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) $m$ 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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #5788 by Brad Dirks on
Comment #5803 by Johan on