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Still in my "commutative algebra marathon", I came across the following exercise:

Any $k$-subalgebra $A$ of $k[x]$ is finitely generated as $k$-algebra; also, if $A\ne k$, then $\dim A=1$.

Although I have found a related answer here, I don't get how to follow the steps mentioned there. Thus, any help to understand the linked answer above is welcome $-$ or a new answer to the original problem.

Thanks.

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Renan Mezabarba
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    It would be useful if you can tell us where exactly you are stuck. (As a matter of fact, I find that answer pretty clear.) – user26857 Apr 04 '15 at 22:13
  • Essentially in all the four steps of the proof:
    1. why $k[x]$ is integral over $k[f]$?
    2. why $k[x]$ is Noetherian as $k[f]$-module?
    3. why $A$ is also Noetherian as $k[f]$-module?
    4. And finally, why this enables us to conclude that $A$ is f.g. as $k$-algebra?

    I believe this is all clear, but Commutative Algebra is not my "native area".

     – Renan Mezabarba Apr 04 '15 at 22:30

1 Answers1

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1) $x$ is integral over $k[f]$ since it is a root of the monic polynomial $g(Y)=f(Y)−f(x)$;
2) the extension $k[f]⊂k[x]$ is integral (see 1)), and finitely generated, so it's finite, and a finitely generated module over a noetherian ring is also noetherian;
3) submodules of noetherian modules are also noetherian;
4) noetherian modules are finitely generated, and since $k[f]$ is of finite type over $k$ ...

user26857
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