac{u-v}{2\sqrt{2}}\otimes\sqrt{2}$$
The multiplication map under this isomorphism translates to a map
$$\Q(\sqrt{2})\oplus\Q(\sqrt{2})\to Q(\sqrt{2})\\ (a,b)\mapsto a$$
Therefore, the kernel of the multiplication map is $J = 0\oplus\Q(\sqrt{2})$ and the separability idempotent is just $e = (1,0)$. Of course, it’s naturally to want to observe this idempotent in its natural habitat of the tensor product $\Q(\sqrt{2})\otimes_Q\Q(\sqrt{2})$. Here it is folks:
$$e = \frac{1}{2}\otimes 1 + \frac{1}{2\sqrt{2}}\otimes\sqrt{2}$$
It’s easy to see that in general $J$ is the ideal generated by the set $\{ 1\otimes a – a\otimes 1 : a\in A\}$. Now just try verifying $Je = 0$ directly! Or not.

# Projectivivity and Compositions of Ring Homomorphisms

Posted by Jason Polak on 26. November 2016 · Write a comment · Categories: commutative-algebra · Tags:

Let $A\to B\to C$ be ring homomorphisms. Consider the following statements:

1. $B$ is a projective $A$-algebra,
2. $C$ is a projective $B$-algebra,
3. $C$ is a projective $A$-algebra.
Question. Which pairs of these statements imply the third?

# Automorphisms of Matrix Rings over Fields are Inner

Posted by Jason Polak on 22. November 2016 · Write a comment · Categories: commutative-algebra · Tags: ,

Let $k$ be a field. The ring $M_n(k)$ of $n\times n$ matrices over $k$ has some automorphisms, given by conjugation by elements of $\GL_n(k)$. These are inner automorphisms, and this action happens to be the adjoint action of $\GL_n$ on its Lie algebra. Are there any other automorphisms? The answer is no, and the reason fits in the framework of automorphisms of central separable algebras. In this post we’ll sketch how this works.

Let $R$ be a commutative ring. An $R$ algebra $A$ is called separable if $A$ is projective as an $A\otimes_R A^{\rm op}$-module and $A$ is called central separable if it is separable and the center of $A$ coincides with $R$.

A typical example is a matrix ring $M_n(R)$ over $R$. In fact, whenever $M$ is a finitely generated projective and faithful module then ${\rm Hom}_R(M,M)$ is a central separable $R$-algebra. So even if you don’t have much intuition for central separable algebras in general, everything we say about them will certainly apply to algebras of the form ${\rm Hom}_R(M,M)$.

For a commutative ring, let’s call a finitely generated projective module of constant rank one a line bundle. The Picard group of a commutative ring $R$ is the group of isomorphism classes of line bundles. The group operation is the tensor product. If $A$ is a central separable algebra, then $A^A = \{ b\in A : ab = ba \forall a\in A\} = R$ is a trivial example of a line bundle.

More generally, $(A_\sigma)^A = \{ b\in A : ab = b\sigma(a)\forall a\in A\}$ is also a line bundle. This gives a well-defined map
$${\rm Aut}_R(A)\to {\rm Pic}(R)$$
that happens to fit into an exact sequence
$$1\to {\rm Inn}_R(A)\to {\rm Aut}_R(A)\to {\rm Pic}(R).$$
Consequently, if ${\rm Pic}(R)$ is trivial, then every $R$-algebra automorphism of $A$ must in fact be inner. For example, this works for fields and principal ideal domains, and hence answers our query that we started with: why is every algebra automorphism of a full matrix ring inner? The construction of the exact sequence, which I will not explain at this point, actually gives a somewhat constructive answer: if $\sigma$ is a given automorphism then $(A_\sigma)^A$ is a free $R$-module on a generator $v$ that is a unit in $A$, and $\sigma(x) = v^{-1}xv$.

# Yet Another Algebra that is not Separable

Posted by Jason Polak on 22. November 2016 · Write a comment · Categories: commutative-algebra · Tags: ,

Let $R$ be a commutative ring and $A$ be an $R$-algebra. We say that $A$ is a separable $R$-algebra if $A$ is projective as an $A\otimes_R A^{\rm op}$-module, where the action of $A\otimes_RA^{\rm op}$ is given by $(a\otimes a’)b = aba’$.

We already showed that the ring of upper triangular matrices over a commutative ring is not separable. This type of example is along the lines of: given a ring $R$, give some examples of $R$-algebras that are not separable.

On the other hand, you could also try constructing examples of non-separable algebras by taking a commutative ring $S$ as asking: what subrings $R$ of $S$ are such that $S$ is not a separable $R$-algebra? One way of producing subrings of $S$ is taking a finite group $G$ of automorphisms of $S$ and setting $R = S^G$. Then $S$ is an $R$-algebra and $G$ is a group of $R$-algebra automorphisms of $S$. Then:

Theorem. Suppose $S$ has no nontrivial idempotents. Then, $S$ is a separable $R = S^G$ algebra if and only if for each maximal ideal $M$ of $S$ and for each nontrivial $\sigma\in G$, there exists an $x\in S$ such that $\sigma(x) – x\not\in M$.

For example, if $k$ is a field then $k[x,y]$ is not separable over $k[x+y,xy]$, where the algebra structure is given via the inclusion map $k[x+y,xy]\to k[x,y]$.

Why? It’s easy to check that $k[x,y]^{\Z/2} = k[x+y,xy]$ where the nontrivial element $\sigma$ of $\Z/2$ permutes $x$ and $y$. Consider the maximal ideal $M = (x,y)$. For any $f\in k[x,y]$, the element $\sigma(f) – f$ is a polynomial without constant term, and hence is in the ideal $(x,y)$. Therefore, by the stated theorem, $k[x,y]$ cannot be separable as a $k[x+y,xy]$ algebra. Of course, there’s nothing special about $\Z/2$ here: this example works for any finite subgroup of permutations of the variables $x_1,\dots,x_n$ acting on $k[x_1,\dots,x_n]$.

