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In this question, I asked about commutative finite-dimensional $\Bbb C$-algebras without nilpotents. Turns out that all such algebras are isomorphic to $\Bbb C$^n.

How does this classification change if we look at commutative finite-dimensional $\Bbb Z$-algebras instead? These can also be thought of as finite-dimensional semiprime rings, or as reduced rings, which are equivalent in the commutative case.

All of the examples I've looked at so far seem to be a product of integral domains - things like $\Bbb Z$, finite fields $GF(p^n)$'s, or ring extensions such as $\Bbb Z[i]$. This leads to the following conjecture:

Is every finitely generated commutative semiprime $\Bbb Z$-algebra a direct product of integral domains?

If not, then in general, is every finite-dimensional commutative semiprime ring a direct product of something? If so, what?

EDIT: this page says that every finite-dimensional commutative semiprime ring injects into a direct product of integral domains, e.g. it is a "subdirect product". The question is now whether or not this reduces to a full direct product in the finitely generated case.

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  • $\begingroup$ Do you require a multiplicative identity? ​ Do you mean finitely-generated instead of finite-dimensional? ​ If no to the latter, then how do you define whether-or-not a $\big[\mathbb{Z}$-algebra whose additive group has non-zero torsion elements$\big]$ is finite-dimensional? ​ ​ ​ ​ $\endgroup$ – user57159 May 8 '17 at 1:00
  • $\begingroup$ Yes, with multiplicative identity, and I do mean finitely generated. Will edit the question for clarity. $\endgroup$ – Mike Battaglia May 8 '17 at 1:06
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    $\begingroup$ $ \mathbf Z[\sqrt{-1}] $ (or more generally any number ring) is a finite dimensional commutative semiprime ring which is not of the form you describe. $\endgroup$ – Starfall May 8 '17 at 1:51
  • $\begingroup$ Good catch. I think I'm going to simplify the conjecture to - every semiprime/reduced ring is a product of integral domains. Will edit. $\endgroup$ – Mike Battaglia May 8 '17 at 1:54
  • $\begingroup$ By finitely generated you mean finitely generated as a module, rather than as an algebra? $\endgroup$ – Eric Wofsey May 8 '17 at 2:41
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Consider the ring $A=\mathbb{Z}[x]/(x^2-1)$. This ring is reduced and not a domain, since $(x+1)(x-1)=0$. However, it has no nontrivial idempotents (as you can check by either a direct computation or by the geometric argument below), so it cannot be a product of domains either.

Geometrically, $\operatorname{Spec} A$ has two irreducible components, one where $x=1$ and another where $x=-1$. However, these components intersect when $2=0$ (i.e., at the maximal ideal $(2,x-1)=(2,x+1)$), so $\operatorname{Spec} A$ is still connected (and thus $A$ is not a product of rings in any nontrivial way). You can think of $\operatorname{Spec} A$ as an arithmetic version of two lines which cross at a point.

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  • $\begingroup$ Ah, OK. I had a comment about this being related to the split-complex numbers, but I see where I went wrong now. You would get idempotents, but you can't express $\frac{1+x}{2}$ with integer coefficients. So this is good, thanks. $\endgroup$ – Mike Battaglia May 8 '17 at 2:55

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