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It's well known that if $K$ is a finitely generated extension of some field $E$, then any intermediate field $F$, $E\subseteq F\subseteq K$, is also finitely generated over $E$.

I'm curious, does the same hold for rings?

Say $S$ is a ring, and $R\supset S$ is a finitely generated extension of $S$. If $T$ is any intermediate ring, is it necessarily true that $T$ is finitely generated over $S$ as a ring?

Is it as simple as saying that for any $t\in T$, $T$ can be generated by the generators of $R$ over $S$? I feel unsure about this statement, since it's not clear to me that the generators of $R$ over $S$ need be in $T$.

If not, what is an example what shows otherwise? Thanks.

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up vote 12 down vote accepted

Let $k$ be a field. Then $k[x, y]$ is a finitely generated extension of $k$, yet the subring $R\subseteq k[x,y]$ generated by $\{xy^i:i\geq0\}$ is not finitely generated.

Life would be very much simpler in some respects if the answer were yes :)

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Thank you for the example Mariano, I suspected we wouldn't be so lucky. :) – Vika Feb 3 '12 at 4:54
Note, since it came up in another question, that $k[x, y]$ is even Noetherian. – Dylan Moreland Feb 5 '12 at 22:45
For future reference, the proof of the fact that the subring $R\subseteq k[x, y]$ is not finitely-generated over $k$ can be found in this thread. – Prism Jun 29 '15 at 7:28

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