Let $k$ be a field, and let $R$ be a subring of $k[X]$ which contains $k$. Does $R$ have to be a finitely generated $k$-algebra? I tried to prove this, but I didn't get anywhere. Would someone kindly give me a hint?
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1There is a Search function on this site which one can use before posting. – user26857 Nov 26 '15 at 21:03
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If $R=k$, there is nothing to prove. If not, pick $f\in R$ non-constant and let $S=k[f]$. If $\deg f=n>0$, $k[X]$ is generated as an $S$-module by $1,X,\ldots, X^{n-1}$ and thus $R$ is a finitely generated $S$-module, being a submodule of a finitely generated module. If $R$ is generated by $g_1,\ldots, g_p$ as an $S$-module, $R=k[f,g_1,\ldots,g_p]$.
Mohan
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1Small remark, but submodule of fg module need not be fg, here it works since the base ring is noetherian. – Academic Nov 10 '24 at 05:19
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Hint: Consider the set $D\subseteq\mathbb{N}$ of degrees $d$ such that $R$ contains a polynomial of degree exactly $d$. This is a submonoid of $\mathbb{N}$. Show that $D$ is finitely generated as a monoid (in fact, every submonoid of $\mathbb{N}$ is finitely generated). Now use this to show $R$ is finitely generated (you may find it helpful to think about the vector spaces $R_n$ of elements of $R$ of degree $\leq n$ and how they change as $n$ grows).
Eric Wofsey
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