What is the smallest variety of algebras containing all fields? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:34:52Z http://mathoverflow.net/feeds/question/91889 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91889/what-is-the-smallest-variety-of-algebras-containing-all-fields What is the smallest variety of algebras containing all fields? Thomas Klimpel 2012-03-22T07:33:31Z 2012-03-23T01:38:04Z <p>A field is a ring whose nonzero elements form a commutative group under multiplication. A field is also a commutative <a href="http://en.wikipedia.org/wiki/Inverse_semigroup" rel="nofollow">inverse semigroup</a> 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 $x = 0$.</p> <p>To simplify the discussion, define an <em>inverse ring</em> 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 <em>inverse ring</em> doesn't exist outside of this question.) Both rings and <em>inverse rings</em> form a <a href="http://en.wikipedia.org/wiki/Variety_%28universal_algebra%29" rel="nofollow">variety of algebras</a>, 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 <em>inverse rings</em> are the smallest variety of algebras containing all fields.</p> <p><strong>Question</strong></p> <blockquote> <p>A direct product of a family of fields is no longer a field. However, it is still a commutative <em>inverse ring</em>. My question is whether every commutative <em>inverse ring</em> is a <a href="http://en.wikipedia.org/wiki/Subdirect_product" rel="nofollow">subdirect product</a> of a family of fields.</p> </blockquote> <p>(Note that <em>subdirect product</em> here must refer to either rings or <em>inverse rings</em>, because the notion of <em>subalgebra</em> isn't defined otherwise. The answer to my question should be independent of which one we choose, but referring to <em>inverse rings</em> would make more sense to me.)</p> <p><strong>Note</strong> This question is identical to this <a href="http://math.stackexchange.com/q/120735/12490" rel="nofollow">question</a> at math.stackexchange.com.</p> http://mathoverflow.net/questions/91889/what-is-the-smallest-variety-of-algebras-containing-all-fields/91897#91897 Answer by Simon Henry for What is the smallest variety of algebras containing all fields? Simon Henry 2012-03-22T09:24:19Z 2012-03-22T09:24:19Z <p>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 :</p> <p>$$A \rightarrow \prod_{\rho \in M(A) } A / \rho$$.</p> <p>Each projection is surjective. the kernel of this map si the jacobson radical $R$ of $A$</p> <p>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 $x=0$.</p> <p>So the previous map is injective, and $A$ is a subdirect product.</p> http://mathoverflow.net/questions/91889/what-is-the-smallest-variety-of-algebras-containing-all-fields/91922#91922 Answer by Benjamin Steinberg for What is the smallest variety of algebras containing all fields? Benjamin Steinberg 2012-03-22T15:22:34Z 2012-03-23T01:38:04Z <p>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. <strike>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. </strike></p> <p><b>Added.</b> 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. </p> <p>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. </p> <p>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. </p> http://mathoverflow.net/questions/91889/what-is-the-smallest-variety-of-algebras-containing-all-fields/91969#91969 Answer by Martin Brandenburg for What is the smallest variety of algebras containing all fields? Martin Brandenburg 2012-03-23T01:37:40Z 2012-03-23T01:37:40Z <p>There is a notion of <a href="http://en.wikipedia.org/wiki/Von_Neumann_regular_ring" rel="nofollow">von Neumann regular rings</a>. For commutative rings $A$, this notion has many equivalent definitions (for proofs see David F. Anderson, <em>Zero-Dimensional Commutative Rings</em>):</p> <p>1) $A$ is zero-dimensional and reduced.</p> <p>2) Every localization of $A$ is a field.</p> <p>3) For every $x \in A$ we have $x^2 | x$.</p> <p>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 $x$.</p> <p>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.</p>