Finite ring extension of local rings

Question

Let $R$ and $S$ be local rings with the maximal ideals $M$ and $N$, respectively. Assume that $R\subset S$ and that $S$ is a finitely generated $R$-module. If there exists a proper ideal $I$ of $R$ such that $I=IS \cap R$ and the canonical image of $R/I$ in $S/IS$ equals $S/IS$, then prove that $R=S$.

I think that I need to do something with Nakayama's lemma but I couldn't get anything so far. Any help would be great.

score 2 · Accepted Answer · edited Aug 18 '15 at 08:32

2

The second hypothesis says $$R+IS/IS=S/IS\iff S=R+IS$$ As $S$ is a finite $R$-module, Nakayama's lemma says there exists an element $a\in I$ such that $(1+a)S\subset R $. As $I$ is contained in the radical of $R$, $1+a$ is a unit in $R$, so really $S\subset R$. As the reverse inclusion is in the hypotheses, this proves $S=R$.

edited Aug 18 '15 at 08:32

user26857

53,190

answered Aug 17 '15 at 23:39

Bernard

179,256

Could you explain how you get the third line by the second hypothesis? – delueze Aug 17 '15 at 23:44
Do you mean the reference to Nakayama's lemma? – Bernard Aug 17 '15 at 23:47
No, how you get the line beginning with $R+IS/IS=IS$ ? Thanks. – delueze Aug 17 '15 at 23:49
Well, the image of $R$ (hence of $R/I$) in $S/IS$ is precisely $(R+IS)/IS$. Isn't that clear? Furthermore, by the Second Isomorphism theorem, it is isomorphic to $R/(IS\cap R)ˆ=R/I$. – Bernard Aug 17 '15 at 23:54
No, it is not clear for me, but thanks, I will try to understand that part, the rest is ok. – delueze Aug 18 '15 at 00:06
I suppose you agree at least that this image is contained in $(R+IS)/IS$? – Bernard Aug 18 '15 at 00:19
I think I got it, thanks though. – delueze Aug 18 '15 at 00:30
1

Why need that detour instead of concluding that $R=S$ directly from $S=R+IS$? – user26857 Aug 18 '15 at 08:34
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@user26857: I wanted to recall the general version of Nakayama's lemma: all other versions are a consequence of this one. But you're right: it could be shortened. – Bernard Aug 18 '15 at 10:02

Finite ring extension of local rings

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