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This question already has an answer here:

So I am to prove that every $\mathbb C$-subalgebra of the ring $\mathbb C[x]$ is a finitely generated $\mathbb C$-algebra.

So.. what I know is:

  • $\mathbb C$ is Noetherian and thus $\mathbb C[x]$ is too.
  • Every finitely generated $\mathbb C$-algebra is a quotient of $\mathbb C[x_1,...,x_n]$.

Well, I am not even sure how to use these two things to begin with. Any help?

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marked as duplicate by user26857 abstract-algebra Aug 15 '16 at 22:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ What is finitely generated and how does it relate to the second point you know? $\endgroup$ – user60589 Aug 15 '16 at 12:48
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For any $\alpha \in A$ we have an algebra map $\varphi_{\alpha} \colon \Bbb C[x] \to \Bbb C[x]$, by evaluating a polynomial in $\Bbb C[x]$ a $\alpha$, i.e. $\varphi(p)=p(a)$. This gives an other $C[x]$-module structure on $\Bbb C[x]$ by defining $p.x=p(a)x$.

Then if $\deg a = n >0$, the monomials $x, \dots, x^{n-1}$ generate $\Bbb C[x]$ in this new module structure. And also $A$ is a submodule of $\Bbb C[x]$ in this structure.

Now a finely generated module over a noetherian ring is noetherian. But a submodule of a noetherian module is finitely generated. So $A \subset \Bbb C[x]$ is finitely generated as $\Bbb C[\alpha]$-module, thus $A$ is finely generated over $\Bbb C$.

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