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Does this make sense as an alternative definition for a finitely-generated field extension?:

A field extension $K \subseteq F$ is finitely-generated if and only if there exists a ring epimorphism $K[t_1 \cdots t_n] \twoheadrightarrow F$.

And similarly, $K \subseteq F$ is a simple extension if and only if there is a ring epimorphism $K[t] \twoheadrightarrow F$.

Epimorphism here is used in the categorical sense, and is in general not going to be surjective.

I'd just like verification on this. It is subtly implied in Aluffi's Algebra Chapter 0 in chapter VIII Remark 1.14. The proof of the Nullstellensatz in section 2.2 relies fundamentally on the implication that if $F$ is a finite-type $K$-algebra, then $F$ must also be finitely-generated as a field extension.

However, the definition for simple and finitely-generated field extensions (in Aluffi, as in other places I've checked) is unsatisfying to me. Namely, to show $K \subseteq F$ is finitely-genreated (or simple) you need to first exhibit a field extension on your own, pick appropriate elements to adjoin, and then show your extension is isomorphic to the "smallest field containing the adjoined elements".

However, having the epimorphism definition would make it easier to reason by diagrams, and it seems like it should trivially imply that fields $F$ which are finite-type $K$-algebras are finitely-generated ring extensions just by "forgetting" you're working in $K$-Alg.

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This is known as Zariski's lemma--a finite type morphism of fields is finite. It is somewhat non-trivial, and can be seen as a weak version of the "general Nullstellensatz" (that Jacobson rings are preserved by finite type extensions). You are correct about simple extensions being those for which $n=1$. I don't know if this really makes things any easier though. Maybe conceptually, but in practice, one rarely thinks about $\mathbb{Q}[i]$ as $\mathbb{Q}[x]/(x^2+1)$. – Alex Youcis Mar 13 '14 at 6:32
Also, you don't mean epimorphism is the category of rings. You mean surjection. The map $\mathbb{Z}\to\mathbb{Q}$ is an epimorphism in the category of rings. – Alex Youcis Mar 13 '14 at 6:34
Ah yes. I remember we had gone over that last year in a class of mine. (You don't really come to understand something until you need it, I guess). I am still curious about this proposed definition. Simple => Exists ring epi is easy, but I'm not sure where I get leverage to show the converse. – Tac-Tics Mar 13 '14 at 6:36
If you're extension is generated by one thing $K=F[\alpha]$, then the morphism $K[t]\to F$ defined by $p\mapsto p(\alpha)$ is surjective. – Alex Youcis Mar 13 '14 at 6:37
And no. I definitely mean epimorphism. If you take $K[t] \rightarrow K(\alpha)$, taking $t \mapsto \alpha$, you might instantiate it with $K = \mathbb{Q}$ and $\alpha$ some transcendental like $\pi$. Then, $\frac{1}{\pi}$ is not in the image of the map. Yet you can check that if two ring homomorphisms agree on the image of this map, they must be equal. – Tac-Tics Mar 13 '14 at 6:39

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