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.


  • 1
    $\begingroup$ 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.) $\endgroup$ – user26857 Apr 4 '15 at 22:13
  • $\begingroup$ 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". $\endgroup$ – Renan Maneli Mezabarba Apr 4 '15 at 22:30

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 $k$ ...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.