# Commutative algebras whose bidual is commutative

Let $k$ be a commutative ring and $A$ a commutative $k$-algebra. Call $D(A) := \mathrm{Hom}_k(A,k)$ the dual of $A$ as a $k$-module, and $DD(A) := \mathrm{Hom}_k(D(A),k)$ the dual of the latter. Let $\Phi\colon A\to DD(A)$ be the canonical map $a \mapsto (u\mapsto u(a))$.

Define a multiplication on $DD(A)$ as follows: if $\xi,\eta \in DD(A)$, define $\xi\bullet\eta$ to be $D(A) \ni u \mapsto \eta(y\mapsto \xi(x\mapsto u(xy))) \in k$. This is clearly $k$-bilinear, and furthermore $\Phi(a) \bullet \eta = \eta \bullet \Phi(a)$ is $u \mapsto \eta(y \mapsto u(ay))$ (for $a \in A$ and $\eta \in DD(A)$); in particular, $\Phi(a)\bullet\Phi(b) = \Phi(ab)$. Clearly this is "the correct" multiplication on $DD(A)$.

I'm sure the following will come as a surprise to others as it did to me: this product is not necessarily commutative. For a counterexample, consider $A = k[t]$ the ring of polynomials over a finite field $k$: then $D(A) = k^{\mathbb{N}}$ as a $k$-vector space, and $DD(A)$ contains at least the elements $\Lambda_{\mathscr{F}}\colon u \mapsto \lim_{\mathscr{F}} u$ where $\mathscr{F}$ is an ultrafilter on $\mathbb{N}$ and their linear combinations (apparently these don't exhaust $DD(A)$: see here; but this doesn't matter); now one can easily check that if $\mathscr{F}$ and $\mathscr{G}$ are ultrafilters on $\mathbb{N}$ then $\Lambda_{\mathscr{F}} \bullet \Lambda_{\mathscr{G}} = \Lambda_{\mathscr{F}+\mathscr{G}}$ where $\mathscr{F}+\mathscr{G} = \{U \subseteq \mathbb{N} : \{j \in \mathbb{N} : U-j \in \mathscr{F}\} \in \mathscr{G}\}$ is the standard addition on $\beta\mathbb{N}$ defined here (§3.2 "Addition on the Stone–Čech compactification of the naturals") and which is not commutative (see, e.g., Hindman & Strauss, Algebra in the Stone-Čech Compactification (1998), §4.2).

So here's my question: What nice conditions on the $k$-algebra $A$ guarantee that $DD(A)$ is commutative? I'm pretty sure that $A$ being finite (i.e., of finite type as a $k$-module) is sufficient, but even this I don't have an appropriate reference for (e.g.: in Vasconcelos, Arithmetic of Blowup Algebras (1994), prop. 1.1.15, the author does not bother to define the multiplication on $DD(A)$).

Contrariwise, does someone have a counterexample to $DD(A)$ being commutative that does not require ultrafilters or some use of the axiom of choice?

Edit: I believe the following gives a positive answer ($DD(A)$ is commutative) when $k$ is a noetherian integral domain and $A$ is a finite $k$-algebra. Indeed, when $k$ is a noetherian integral domain with fraction field $F$, if $M$ is a $k$-module of finite type, then we can write a presentation $k^s \to k^r \to M \to 0$ (with $r,s$ natural numbers), and by comparing the obvious $0 \to D_k(M) \otimes_k F \to F^r \to F^s$ and $0 \to D_F(M \otimes_k F) \to F^r \to F^s$ (where $D_k(M) := \mathrm{Hom}_k(M,k)$ as a $k$-module), we see that the natural map $D_k(M) \otimes_k F \to D_F(M \otimes_k F)$ is an isomorphism — and also, $D_k(M)$ is a $k$-submodule of this. Dualizing twice (and using the fact that $D_k(M)$ is a $k$-module of finite type, being a submodule of $k^r$), we see that $D_k D_k(M) \otimes_k F$ is isomorphic to $D_F D_F(M\otimes_k F) = M\otimes_k F$ (finite dimensional vector space over a field!), and $D_k D_k(M)$ is a $k$-submodule of it. Now if $M = A$ is a finite $k$-algebra, one can check that the multiplication on $D_k D_k (A)$ is indeed the one obtained by restricting the multiplication on $D_F D_F(A\otimes_k F) = A\otimes_k F$ to it: but this multiplication is commutative. So $D_k D_k (A)$ is a commutative $k$-algebra (indeed, a subalgebra of $A\otimes_k F$).

