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](https://mathoverflow.net/questions/9961/colimits-of-schemes).



Pushouts of noetherian rings
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)](http://stacks.math.columbia.edu/tag/0BBZ) (a form of Krull's Hauptidealsatz) shows that $\mathfrak p$ cannot be generated by $n$ elements. $\square$


