Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In Pete Clark's commutative algebra lecture notes which can be found here. He proves the following lemma (14.12)

Let $R$ be a local ring with maximal ideal $\mathfrak{p}$ and $S/R$ an integral ring extension. Then the pushed forward ideal $\mathfrak{p}S$ is proper.

The majority of the proof is clear to me except the induction argument. The proof argues by contradiction that if not we may find $p_i \in \mathfrak p$ and $s_i \in S$ such that $$1=\sum_i p_i s_i.$$ So the pushforward is already not proper in the ring $R[s_1,\dots,s_d]$. Clark says that by induction we need only consider the case $S=R[s]$. I understand the proof in this case, but it's not clear to me how to extend it to say $R[s_1,s_2]$. The argument uses very directly the fact that $R$ is local and if I wanted to extend it in a natural way I would need the stronger statement that $R[s]$ is a local ring. Is there some easy way to extend this argument in general that I'm missing?

share|improve this question
    
"pushed forward ideal" is called "the extension of ideal" in Commutative Algebra. (See Atiyah & MacDonald.) –  user26857 Jan 10 '13 at 11:13
    
Furthermore, the proof of LO in the book of Clark is unnecessary complicated. Look at the proof of the same result in Atiyah & MacDonald, Theorem 5.10. It's almost an one-liner proof (forget about the diagram)! –  user26857 Jan 10 '13 at 11:21
    
@YACP, I agree that the proof is much simpler. Although could you clarify why it's safe to assume the localization maps are injective? –  JSchlather Jan 10 '13 at 11:39
    
The localization preserves injectivity. –  user26857 Jan 10 '13 at 11:42
add comment

1 Answer

up vote 4 down vote accepted

Let's try to extend to $R[s_1,s_2]$. What we know is that $pR[s_1]$ is proper in $R[s_1]$. Let $m$ be a maximal ideal in $R[s_1]$ that contains $pR[s_1]$, and consider $R[s_1]_m \to R_[s_1,s_2]_m$. Using $d=1$ case again we see that $mR[s_1,s_2]_m$ is proper in $R[s_1,s_2]_m$. This implies that $mR[s_1,s_2]$ is proper in $R[s_1,s_2]$, so $pR[s_1,s_2]$ is also proper.

The same argument will work for any $d$.

share|improve this answer
    
Ah of course, pass down to localization. Thanks. –  JSchlather Jan 10 '13 at 8:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.