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 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 $M_3(T_n)$:
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.