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 commutativeinverse ringis 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.