Definition 10.40.1. Let $R$ be a ring and let $M$ be an $R$-module. The support of $M$ is the set
\[ \text{Supp}(M) = \{ \mathfrak p \in \mathop{\mathrm{Spec}}(R) \mid M_{\mathfrak p} \not= 0 \} \]
10.40 Supports and annihilators
Some very basic definitions and lemmas.
Lemma 10.40.2. Let $R$ be a ring. Let $M$ be an $R$-module. Then
\[ M = (0) \Leftrightarrow \text{Supp}(M) = \emptyset . \]
Proof. Actually, Lemma 10.23.1 even shows that $\text{Supp}(M)$ always contains a maximal ideal if $M$ is not zero. $\square$
Definition 10.40.3. Let $R$ be a ring. Let $M$ be an $R$-module.
Given an element $m \in M$ the annihilator of $m$ is the ideal
The annihilator of $M$ is the ideal
\[ \text{Ann}_ R(m) = \text{Ann}(m) = \{ f \in R \mid fm = 0\} . \]
\[ \text{Ann}_ R(M) = \text{Ann}(M) = \{ f \in R \mid fm = 0\ \forall m \in M\} . \]
Lemma 10.40.4. Let $R \to S$ be a flat ring map. Let $M$ be an $R$-module and $m \in M$. Then $\text{Ann}_ R(m) S = \text{Ann}_ S(m \otimes 1)$. If $M$ is a finite $R$-module, then $\text{Ann}_ R(M) S = \text{Ann}_ S(M \otimes _ R S)$.
Proof. Set $I = \text{Ann}_ R(m)$. By definition there is an exact sequence $0 \to I \to R \to M$ where the map $R \to M$ sends $f$ to $fm$. Using flatness we obtain an exact sequence $0 \to I \otimes _ R S \to S \to M \otimes _ R S$ which proves the first assertion. If $m_1, \ldots , m_ n$ is a set of generators of $M$ then $\text{Ann}_ R(M) = \bigcap \text{Ann}_ R(m_ i)$. Similarly $\text{Ann}_ S(M \otimes _ R S) = \bigcap \text{Ann}_ S(m_ i \otimes 1)$. Set $I_ i = \text{Ann}_ R(m_ i)$. Then it suffices to show that $\bigcap _{i = 1, \ldots , n} (I_ i S) = (\bigcap _{i = 1, \ldots , n} I_ i)S$. This is Lemma 10.39.2. $\square$
Lemma 10.40.5. Let $R$ be a ring and let $M$ be an $R$-module. If $M$ is finite, then $\text{Supp}(M)$ is closed. More precisely, if $I = \text{Ann}(M)$ is the annihilator of $M$, then $V(I) = \text{Supp}(M)$.
Proof. We will show that $V(I) = \text{Supp}(M)$.
Suppose $\mathfrak p \in \text{Supp}(M)$. Then $M_{\mathfrak p} \not= 0$. Choose an element $m \in M$ whose image in $M_\mathfrak p$ is nonzero. Then the annihilator of $m$ is contained in $\mathfrak p$ by construction of the localization $M_\mathfrak p$. Hence a fortiori $I = \text{Ann}(M)$ must be contained in $\mathfrak p$.
Conversely, suppose that $\mathfrak p \not\in \text{Supp}(M)$. Then $M_{\mathfrak p} = 0$. Let $x_1, \ldots , x_ r \in M$ be generators. By Lemma 10.9.9 there exists an $f \in R$, $f\not\in \mathfrak p$ such that $x_ i/1 = 0$ in $M_ f$. Hence $f^{n_ i} x_ i = 0$ for some $n_ i \geq 1$. Hence $f^ nM = 0$ for $n = \max \{ n_ i\} $ as desired. $\square$
Lemma 10.40.6. Let $R \to R'$ be a ring map and let $M$ be a finite $R$-module. Then $\text{Supp}(M \otimes _ R R')$ is the inverse image of $\text{Supp}(M)$.
