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There is this exercise in the book on commutative algebra I'm reading:

Let $K$ be a field and $A$ an integral domain which is a finitely generated $K$-algebra. Let $\mathfrak a \subset A$ be an ideal and $$\mathrm{ht}(\mathfrak a):=\min_{\mathfrak p \in V(\mathfrak a)}\mathrm{ht}(\mathfrak p)$$ be the height of $\mathfrak a$. Show that $$\dim(A/\mathfrak a) + \mathrm{ht}(\mathfrak a) = \dim(A).$$ I know this is true for prime ideals, but how can I show the more general case?

user26857
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kade
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1 Answers1

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This is something you should have been able to do after my comment, but let me give a few more details.

For convenience, let $d(I)=\dim A/I$ and $h(I)$ be the height of $I$ for any ideal $I\subset A$.

Choose a prime $P$ containing $\mathfrak{a}$ such that $h(P)=h(\mathfrak{a})$. So, by definition, one has $h(P)\leq h(Q)$ for any prime ideal $Q$ containing $\mathfrak{a}$. Since $h(P)+d(P)=h(Q)+d(Q)$, we see that $d(P)\geq d(Q)$. But, $ d(\mathfrak{a})=\max\{d(Q)|\mathfrak{a}\subset Q\,\text{prime}\}$ and thus $ d(\mathfrak{a})=d(P)$. So, $h(\mathfrak{a})+d(\mathfrak{a})=h(P)+d(P)=\dim A$.

Mohan
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