TITLE: Commutative algebras whose bidual is commutative
QUESTION [11 upvotes]: 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?"

REPLY [3 votes]: Not an answer, but:
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 $A^{\ast \ast}$.