Annihilator (ring theory)

Throughout this section, let ${\displaystyle R}$ be a commutative ring and ${\displaystyle M}$ a finitely generated (for short, finite) ${\displaystyle R}$-module.

Recall that the support of a module is defined as

${\displaystyle \operatorname {Supp} M=\{{\mathfrak {p}}\in \operatorname {Spec} R\mid M_{\mathfrak {p}}\neq 0\}.}$

Then, when the module is finitely generated, there is the relation

${\displaystyle V(\operatorname {Ann} _{R}(M))=\operatorname {Supp} M}$,

where ${\displaystyle V(\cdot )}$ is the set of prime ideals containing the subset.[4]

Given a short exact sequence of modules,

${\displaystyle 0\to M'\to M\to M''\to 0,}$

the support property

${\displaystyle \operatorname {Supp} M=\operatorname {Supp} M'\cup \operatorname {Supp} M'',}$[5]

together with the relation with the annihilator implies

${\displaystyle V(\operatorname {Ann} _{R}(M))=V(\operatorname {Ann} _{R}(M'))\cup V(\operatorname {Ann} _{R}(M'')).}$

More specifically, we have the relations

${\displaystyle \operatorname {Ann} _{R}(M')\cap \operatorname {Ann} _{R}(M'')\supseteq \operatorname {Ann} _{R}(M)\supseteq \operatorname {Ann} _{R}(M')\operatorname {Ann} _{R}(M'').}$

If the sequence splits then the inequality on the left is always an equality. In fact this holds for arbitrary direct sums of modules, as

${\displaystyle \operatorname {Ann} _{R}\left(\bigoplus _{i\in I}M_{i}\right)=\bigcap _{i\in I}\operatorname {Ann} _{R}(M_{i}).}$

Given an ideal ${\displaystyle I\subseteq R}$ and let ${\displaystyle M}$ be a finite module, then there is the relation

${\displaystyle {\text{Supp}}(M/IM)=\operatorname {Supp} M\cap V(I)}$

on the support. Using the relation to support, this gives the relation with the annihilator[6]

${\displaystyle V({\text{Ann}}_{R}(M/IM))=V({\text{Ann}}_{R}(M))\cap V(I).}$

Over ${\displaystyle \mathbb {Z} }$ any finitely generated module is completely classified as the direct sum of its free part with its torsion part from the fundamental theorem of abelian groups. Then, the annihilator of a finite module is non-trivial only if it is entirely torsion. This is because

${\displaystyle {\text{Ann}}_{\mathbb {Z} }(\mathbb {Z} ^{\oplus k})=\{0\}=(0)}$

since the only element killing each of the ${\displaystyle \mathbb {Z} }$ is ${\displaystyle 0}$. For example, the annihilator of ${\displaystyle \mathbb {Z} /2\oplus \mathbb {Z} /3}$ is

${\displaystyle {\text{Ann}}_{\mathbb {Z} }(\mathbb {Z} /2\oplus \mathbb {Z} /3)=(6)=({\text{lcm}}(2,3)),}$

the ideal generated by ${\displaystyle (6)}$. In fact the annihilator of a torsion module

${\displaystyle M\cong \bigoplus _{i=1}^{n}(\mathbb {Z} /a_{i})^{\oplus k_{i}}}$

is isomorphic to the ideal generated by their least common multiple, ${\displaystyle (\operatorname {lcm} (a_{1},\ldots ,a_{n}))}$. This shows the annihilators can be easily be classified over the integers.

In fact, there is a similar computation that can be done for any finite module over a commutative ring ${\displaystyle R}$. Recall that the definition of finiteness of ${\displaystyle M}$ implies there exists a right-exact sequence, called a presentation, given by

${\displaystyle R^{\oplus l}\xrightarrow {\phi } R^{\oplus k}\to M\to 0}$

where ${\displaystyle \phi }$ is in ${\displaystyle {\text{Mat}}_{k,l}(R)}$. Writing ${\displaystyle \phi }$ explicitly as a matrix gives it as

${\displaystyle \phi ={\begin{bmatrix}\phi _{1,1}&\cdots &\phi _{1,n}\\\vdots &&\vdots \\\phi _{n,1}&\cdots &\phi _{n,n}\end{bmatrix}}}$

hence ${\displaystyle M}$ has the direct sum decomposition

${\displaystyle M=\bigoplus _{i=1}^{k}{\frac {R}{(\phi _{i,1}(1),\ldots ,\phi _{i,n}(1))}}}$

If we write each of these ideals as

${\displaystyle I_{i}=(\phi _{i,1}(1),\ldots ,\phi _{i,n}(1))}$

then the ideal ${\displaystyle I}$ given by

${\displaystyle V(I)=\bigcup _{i=1}^{n}V(I_{i})}$

presents the annihilator.

Over the commutative ring ${\displaystyle k[x,y]}$ for a field ${\displaystyle k}$, the annihilator of the module

${\displaystyle M={\frac {k[x,y]}{(x^{2}-y)}}\oplus {\frac {k[x,y]}{(y-3)}}}$

is given by the ideal

${\displaystyle {\text{Ann}}_{k[x,y]}(M)=((x^{2}-y)(y-3)).}$