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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.)

Edit The original question wasn't restricted to fields, and asked how to prove $HSP(K)=ISP(K)$. The current question asks whether $HSP(K)=P_S(K)$ for the class $K$ of fields (interpreted as algebras by extending the multiplicative inverse operation to a global operation). Because the original question didn't received any feedback, I decided to narrow down the question and ask about a stronger statement ($HSP(K)=P_S(K)$ implies $HSP(K)=ISP(K)$) without explicitly asking about a proof.

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Reposted as mathoverflow.net/questions/91889/… at mathoverflow. The question has been answered there with 'yes', including a short proof... –  Thomas Klimpel Mar 22 '12 at 16:53
    
Are you sure that inverse rings form a variety of algebras? The definition appears to include the assertion that inverses are unique, which isn't (obviously) described by an identity. –  Hurkyl Oct 27 '12 at 22:47
    
@Hurkyl Good observation that this a non-trivial (even so well known) fact. I tend to verify at mathematical structures whether I got such properties right. There is written "Classtype: variety" and the important identity (ensuring uniqueness) is "idempotents commute: $xx^{-1}y^{-1}y=y^{-1}yxx^{-1}$". –  Thomas Klimpel Oct 28 '12 at 7:20

2 Answers 2

While I don't know enough category theory to be certain if there is a well-defined notion of "smallest" for varieties, it is clear that any class containing all fields is a proper class; Löwenheim-Skolem in ZFC verifies that for any cardinality $ \kappa $ there exists a field $ F $ with $ card(F)=\kappa $.

It is even possible to show directly (but in a set-theoretically less well defined way) that there are fields with arbitrary cardinality: Given an arbitrary integral domain $ R $, consider $ R[\kappa] $, the ring of (finite) polynomials in $ \kappa $ variables is also an integral domain, and thus it can be made into the fraction field $ R(\kappa) $.

And thus "small" in the sense of cardinality is not well-defined.

That said, the surreal numbers come to mind as possibly relevant; while not a field themselves as they are not a set when considered over ZFC, they might be related to the class you're looking for.

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I think you've misunderstood the question. Small refers not to the size of the class but to containment. –  Zhen Lin Mar 17 '12 at 11:41
    
Actually, the original question just wrote $HSP(K)$ for the smallest variety of algebras containing the class $K$ of algebras. However, I rewrote the question to omit as much universal algebra notation as possible, to make it more accessible. I admit that universal algebra might be a bit more sloppy with respect to sets and classes than category theory, because it's more concerned with other questions and doesn't try to provide an alternative foundation of mathematics. –  Thomas Klimpel Mar 17 '12 at 12:12
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The following answer is based on the answer from Simon Henry at MO (to have an accepted answer, and to check again whether I still understand the proof).

The first step is to take an arbitrary commutative inverse ring $A$, let $M(A)$ be the set of maximal ideals of $A$ and look at the map $$A \rightarrow \prod_{\rho\in M(A)}A/\rho $$ Each projection is surjective, all $A/\rho$ are fields, so we just need to check that this map is injective.

This construction with the maximal ideals is precisely what I was looking for when I wrote: "I wonder whether there isn't a much simpler proof, somehow related to the fact that the subalgebras of the algebras in K have only trivial congruences." in my initial question.

In order to check whether this map is injective, we have to show that its kernel is $\{0\}$. The kernel is the intersection of all maximal ideals of $A$, which is one of the characterizations of the Jacobson radical $J(A)$. Let $x\in J(A)$ then $(1−xy)$ is invertible for all $y$, because it can be shown that "$x$ is in $J(A)$ if and only if $yx$ is left quasiregular for all $y$ in $A$." For $y=x^{-1}$ we have $y(1-xy)=y-x^{-1}xx^{-1}=y-x^{-1}=0$ so that we can conclude $y=0$. But $x=y^{-1}=0^{-1}=0$ and hence $J(A)=\{0\}$.

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