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Let $k$ be a field, and let $R$ be a finitely generated $k$-algebra. (If it helps, you may assume $R$ is an integral domain.) Let $I$ be an ideal of finite colength. Note that $A:=k+I$ is a subring of $R$. Indeed, it is the pullback of the diagram $k \hookrightarrow R/I \twoheadleftarrow R$. Here's my question: $$\text{Is }A \text{ finitely generated as a } k\text{-algebra?}$$

By the Eakin-Nagata theorem, I know that $A$ is Noetherian, since the Noetherian ring $R$ is module-finite over it (as it is module-generated over $A$ by any lifting of any $k$-basis of $R/I$). Also, without the condition on colength or something like it, this tends to be false (and maybe always is). Witness that when $R=k[x,y]$ and $I=(x)$, $k+I \cong k[x,xy,xy^2,xy^3,\ldots]$ is not even Noetherian.

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    $\begingroup$ But notice the silly example where $R=k[x,y]/(xy)$, $I=xR$. Here $I$ is not of finite colength and yet $A$ is finitely generated as a $k$-algebra. $\endgroup$ – Wilberd van der Kallen Jun 1 '15 at 14:16
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Yes. Both $R$ and $A=k+I$ are filtered by powers of $I$ and we may look at the associated graded rings. Now $I/I^2$ is a finitely generated $R/I$ module and $R/I$ is a finitely generated vector space, by Zariski's lemma on finitely generated $k$ algebras that are fields. So $I/I^2$ is a finitely generated vector space and the associated graded of $A=k+I$ is a finitely generated $k$ algebra. But that is not enough. It only shows that the hypotheses are reasonable.

Choose $v_1$, $\dots$, $v_m$ in $R$ so that their images together with $1$ form a basis of the vector space $R/I$. Then $R$ is generated as an $A$-module by $1$, $v_1,\dots,v_m$. So $A$ is a finitely generated $k$-algebra by the Artin-Tate lemma in wikipedia.

One may argue more directly. Every $r\in R$ may be written as a $k$ linear combination of $1$, $v_1,\dots,v_m$ plus an element, say $f(r)$, in $I$. Now take a generating set $y_1,\dots,y_m$ of the $k$-algebra $R$ and choose $x_1$, $\dots$, $x_n$ in $I$ so that the $f(y_i)$ and the $f(v_iv_j)$ are amongst the $x_j$.

We claim that every element of $R$ can be written as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ plus a polynomial in the $x_j$, $v_ix_j$. Indeed it is easy to check that the set of elements that can be written this way is invariant under multiplication by the $v_i$ and the $x_j$, hence also by the $y_j$.

If an element of $A$ is written as a $k$ linear combination of the generators $1$, $v_1$, $\dots$, $v_m$ plus a polynomial in the $x_j$, $v_ix_j$, then it must be a polynomial in the $x_j$, $v_ix_j$.

So the $x_j$ together with the $v_ix_j$ generate the $k$ algebra $A=1+I$.

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  • $\begingroup$ I'm confused. If you are claiming that whenever the associated graded ring of a ring $B$ is finitely generated over $k$, so is $B$ itself, this is wrong (e.g. $B=k[\![x]\!]$). $\endgroup$ – Neil Epstein May 28 '15 at 18:47
  • $\begingroup$ Also, I don't understand the explicit construction you give afterwards. How do you know you can pick $v_i$s and $x_j$s that satisfy the conditions (under the clauses starting with "so that" in the second sentence of the second paragraph) which you specify? $\endgroup$ – Neil Epstein May 28 '15 at 19:09
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    $\begingroup$ @Neil Epstein. It is not true that $R$ is module-generated over $A$ by any lifting of any $k$-basis of $R/I$. But if 1 is in the lifting then $R$ is indeed module generated by the lifting and the finite generation of $A$ follows from the wikipedia Artin-Tate lemma. $\endgroup$ – Wilberd van der Kallen May 31 '15 at 17:39
  • $\begingroup$ That's exactly what I was looking for but was not aware of (i.e. the Artin-Tate lemma). Thank you! $\endgroup$ – Neil Epstein Jun 1 '15 at 18:41

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