Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Matsumura is proving that $\dim A[X]=\dim A+1$. Using a theorem proved previously he proves that if $P$ is a prime ideal of $A[X]$ and $p=P\cap A$ then $\mathrm{ht}\;P=\mathrm{ht}\;p+1$, but it doesn't seem obvious to me that from this last equality we get $\dim A[X]=\dim A+1$, could you explain to me how to get to the equality that I want to prove?

share|cite|improve this question
Which Matsumura, Commutative Ring Theory or Commutative Algebra? – Zev Chonoles Jan 26 '13 at 5:44
@ZevChonoles Commutative Ring Theory – Makoto Kato Jan 28 '13 at 12:42
ht $P =$ ht $p + 1$ should be ht $P \le$ ht $p + 1$. – Makoto Kato Jan 28 '13 at 12:43
up vote 4 down vote accepted

Let $B = A[X]$. Let dim $A = n$. If $p$ is a prime ideal of $A$, then $pB + XB$ is a prime ideal of $B$. Hence dim $B \ge n + 1$. Let $P$ be a prime ideal of $B$. Let $p = A \cap P$. Since ht $P \le$ ht $p + 1$, ht $P \le n + 1$. Hence dim $B \le n + 1$

share|cite|improve this answer

Your Answer


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.