Serre's inequality on height

In algebra, specifically in the theory of commutative rings, Serre's inequality on height states: given a (Noetherian) regular ring A and a pair of prime ideals ${\displaystyle {\mathfrak {p}},{\mathfrak {q}}}$ in it, for each prime ideal ${\displaystyle {\mathfrak {r}}}$ that is a minimal prime ideal over the sum ${\displaystyle {\mathfrak {p}}+{\mathfrak {q}}}$, the following inequality on heights holds:^[1][2]

${\displaystyle \operatorname {ht} ({\mathfrak {r}})\leq \operatorname {ht} ({\mathfrak {p}})+\operatorname {ht} ({\mathfrak {q}}).}$

Without the assumption on regularity, the inequality can fail; see scheme-theoretic intersection#Proper intersection.

Sketch of Proof

Serre gives the following proof of the inequality, based on the validity of Serre's multiplicity conjectures for formal power series ring over a complete discrete valuation ring.^[3]

By replacing ${\displaystyle A}$ by the localization at ${\displaystyle {\mathfrak {r}}}$, we assume ${\displaystyle (A,{\mathfrak {r}})}$ is a local ring. Then the inequality is equivalent to the following inequality: for finite ${\displaystyle A}$-modules ${\displaystyle M,N}$ such that ${\displaystyle M\otimes _{A}N}$ has finite length,

${\displaystyle \dim _{A}M+\dim _{A}N\leq \dim A}$

where ${\displaystyle \dim _{A}M=\dim(A/\operatorname {Ann} _{A}(M))}$ = the dimension of the support of ${\displaystyle M}$ and similar for ${\displaystyle \dim _{A}N}$. To show the above inequality, we can assume ${\displaystyle A}$ is complete. Then by Cohen's structure theorem, we can write ${\displaystyle A=A_{1}/a_{1}A_{1}}$ where ${\displaystyle A_{1}}$ is a formal power series ring over a complete discrete valuation ring and ${\displaystyle a_{1}}$ is a nonzero element in ${\displaystyle A_{1}}$. Now, an argument with the Tor spectral sequence shows that ${\displaystyle \chi ^{A_{1}}(M,N)=0}$. Then one of Serre's conjectures says ${\displaystyle \dim _{A_{1}}M+\dim _{A_{1}}N<\dim A_{1}}$, which in turn gives the asserted inequality. ${\displaystyle \square }$