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I know that for prime ideals $\text{ht}(P)+\dim(A/P)\leq \dim A$. I know that the result holds for any ideal $I$. Any hint on how to show this? I tried taking a minimal prime of $I$, and trying to estimate, but didn't get too far, since I don't know much on the structure of $\mathrm{Spec}(A)$. Thanks!

user26857
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John T.
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  • Hint: Take a maximal chain of primes contained in I and concatenate it with a chain of primes containing I, coming from a maximal chain of primes in A/I. – m.s Jun 14 '17 at 05:59
  • How would I use the definition using that the height of $I$ is the inf of the height of minimal overprimes of $I$? – John T. Jun 14 '17 at 06:34
  • Isn't obvious that $\dim A/I=\sup\{\dim A/P: P\supseteq I\}$? Then choose a prime $P$ where the equality holds. – user26857 Jun 14 '17 at 08:54

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