I should also mention that $k[X] := k[x_1,\dots,x_n]$ is not separable as a $k$-algebra either, even though it is “classically separable” (in the sense that $K[X]$ has zero Jacobson radical for every extension $K/k$). That $k[X]$ is not separable in our sense follows because in general, if $A$ is both $R$-separable and $R$-projective, then it must also be finitely generated as an $R$-module.

# Azumaya’s Theorem

Posted by Jason Polak on 20. November 2016 · Write a comment · Categories: commutative-algebra · Tags:

In the last post, we saw that an upper triangular $n\times n$ matrix ring $T_n$ over a commutative ring $R$ for $n \geq 2$ is not a separable $R$-algebra. We did this by invoking the commutator theorem: if $A$ is a central separable algebra and $B$ is a separable subalgebra then $C = A^B$ is also separable and $A^C = B$. The notation $A^B$ means $\{ a\in A : ab = ba~\forall b\in B\}$.

For $M_n(R)$, we have $M_n(R)^{T_n} = R$, whereas $M_n(R)^R = M_n(R)$. Therefore, $T_n$ cannot be separable.

Instead of using the commutator theorem, we could use Azumaya’s theorem:

Theorem (Azumaya). Let $A$ be a finitely generated faithful $R$-algebra with generating set $a_1,\dots,a_n$. Then $A$ is a central separable free $R$-algebra with basis $a_1,\dots,a_n$ if and only if the matrix $(a_{ij}) = (a_ia_j)$ is invertible in $M_n(A)$.

For example, take the upper triangular matrices $T_n$. For expository purposes, let’s suppose $n = 2$, but the general case is only more difficult to write down. So $T_2$ is certainly a free module, with basis $e_{1,1},e_{1,2},e_{2,2}$ where $e_{ij}$ is the matrix whose only nonzero entry is $1$ in the $i,j$-spot. So $T_n$ is a free $R$-module. Since the center of $T_n$ is $R$, Azumaya’s theorem tells us that $T_n$ is separable if and only if the following matrix is invertible in $M_3(T_n)$:

So, we need to check whether this matrix is invertible in $M_3(T_2)$. Now matrix multiplication here is the same thing as if were were just multiplying in $M_9(R)$, so if the determinant in $M_9(R)$ is not a unit, then it certainly cannot be invertible in $M_3(T_2)$. And by inspection, the determinant of this matrix, considered as a matrix in $M_9(R)$, is zero. Therefore, it is not invertible in $M_3(T_2)$ and $T_2$ as a result is not separable.

# What is a Hilbert Ring?

Posted by Jason Polak on 30. October 2016 · Write a comment · Categories: commutative-algebra · Tags:

In a small way, the next post will use the concept of a Hilbert ring. It’s something I’ve talked about before on this blog somewhere, but here I will summarize the essential facts of Hilbert rings so that the next post can be completely self-contained.

Now, a Hilbert ring is one of those really neat ideas, because you can reduce the question “what the devil is the Jacobson radical?” to “are there any nilpotents?”. Moreover, Hilbert rings are very common. Let me explain.

We will stay entirely in the realm of commutative rings.

Definition. A $G$-ideal $I$ in a ring $R$ is a prime ideal such that the fraction field of $R/I$ is finitely generated as an $R/I$ algebra.

So, a $G$-ideal is a special kind of prime ideal. It turns out that an ideal $I$ is a $G$-ideal in a ring $R$ if and only if it can be written as $M\cap R$ where $M$ is a maximal ideal of $R[x]$. One fact that makes $G$-ideals so important is:

Theorem. In a commutative ring $R$, the nilradical of $R$ is the intersection of all the $G$-ideals of $R$.

Maximal ideals are of course examples of $G$-ideals, because the quotient of a ring by a maximal ideal is a field. This motivates the following definition:

Definition. A Hilbert ring is a commutative ring in which every $G$-ideal is maximal.

This definition wouldn’t be much use if Hilbert rings were rare. But, they are actually quite common. For example, the ring of integers $\Z$ is a Hilbert ring. Any field is of course a Hilbert ring. More importantly:

1. The homomorphic image of a Hilbert ring is a Hilbert ring.
2. A ring $R$ is a Hilbert ring if and only if $R[x]$ is a Hilbert ring.

Hence, in particular, the finitely generated $\Z$ algebras and $k$-algebras where $k$ is a field are all Hilbert rings. This includes the coordinate rings of affine varieties over a field. So, for any of these Hilbert rings, the Jacobson radical is the nilradical!

# Two Versions of Nakayama’s Lemma

Posted by Jason Polak on 13. October 2016 · Write a comment · Categories: commutative-algebra

Nakayama’s lemma probably comes in as many flavours as ice cream.

In this post we’ll review some forms of them, and deduce some consequences. Before we continue, recall that in an associative ring $R$ with unity, the Jacobson radical is the intersection of all the left maximal ideals of $R$. It is easy to show that this ideal coincides with the intersection of all the right maximal ideals of $R$; thus, the Jacobson radical is a two-sided ideal.

The form of Nakayama’s lemma I like best is:

Nakayama’s Lemma #1. Let $R$ be a ring, $I\subset R$ an ideal contained in the Jacobson radical of $R$, and $M$ a finitely-generated left $R$-module. If $IM = M$ then $M = 0$.

# Self Injective Integral Domains are Fields: Two Proofs

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.

# Submodules of the Form R/P

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?