Proof. Let $\mathfrak p \in \text{Supp}(M)$. By Nakayama's lemma (Lemma 10.20.1) we see that
\[ M \otimes _ R \kappa (\mathfrak p) = M_\mathfrak p/\mathfrak p M_\mathfrak p \]
is a nonzero $\kappa (\mathfrak p)$ vector space. Hence for every prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$ we see that
\[ (M \otimes _ R R')_{\mathfrak p'}/\mathfrak p' (M \otimes _ R R')_{\mathfrak p'} = (M \otimes _ R R') \otimes _{R'} \kappa (\mathfrak p') = M \otimes _ R \kappa (\mathfrak p) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p') \]
is nonzero. This implies $\mathfrak p' \in \text{Supp}(M \otimes _ R R')$. For the converse, if $\mathfrak p' \subset R'$ is a prime lying over an arbitrary prime $\mathfrak p \subset R$, then
\[ (M \otimes _ R R')_{\mathfrak p'} = M_\mathfrak p \otimes _{R_\mathfrak p} R'_{\mathfrak p'}. \]
Hence if $\mathfrak p' \in \text{Supp}(M \otimes _ R R')$ lies over the prime $\mathfrak p \subset R$, then $\mathfrak p \in \text{Supp}(M)$. $\square$
Lemma 10.40.7. Let $R$ be a ring, let $M$ be an $R$-module, and let $m \in M$. Then $\mathfrak p \in V(\text{Ann}(m))$ if and only if $m$ does not map to zero in $M_\mathfrak p$.
Proof. We may replace $M$ by $Rm \subset M$. Then (1) $\text{Ann}(m) = \text{Ann}(M)$ and (2) $x$ does not map to zero in $M_\mathfrak p$ if and only if $\mathfrak p \in \text{Supp}(M)$. The result now follows from Lemma 10.40.5. $\square$
Lemma 10.40.8. Let $R$ be a ring and let $M$ be an $R$-module. If $M$ is a finitely presented $R$-module, then $\text{Supp}(M)$ is a closed subset of $\mathop{\mathrm{Spec}}(R)$ whose complement is quasi-compact.
Proof. Choose a presentation
\[ R^{\oplus m} \longrightarrow R^{\oplus n} \longrightarrow M \to 0 \]
Let $A \in \text{Mat}(n \times m, R)$ be the matrix of the first map. By Nakayama's Lemma 10.20.1 we see that
\[ M_{\mathfrak p} \not= 0 \Leftrightarrow M \otimes \kappa (\mathfrak p) \not= 0 \Leftrightarrow \text{rank}(A \bmod \mathfrak p) < n. \]
Hence, if $I$ is the ideal of $R$ generated by the $n \times n$ minors of $A$, then $\text{Supp}(M) = V(I)$. Since $I$ is finitely generated, say $I = (f_1, \ldots , f_ t)$, we see that $\mathop{\mathrm{Spec}}(R) \setminus V(I)$ is a finite union of the standard opens $D(f_ i)$, hence quasi-compact. $\square$
Lemma 10.40.9. Let $R$ be a ring and let $M$ be an $R$-module.
If $M$ is finite then the support of $M/IM$ is $\text{Supp}(M) \cap V(I)$.
If $N \subset M$, then $\text{Supp}(N) \subset \text{Supp}(M)$.
If $Q$ is a quotient module of $M$ then $\text{Supp}(Q) \subset \text{Supp}(M)$.
If $0 \to N \to M \to Q \to 0$ is a short exact sequence then $\text{Supp}(M) = \text{Supp}(Q) \cup \text{Supp}(N)$.
Proof. The functors $M \mapsto M_{\mathfrak p}$ are exact. This immediately implies all but the first assertion. For the first assertion we need to show that $M_\mathfrak p \not= 0$ and $I \subset \mathfrak p$ implies $(M/IM)_{\mathfrak p} = M_\mathfrak p/IM_\mathfrak p \not= 0$. This follows from Nakayama's Lemma 10.20.1. $\square$
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