# Support of a module

In commutative algebra, the support of a module M over a commutative ring A is the set of all prime ideals ${\displaystyle {\mathfrak {p}}}$ of A such that ${\displaystyle M_{\mathfrak {p}}\neq 0}$.[1] It is denoted by ${\displaystyle \operatorname {Supp} (M)}$. In particular, ${\displaystyle M=0}$ if and only if its support is empty.

• Let ${\displaystyle 0\to M'\to M\to M''\to 0}$ be an exact sequence of A-modules. Then
${\displaystyle \operatorname {Supp} (M)=\operatorname {Supp} (M')\cup \operatorname {Supp} (M'').}$
• If ${\displaystyle M}$ is a sum of submodules ${\displaystyle M_{\lambda }}$, then ${\displaystyle \operatorname {Supp} (M)=\cup _{\lambda }\operatorname {supp} (M_{\lambda }).}$
• If ${\displaystyle M}$ is a finitely generated A-module, then ${\displaystyle \operatorname {Supp} (M)}$ is the set of all prime ideals containing the annihilator of M. In particular, it is closed.
• If ${\displaystyle M,N}$ are finitely generated A-modules, then
${\displaystyle \operatorname {Supp} (M\otimes _{A}N)=\operatorname {Supp} (M)\cap \operatorname {Supp} (N).}$
• If ${\displaystyle M}$ is a finitely generated A-module and I is an ideal of A, then ${\displaystyle \operatorname {Supp} (M/IM)}$ is the set of all prime ideals containing ${\displaystyle I+\operatorname {Ann} (M).}$ This is ${\displaystyle V(I)\cap \operatorname {Supp} (M)}$.