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?
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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 $B \le n + 1$ |
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