This is the case I was interested in, but I'm leaving the question open, since maybe someone can say something more general or more interesting about the question.

Edit 2: I'm told that the product I defined is known (at least in the context of normed algebras) as the Arens multiplication [note: the EoM article contains a number of typos / missing symbols: beware; here is another text defining the Arens product]. So my question could be rephrased as: "In the context of pure algebra, when is a commutative algebra Arens-regular?"

• If I remember correctly, without AC it is possible that every vector space is reflexive that is $V\cong DD(V)$ via the canonical homomorphism. A choice-free counterexample would therefore have to be other a ring that is not a ring (and might not exist at all for all I know ...) – Johannes Hahn Jan 29 '14 at 16:30
• @JohannesHahn: I don't think it's possible for every vector space to be reflexive. If $V=\bigoplus k$ is a countable direct sum of one-dimensional spaces, and $V$ is reflexive, then the dual of $W=\prod k/\bigoplus k$ is zero, so $W$ is not reflexive. I'm not sure this invalidates your point, though. – Jeremy Rickard Jan 29 '14 at 19:05
• @JeremyRickard Without some form of AC, $\prod k$ might be zero itself. – Johannes Hahn Jan 29 '14 at 20:21
• @Johannes: I don't understand: there are plenty of explicit nonzero elements in $\prod_{i\in I}k$ for any infinite set $I$, e.g., there is the diagonal embedding of $k$; there is the inclusion of $\bigoplus k$ which also contains plenty of explicit elements. – YCor Jan 30 '14 at 15:43
• Had you had a look at the paper On the second conjugate space of a Banach algebra as an algebra, projecteuclid.org/euclid.pjm/1103037121, by Paul Civin and Bertram Yood. They seem to give a fairly detailed study of the question. In particular, they give conditions under which the analogue of this algebra (in the Banach category) is not commutative. (It is almost never.) – ACL Jun 9 '14 at 22:16

It shouldn't be surprising that this product isn't commutative: your definition has no obvious symmetry between $\xi$ and $\eta$, and so to me it's quite unclear that it's "the correct" multiplication on the double dual. Furthermore an attempt at a more symmetric construction fails. Namely, let's think about what happens when we dualize a multiplication $A \otimes A \to A$. We get a diagram
$$A^{\ast} \to (A \otimes A)^{\ast} \leftarrow A^{\ast} \otimes A^{\ast}$$
where the second arrow is not an isomorphism in general and hence is pointing in the wrong direction to give us a comultiplication on $A^{\ast}$. Dualizing a second time gives us a diagram
$$A^{\ast \ast} \leftarrow (A \otimes A)^{\ast \ast} \to (A^{\ast} \otimes A^{\ast})^{\ast} \leftarrow A^{\ast \ast} \otimes A^{\ast \ast}$$
where again the middle arrow is not an isomorphism in general and hence is pointing in the wrong direction to give us a multiplication on $A^{\ast \ast}$.
• Yes, when I first tried to write a multiplication on the bi-dual (I was trying to work out an explicit example), I failed for the reason you point out. But once you demand that multiplication by $\Phi(a)$ should be the bi-transpose of multiplication by $a$, there's basically only one thing you can write down (maybe linear logicians could prove that it's unique in some deeper sense). And since it's simple enough, associative, and relates to $\Phi$ as one would like, I think I can safely say it's "the correct" formula. (Unless someone can suggest another one! :-) – Gro-Tsen Jan 30 '14 at 15:35