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.

natural/usefulexamples of non-noetherian schemes ($\widehat{A} \otimes_ A \widehat{A}$); you'll get over it. $\endgroup$ – Boyarsky Jun 28 '10 at 11:57alwaysa scheme.) $\endgroup$ – Boyarsky Jun 28 '10 at 12:436more comments