A field is a ring whose nonzero elements form a commutative group under multiplication. A field is also a commutative [inverse semigroup](http://en.wikipedia.org/wiki/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](http://en.wikipedia.org/wiki/Variety_%28universal_algebra%29), 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**



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> 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](http://en.wikipedia.org/wiki/Subdirect_product) of a family of fields.
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(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](https://math.stackexchange.com/q/120735/12490) at math.stackexchange.com.



What is the smallest variety of algebras containing all fields?
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.


