12
$\begingroup$

A commutative ring $R$ is reduced if $r^2=0 \Rightarrow r=0$ holds for all $r \in R$. Commutative rings are precisely the commutative algebra objects in the symmetric monoidal category $(\mathsf{Ab},\otimes)$. This leads to the following question: Has any notion of "reduced commutative algebra objects" been studied for general linear symmetric monoidal categories (perhaps with some additional properties)?

I would like to require that this notion should be naturally definable in the language of linear symmetric monoidal categories. (In particular, saying that an induced ringed space or stack is reduced would not qualify as an answer.) Ideally it should also be first-order and constructive; this rules out the condition that the intersection of prime ideals is zero.

If $X$ is a well-behaved ringed space, then $\mathcal{O}_X$ should turn out to be reduced according to the categorical notion if and only if it is in the usual sense, i.e. for all opens $U \subseteq X$ the ring $\Gamma(U,\mathcal{O}_X)$ is reduced. This rules out the naive definition that $R$ is reduced if $r^2=0 \Rightarrow r=0$ holds for morphisms $r : 1 \to R$. Also notice that for arbitrary morphisms $r : T \to R$ (replacing $r^2$ by $r \otimes r$) the property would be far too strong. If there is no such definition, I wonder which additional structure do we need in order to give such a definition. Presumably, $\mathsf{Ab}$ has this extra structure.

An idea would be to define $R$ reduced if for all invertible objects $\mathcal{L}$ and all morphisms $r : \mathcal{L} \to R$ with $r \otimes r = 0$ we have $r=0$. But this probably only gives the correct answer for ringed spaces where there are enough invertible objects.

$\endgroup$
  • 2
    $\begingroup$ I don't know if it helps here, but a commutative ring is reduced iff it's a subring of a product of fields. $\endgroup$ – YCor Nov 9 '16 at 16:08
3
$\begingroup$

One can use idals. The unit $1_{\mathcal{C}} \in \mathcal{C}$ is called reduced if for all idals $e : I \to 1_{\mathcal{C}}$ (i.e. morphisms $e$ satisfying $e \otimes I = I \otimes e$) with $e \otimes e = 0$ one has $e=0$. If $\mathcal{C}=\mathsf{Mod}(X)$ for some ringed space $X$, this gives the usual notion. It also gives the usual notion if $\mathcal{C}=\mathsf{Qcoh}(X)$ for a quasi-separated scheme $X$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.