What is the smallest variety of algebras containing all fields? A field is a ring whose nonzero elements form a commutative group under multiplication. A field is also a commutative inverse semigroup with respect to multiplication. The unique multiplicative inverse $y$ of an element $x$ (in the sense that $xyx=x$ and $yxy=y$) is $y=x^{-1}$ if $x \neq 0$ and $y=0$ if $x = 0$.
To simplify the discussion, define an inverse ring to be a ring which is an inverse semigroup with respect to multiplication. Denote the multiplicative inverse operation by $()^{-1}$. (Warning: The notion of an inverse ring doesn't exist outside of this question.) Both rings and inverse rings form a variety of algebras, i.e. they can be defined by a set of operations ($+$, $*$, $-()$, $()^{-1}$, $0$, $1$ in this case) together with set of identities satisfied by these operations. I think that the commutative inverse rings are the smallest variety of algebras containing all fields.
Question

A direct product of a family of fields is no longer a field. However, it is still a commutative inverse ring. My question is whether every commutative inverse ring is a subdirect product of a family of fields.

(Note that subdirect product here must refer to either rings or inverse rings, because the notion of subalgebra isn't defined otherwise. The answer to my question should be independent of which one we choose, but referring to inverse rings would make more sense to me.)
Note This question is identical to this question at math.stackexchange.com.
 A: Commutative inverse semigroups are examples of completely regular semigroups, that is, semigroups where each element belongs to a subsemigroup which is a group. It is an old result that any ring whose multiplicative reduct is completely regular is a subdirect product of division rings. I don't have the reference off hand, but I can find it. Basically one shows the Jacobson radical is trivial and then one observes that a primitive ring which is completely regular must be a division ring. A key step is to show the idempotents are central and so you have a possibly noncommutative inverse ring. 
Added. I haven't quite found the old reference yet, but here is the proof. If R is a ring whose multiplicative reduct is completely regular, then it is von Neumann regularb and so its Jacobson radical is trivial. Thus it suffices to handle the primitive case, so assume R has a faithful simple module.  Clearly 1 and 0 are then the only central idempotents of R. Now we show all idempotents of R are central. By Clifford's structure theorem for completely regular semigroups, R is a semilattice of completely simple semigroups. So it suffices to show these completely simple semigroups are groups and then R will be an inverse semigroup with central idempotents. For this it suffices to show $\mathcal R$-equivalent and $\mathcal L$-equivalent idempotents e,f are equal. By symmetry we assume eR=fR. Then $(e-f)^2=0$ and hence e-f=0 since R is completely regular. 
Thus R is a completely regular inverse monoid whose only idempotents are 0,1 and so R-{0} is a group, i.e., R is a division ring.  
I believe the old paper I can't find shows a ring R is completely regular iff it is what ring theorists call strongly regular. 
A: There is a notion of von Neumann regular rings. For commutative rings $A$, this notion has many equivalent definitions (for proofs see David F. Anderson, Zero-Dimensional Commutative Rings):
1) $A$ is zero-dimensional and reduced.
2) Every localization of $A$ is a field.
3) For every $x \in A$ we have $x^2 | x$.
4) For every $x \in A$ there is a unique $y \in A$ with $x = x^2 y$ and $y = y^2 x$; this $y$ is called the weak inverse of $x$.
So this is what you have called an inverse ring. The resulting variety is the smallest one containing all fields, since every reduced ring embeds into a product of fields.
A: If I'm not mistaken, your answer is 'yes' : Let $M(A)$ be the set of maximal ideal of your commutative inverse ring $A$. Then you have a map :
$$A \rightarrow \prod_{\rho \in M(A) } A / \rho $$.
Each projection is surjective. the kernel of this map si the jacobson radical $R$ of $A$
so let $x$ be in $R$ then $(1-xy)$ is invertible for all in $y$, in particular for $y$ the multiplicative inverse of $x$. but as $y(1-xy) =0$, then automaticaly $y=0$, and hence $x=0$.
So the previous map is injective, and $A$ is a subdirect product.
