Editing Annihilator (ring theory) (section)

==Properties==
If ''S'' is a subset of a left ''R''-module ''M'', then Ann(''S'') is a left [[Ideal (ring theory)#Definitions|ideal]] of ''R''.<ref>Proof: If ''a'' and ''b'' both annihilate ''S'', then for each ''s'' in ''S'', (''a''&nbsp;+&nbsp;''b'')''s'' = ''as''&nbsp;+&nbsp;''bs'' = 0, and for any ''r'' in ''R'', (''ra'')''s'' = ''r''(''as'') = ''r''0 = 0.</ref>

If ''S'' is a [[Module_(mathematics)#Submodules_and_homomorphisms|submodule]] of ''M'', then Ann<sub>''R''</sub>(''S'') is even a two-sided ideal: (''ac'')''s'' = ''a''(''cs'') = 0, since ''cs'' is another element of ''S''.<ref>Pierce (1982), p. 23, Lemma b, item (i).</ref>

If ''S'' is a subset of ''M'' and ''N'' is the submodule of ''M'' generated by ''S'', then in general Ann<sub>''R''</sub>(''N'') is a subset of Ann<sub>''R''</sub>(''S''), but they are not necessarily equal.  If ''R'' is [[Commutative ring|commutative]], then the equality holds.

''M'' may be also viewed as an ''R''/Ann<sub>''R''</sub>(''M'')-module using the action <math>\overline{r}m:=rm\,</math>.  Incidentally, it is not always possible to make an ''R''-module into an ''R''/''I''-module this way, but if the ideal ''I'' is a subset of the annihilator of ''M'', then this action is well-defined. Considered as an ''R''/Ann<sub>''R''</sub>(''M'')-module, ''M'' is automatically a faithful module.

=== For commutative rings ===
Throughout this section, let <math>R</math> be a commutative ring and <math>M</math> a [[finitely generated module|finitely generated]] <math>R</math>-module.

==== Relation to support ====
The [[support of a module]] is defined as
:<math>\operatorname{Supp}M = \{ \mathfrak{p} \in \operatorname{Spec}R \mid M_\mathfrak{p} \neq 0 \}.</math>
Then, when the module is finitely generated, there is the relation
:<math>V(\operatorname{Ann}_R(M)) = \operatorname{Supp}M</math>,
where <math>V(\cdot)</math> is the set of [[prime ideal]]s containing the subset.<ref>{{Cite web|title=Lemma 10.39.5 (00L2)—The Stacks project|url=https://stacks.math.columbia.edu/tag/00L2|website=stacks.math.columbia.edu|access-date=2020-05-13}}</ref>

==== Short exact sequences ====
Given a [[short exact sequence]] of modules,
:<math>0 \to M' \to M \to M'' \to 0,</math>
the support property
:<math>\operatorname{Supp}M = \operatorname{Supp}M' \cup \operatorname{Supp}M'',</math><ref>{{Cite web|title=Lemma 10.39.9 (00L3)—The Stacks project|url=https://stacks.math.columbia.edu/tag/00L3|website=stacks.math.columbia.edu|access-date=2020-05-13}}</ref>
together with the relation with the annihilator implies
:<math>V(\operatorname{Ann}_R(M)) = V(\operatorname{Ann}_R(M')) \cup V(\operatorname{Ann}_R(M'')).</math>
More specifically, the relations
:<math>\operatorname{Ann}_R(M') \cap \operatorname{Ann}_R(M'') \supseteq \operatorname{Ann}_R(M) \supseteq \operatorname{Ann}_R(M') \operatorname{Ann}_R(M''). </math>
If the sequence splits then the inequality on the left is always an equality. This holds for arbitrary [[direct sum of modules|direct sums]] of modules, as
:<math>\operatorname{Ann}_R\left( \bigoplus_{i\in I} M_i \right) = \bigcap_{i\in I} \operatorname{Ann}_R(M_i).</math>

==== Quotient modules and annihilators ====
Given an ideal <math>I \subseteq R</math> and let <math>M</math> be a finitely generated module, then there is the relation
:<math>\text{Supp}(M/IM) = \operatorname{Supp}M \cap V(I)</math>
on the support. Using the relation to support, this gives the relation with the annihilator<ref>{{Cite web|title=Lemma 10.39.9 (00L3)—The Stacks project|url=https://stacks.math.columbia.edu/tag/00L3|website=stacks.math.columbia.edu|access-date=2020-05-13}}</ref>
:<math>V(\text{Ann}_R(M/IM)) = V(\text{Ann}_R(M)) \cap V(I).</math>