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.

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?

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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 $\sigma(x) = v^{-1}xv$.

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.

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 $M_3(T_n)$:
matrices
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.

Posted by Jason Polak on 13. November 2016 · Write a comment · Categories: math

A draft of “Derived Categories — a Textbook” by Amnon Yekutieli has appeared on the arXiv recently. Here is the abstract:

This is the first part of a book on derived categories. The purpose of the book is to provide solid foundations for the theory of derived categories, and to present several applications of this theory to algebra and algebraic geometry. The emphasis is on constructions and examples, rather than on axiomatics.

The second part of the book is expected to be finished in 2017.

Posted by Jason Polak on 13. November 2016 · Write a comment · Categories: math

Let $E/F$ be a finite extension of fields. Not too long ago we saw that the $E/F$ is separable if and only if for every extension $L/F$, the ring $E\otimes_F L$ has zero Jacobson radical.

The notion of separable algebra generalizes this concept to any $R$-algebra where $R$ is a commutative ring.

Definition. 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 $A\otimes_R A^{o}$ module, where $A^o$ denotes the opposite $R$-algebra.

The best kind of separable algebra is the central separable $R$-algebra. These are the separable $R$-algebras whose center is $R$. You can always put yourself in this case because a separable algebra is a central separable algebra over its center.

Here’s an example of a central separable algebra: if $R$ is a commutative ring then the $n\times n$ matrix ring $M_n(R)$ is a central separable $R$-algebra. What’s an example of an algebra that is not separable?

To explain how to construct them, suppose $A$ is an $R$-algebra and $B\subseteq A$ is an $R$-subalgebra. Let’s write $A^B = \{ a\in A : ab = ba \forall b\in B\}$. In words, $A^B$ is the set of elements of $A$ that commute with $B$. The notation is perhaps jarring but convenient. If $A$ happens to be central separable and $B$ is separable, then it’s a theorem that $C = A^B$ is also separable and $A^C = B$.

This gives us a way to find algebras that are not separable, as subalgebras of matrix algebras. For example, a natural guess might be $T$, the upper triangular $n\times n$, $n > 1$ matrix algebra over a commutative ring $R$. Then $T$ is a subalgebra of $M$, the full $n\times n$ matrix ring.

We claim that $T$ is not a separable $R$-algebra. Indeed, it is easy to compute that $M^T = R$, where $R$ here is embedded into $M$ as the diagonal subalgebra. If $T$ were separable, then as we have remarked, we would have $M^R = T$. But this is false: $M^R = M$. This argument applies more generally: any proper subalgebra of $M$ whose centralizer in $M$ is just $R$ cannot be separable.

Posted by Jason Polak on 30. October 2016 · Write a comment · Categories: math

Let $F$ be a field. We say that a polynomial $f\in F[x]$ is separable if $f$ has no multiple roots in an algebraic closure of $F$. Let $E/F$ be a finite extension. We say that $E$ is a separable extension of $F$ if every element in $E$ is the root of a separable polynomial with coefficients in $F$. If $F$ is perfect or a field of characteristic zero, then any finite extension of $F$ is separable. On the other hand, the extension of $\F_p((t))$ obtained by adjoining a root of $x^p – t$ is not separable, since its factorisation over an algebraic closure is $(x-t^{1/p})^p$ where $t^{1/p}$ is a $p$-th root of $t$.

On the other hand, given an arbitrary ring $R$, the Jacobson radical of $R$ which I’ll denote $J(R)$, is the intersection of all the left maximal ideals of $R$. It turns out that $J(R)$ is also the intersection of all the right maximal ideals of $R$ and is therefore a two-sided ideal of $R$.

What does the Jacobson radical have to do with separable field extensions?!

It turns out that we can rephrase separability using the Jacobson radical. Suppose that $E$ can be obtained from $F$ by adjoining a root of a separable polynomial $f$, so that $E \cong F[x]/f$. If $L/F$ is any extension of fields, we claim that $E\otimes_F L$ has zero Jacobson radical. To see this, we can assume that $L$ splits $f$ because the Jacobson radical can always be computed by base change to a finite extension. More precisely:

Theorem. If $R$ is a $k$-algebra and $K/k$ is an algebraic extension of fields then $J(R) = R\cap J(R\otimes_k K)$.

So, if $L$ splits $f$ then $E\otimes_F L^n$ where $n$ is the degree of $f$, since $f$ is separable. And, of course $L^n$ has zero Jacobson radical. It’s easy to see by taking a series of separable extensions that $J(E\otimes_F L) = 0$ whenever $E/F$ is a finite separable extension. To summarize:

Theorem.If $E/F$ is a finite separable extension of fields and $L/F$ is any field extension, then $E\otimes_F L$ has zero Jacobson radical.

What about the converse? If $E/F$ is a finite extension such that $E\otimes_F L$ has zero Jacobson radical, is $E/F$ a separable extension? The answer is yes. To see this, let’s suppose that $E/F$ is not separable. Then I claim that $E\otimes_F E$ has nonzero Jacobson radical! To prove this, it suffices to show that $E\otimes_F E$ contains a nonzero nilpotent element. That’s because $E\otimes_F E$ is finitely generated as an $F$-algebra, and hence is a Hilbert ring, so the nilradical can actually be computed as the intersection of all the maximal ideals – in other words the Jacobson radical is actually the nilradical.

Anyway, if we can prove that $E\otimes_F E$ contains a nilpotent element when $E$ is obtained from $F$ by adjoining a single purely inseparable element, then we are done, because if $L/F$ is any intermediate extension then $L\otimes_F L\to E\otimes_F E$ is an injection.

So assume $E = F[x]/(x^{p^k} – y)$. Then $E\otimes_F E\cong E[x]/(x-y^{p^{-k}})^{p^k}$. Thus, in the ring $E\otimes_F E$, the element corresponding to $(x- y^{p^{-k}})$ is a nonzero nilpotent element. Therefore, if $E/F$ is not separable, then $E\otimes_F E$ has nonzero Jacobson radical. In summary, we have proved:

Theorem. Let $E/F$ be a finite extension of fields. Then $E$ is separable over $F$ if and only if for each extension $L/F$ of fields, the Jacobson radical of $E\otimes_F L$ is zero.
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!

Posted by Jason Polak on 20. October 2016 · Write a comment · Categories: math

Let $R$ be a commutative ring, and $M_n(R)$ be the ring of $n\times n$ matrices with coefficients in $R$. Did you know that the center of $M_n(R)$ is the set of scalar matrices? How can we prove this? Should we use matrix multiplication? No way! Let’s use Morita theory instead!

Morita theory works even if $R$ is not commutative. I’ll state the gist of Morita theory, more of which can be found in the post “The Double Dual and Morita Duality”. Let $M$ a left $R$-module. Write $M^* = {\rm Hom}(M,R)$ for the dual of $M$. We have an evaluation homomorphism $M^*\otimes_R M\to R$ given on pure tensors by $f\otimes m\mapsto f(m)$. The image of this homomorphism is an ideal of $R$ called the trace ideal of $M$. If this ideal is all of $R$, then $M$ is called a generator. If in addition to being a generator, $M$ is finitely generated and projective, then $M$ is called a progenerator.
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