Does the category of noetherian commutative rings have pushouts?

Background: If $X/S$ is an abelian scheme, then the relative Picard functor $\mathrm{Pic}_{X/S}$ is only defined on the category of locally noetherian $S$-schemes (as far as I know). It is a group functor and in some situations it is representable. We then get a group object in the category of locally noetherian $S$-schemes, and I ask myself if it has a multiplication morphism. [Edit: Boyarsky has mentioned in the comments how to deal with this.]

Observe that the tensor product of noetherian commutative rings does not have to be noetherian (isn't this ugly?). Even for fields there is a counterexample: Let $L/K$ be a purely transcendental field extension of infinite transcendence degree. Then $\Omega^1_{L/K}$ is infinite-dimensional, from which you can concluce that the kernel of $L \otimes_K L \to L, a \otimes b \mapsto ab$ is not finitely generated. Thus $L \otimes_K L$ is not noetherian.

Of course, this does not disprove that $L \leftarrow K \rightarrow L$ has a pushout in the category of noetherian commutative rings. How can this be done? The question has a similar spirit as this one.

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    $\begingroup$ (i) The "locally noetherian" restriction is easy to remove via the limit formalism of EGA IV$_3$, sections 8--12. (ii) It is not "ugly" that the noetherian property is generally lost of the formation of fiber products of noetherian schemes; what is ugly/unnatural is to insist on noetherian hypotheses when unnecessary. As Deligne once said, Grothendieck taught us that it is better to have a category with some "bad" objects and good operations. (iii) Faithfully flat descent provides natural/useful examples of non-noetherian schemes ($\widehat{A} \otimes_ A \widehat{A}$); you'll get over it. $\endgroup$ – Boyarsky Jun 28 '10 at 11:57
  • $\begingroup$ By the way, the Picard functor is locally of finite presentation, and clearly the (locally) noetherian property is preserved under fiber products when at least one of the structure maps is locally of finite type. So the assertion that your Picard scheme is a group scheme makes perfectly good sense within the category of locally noetherian schemes if you insist on $S$ being locally noetherian and $X$ projective. (What lies deeper is that the Picard functor of an abelian scheme is an algebraic space when dropping projective hypotheses, and its relative identity component is always a scheme.) $\endgroup$ – Boyarsky Jun 28 '10 at 12:43
  • $\begingroup$ If R-->S is universal for maps from a given ring R to noetherian rings, then by using maps into fields you can show that every prime ideal of R is the contraction of a unique prime ideal of S. If you could also show that this bijection is an order isomorphism, then you could conclude that R has no infinite ascending chain of prime ideals, which would rule out the existence of such a universal map in the case you are looking at ($R=L\otimes_KL$). $\endgroup$ – Tom Goodwillie Jun 28 '10 at 14:50
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    $\begingroup$ @Tom: there are non-noetherian valuation rings of finite Krull dimension (even spectrum homeomorphic to $[0,n]$ with the order topology), so one cannot infer that even a domain is not noetherian merely from topological considerations with Spec. So your second sentence is unclear. $\endgroup$ – Boyarsky Jun 28 '10 at 15:04
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    $\begingroup$ @Boyarsky, I like that Deligne quote. Do you have a reference? (Please email it too!) $\endgroup$ – Ravi Vakil Apr 8 '11 at 17:22

Let $k$ be a field, and let $\ell = k(x_1,x_2,\ldots)$ be the fraction field of $k[x_1,x_2,\ldots]$. Then $\ell \otimes_k \ell$ is the localisation of $k[x_1,x_2,\ldots][y_1,y_2,\ldots]$ at the multiplicative set $$S = \left\{fg\ \bigg|\ \begin{array}{ll}f \in k[x_1,x_2,\ldots]\setminus\{0\},\\g \in k[y_1,y_2,\ldots] \setminus\{0\}\end{array}\right\}.$$

Claim. The pushout $\ell \coprod_k \ell$ in the category of Noetherian rings does not exist.

Proof. If it does, it has natural maps $\iota_i \colon \ell \to \ell \coprod_k \ell$, hence a natural map $\phi \colon \ell \otimes_k \ell \to \ell \coprod_k \ell$. Consider the ideal $I = (x_1-y_1,x_2-y_2,\ldots) \subseteq \ell \otimes_k \ell$. Since $\ell \coprod_k \ell$ is Noetherian, the ideal in $\ell \coprod_k \ell$ generated by $\phi(I)$ is finitely generated. But if $\phi(I) (\ell \otimes_k \ell)$ is generated by $n$ elements, then the same is true for $\psi(I)A$ for any morphism $\psi \colon \ell \otimes_k \ell \to A$ for a Noetherian ring $A$. On the other hand, consider the ring \begin{align*} A &= (\ell \otimes_k \ell)/(x_{n+2}-y_{n+2},x_{n+3}-y_{n+3},\ldots)\\ &\cong k(z_{n+2},\ldots)[x_1,\ldots,x_{n+1},y_1,\ldots,y_{n+1}][T^{-1}]\\ &\cong k(z_{n+2},\ldots)(x_1,\ldots,x_{n+1}) \underset{{k(z_{n+2},\ldots)}}{\otimes} k(z_{n+2},\ldots)(y_1,\ldots,y_{n+1}), \end{align*} where $$T = \left\{fg\ \bigg|\ \begin{array}{ll}f \in k(z_{n+2},\ldots)[x_1,\ldots,x_{n+1}]\setminus\{0\},\\g \in k(z_{n+2},\ldots)[y_1,\ldots,y_{n+1}] \setminus\{0\}\end{array}\right\}.$$ Then $\mathfrak p = \psi(I)A = (x_1-y_1,\ldots,x_{n+1}-y_{n+1})$ is a prime ideal with $\operatorname{ht}(\mathfrak p) \geq n+1$, since we have a chain of prime ideals $$0 \subsetneq (x_1-y_1) \subsetneq \ldots \subsetneq (x_1-y_1,\ldots,x_{n+1}-y_{n+1}).$$ But now Tag 0BBZ (1) (a form of Krull's Hauptidealsatz) shows that $\mathfrak p$ cannot be generated by $n$ elements. $\square$

  • $\begingroup$ I just realised that different arguments are given here. $\endgroup$ – R. van Dobben de Bruyn Oct 24 '19 at 17:27
  • $\begingroup$ Oh, I just saw your answer now. $\endgroup$ – Martin Brandenburg Feb 23 '20 at 15:07
  • $\begingroup$ In the linked question, François Brunault shows that $\overline{\mathbb{Q}} \sqcup \overline{\mathbb{Q}}$ does not exist in the category of Noetherian commutative rings. $\endgroup$ – Martin Brandenburg Feb 23 '20 at 15:08
  • $\begingroup$ I like your example (and proof) much more, though :) $\endgroup$ – Martin Brandenburg Feb 23 '20 at 15:42

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