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Jacob Lurie made the following claim during his lecture:

If $R\rightarrow R'$ is a morphism that makes $R'$ a finitely generated $R$-module (in particular, integral over $R$). Let $m'\subset R'$ be maximal. Let $m$ be the pull back to $R$, which is also maximal as $R'\rightarrow R$ is integral. Let $M$ be a finitely generated $R$' module, hence also a finitely generated $R$-module.

Lurie sets out to prove that $\dim suppM_{m}\ge \dim supp M_{m'}$ here $M_{m}$ is an $R_{m}$ module, $M_{m'}$ is an $R'_{m'}$ module. He showed that:

Consider $R/m\rightarrow R'/mR'\rightarrow R'/m'$. Then we see $R'/mR'$ is a finite $R/m$ module, so a finite $R/m$ vector space. Now $R'/mR'$ is of finite length as an $R/m$ module, in particular an artian ring. So it is a product of local artinian rings. These rings are localizations of $R'/mR'$ at ideals of $R'$ lying over $m$. One of the ideals is $m'$. So in particular $$R'/mR\cong R'/m'\times \text{other factors}$$So he has $m'^{c}R'_{m'}\subset mR'_{m}$ for some $c$ since the nilradical of an artinian ring being nilpotent. But I do not see how this helps to prove the claim $\dim suppM_{m}\ge \dim supp M_{m'}$.

A related question is as follows: Let $R'=k[x_{1},...,x_{n}]/p$, Noether normalization says that there exists injective map $k[y_{1},...,y_{a}]\rightarrow R'$. The claim is that $\dim R'_{m}=a$ regardless of maximal ideal $m$ we choose, and I do not know why this is true.

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up vote 2 down vote accepted

There are lots of ways (all more or less related) to prove the claimed inequality. I'm not sure if I can completely reconstruct the details of Jacob's approach from what you've written, and you should probably just ask him, or someone else in the class, for the details. But here is an attempt:

First of all, the conclusions you are making from the argument about Artinian rings don't seem to quite make sense (e.g. $R'/mR$ won't be a ring, but just an $R$-module, unless $R = R'$; probably you mean $R'/mR'$; but then what is meant by $R'/m$ on the right hand side of the purported isomorphism?).

The correct conclusion is that $R'_{m'}/m R'_{m'}$ is a direct factor of $R'/m R'$, and hence that $R'_{m'}$ is a direct factor of $R'_{m}$. Tensoring both sides with $M$ over $R'$, we find that $M_{m'}$ is a direct factor of $M_m$, and hence the support of the former is contained in the support of the latter. The statement about dimensions immediately follows.

If you've not thought about this kind of thing before, you might want to consider some simples examples, such as $R = \mathbb Z$, $R' = \mathbb Z[i]$, $m = (5)$, and $m' = (2+i)$. Now try to find examples of $M$'s for which the inequality is actually an equality, and others for which the inequality is strict.

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Dear Emerton: I have added detail. It is from Matthew's notes (, page 122. I need to read your answer carefully. Yes I was confused with this passage and cannot really follow it. – Bombyx mori Jan 28 '13 at 5:31
Sorry there was a typo, $R'/m'$ is on the right hand side. – Bombyx mori Jan 28 '13 at 5:36
Dear Emerton: The map from $R'_{m'}/mR'_{m'}$ seems to be $\frac{r}{s}\rightarrow \frac{r\pmod{m}}{s}$, but I do not see why the image lies in $R'/mR'$ or at least why it is in $R'$. Sorry for such elementary level question. – Bombyx mori Jan 28 '13 at 6:38
Do you mean $R'_{m'}/mR'_{m'}$ is a direct factor of $R'_{m}/mR'_{m}$ instead? – Bombyx mori Jan 28 '13 at 6:40
@user: Dear user, $R'/mR'$ is an Artinian ring, and so is a product of local rings, namely its localizations at its various maximal ideals. These exactly correspond to the maximal ideals of $R'$ containing $m$, and so one of them is $m'$. Hence $R'_{m'}/mR_{m'}$ (which is the same thing as $R'/mR'$ localized at $m'$) is one of the local direct factors of $R'/mR'$. Also, $R'/mR'$ and $R'_m/mR'_m$ are the same thing (the canonical map from the first to the second is an isomorphism), since $m$ is maximal. Regards, – Matt E Jan 28 '13 at 16:42

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