The Stacks project

Proof. Actually, Lemma 10.23.1 even shows that $\text{Supp}(M)$ always contains a maximal ideal if $M$ is not zero. $\square$

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. $\square$

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. $\square$

Proof. Let $\mathfrak p \in \text{Supp}(M)$. By Nakayama's lemma (Lemma 10.20.1) we see that

is a nonzero $\kappa (\mathfrak p)$ vector space. Hence for every prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$ we see that

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

Hence if $\mathfrak p' \in \text{Supp}(M \otimes _ R R')$ lies over the prime $\mathfrak p \subset R$, then $\mathfrak p \in \text{Supp}(M)$. $\square$

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. $\square$

Proof. Choose a presentation

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

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. $\square$

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. $\square$