POTW Support of a Finitely Generated Module

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If $R$ is a commutative ring with $1$ and $M$ is an $R$-module, the support of $M$ is defined as $\operatorname{Supp}M = \{\mathfrak{p}\in \operatorname{Spec} R\mid M_\mathfrak{p} \neq 0\}$. Show that if $\phi : R \to S$ is a ring homomorphism and $M$ is a finitely generated $R$-module, then $$\operatorname{Supp}(S\otimes_R M) = (\phi^*)^{-1}(\operatorname{Supp} M)$$ where $\phi^* : \operatorname{Spec} S \to \operatorname{Spec} R$ is defined by $\phi^*(\mathfrak{q}) = \phi^{-1}(\mathfrak{q})$.

 

[bob012345]

If $R$ is a commutative ring with $1$ and $M$ is an $R$-module, the support of $M$ is defined as $\operatorname{Supp}M = \{\mathfrak{p}\in \operatorname{Spec} R\mid M_\mathfrak{p} \neq 0\}$. Show that if $\phi : R \to S$ is a ring homomorphism and $M$ is a finitely generated $R$-module, then $$\operatorname{Supp}(S\otimes_R M) = (\phi^*)^{-1}(\operatorname{Supp} M)$$ where $\phi^* : \operatorname{Spec} S \to \operatorname{Spec} R$ is defined by $\phi^*(\mathfrak{q}) = \phi^{-1}(\mathfrak{q})$.

Just curious, what level and course would a student be exposed to this material? Thanks.

 

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A course in abstract algebra that covers commutative rings and modules. The concepts in this POTW stem from commutative algebra. I would say the material here is at the level of the 3rd/4th year undergraduate.

If anyone would like hints or clarification on the problem, please feel free to ask me.

 

[jbergman]

A course in abstract algebra that covers commutative rings and modules. The concepts in this POTW stem from commutative algebra. I would say the material here is at the level of the 3rd/4th year undergraduate.

If anyone would like hints or clarification on the problem, please feel free to ask me.

Is $\mathfrak{p}$ a prime ideal? And what does $M_{\mathfrak{p}}$ mean? I studied some of this a while ago I am Aluffi and Jacobsen but have forgotten most of it.

 

[julian]

Is $\mathfrak{p}$ a prime ideal? And what does $M_{\mathfrak{p}}$ mean? I studied some of this a while ago I am Aluffi and Jacobsen but have forgotten most of it.

I think $\mathfrak{p}$ is a prime ideal.

Reminder:

Definition: Multiplicatively closed
Let $R$ be a ring. A subset $S \subset R$ is called multiplicatively closed if $1 \in S$, and $ab \in S$ for all $a,b \in S$.

Definition: Localisation of a ring
Let $S$ be a multiplicatively closed subset of a ring $R$. Then
$$
(a,s) \sim (a',s') : \Leftrightarrow \text{ there is an element } u \in S \text{ such that } u (as' - a's) = 0
$$
is an equivalence relation on $R \times S$. We denote the equivalence class of $(a,s) \in R \times S$ by $\frac{a}{s}$. The set of all equivalence classes
$$
S^{-1} R := \left\{ \frac{a}{s} : a \in R , s \in S \right\}
$$
is then called the localization of $R$ at the multiplicatively closed set $S$. It is a ring together with the addition and scalar multiplication
$$
\frac{a}{s} + \frac{a'}{s'} := \frac{as' +a's}{ss'} \quad \text{and} \quad \frac{a}{s} \cdot \frac{a'}{s'} = \frac{aa'}{ss'} .
$$

Complement of a ring with respect to a prime ideal is a multiplicatively closed set. Proof:
We must have $1 \in R / \mathfrak{p}$ because $\mathfrak{p}$ is proper being a prime ideal. For if $1 \in \mathfrak{p}$, then would have $r = r1 \in \mathfrak{p}$ for all $r \in R$, implying $R = \mathfrak{p}$. Contradicting that $\mathfrak{p}$ is proper.
Suppose $a,b \in R / \mathfrak{p}$ but $ab \not\in R / \mathfrak{p}$. This means that $a,b \not\in \mathfrak{p}$ but $ab \in \mathfrak{p}$. This contradicts the definition of a prime ideal, and therefore we must have $ab \in R / \mathfrak{p}$.

If $S = R / \mathfrak{p}$ is the complement of a prime ideal $\mathfrak{p}$, then $S^{-1} R$ is denoted $R_\mathfrak{p}$.Definition: Localisation of a module
Let $S$ be a multiplicatively closed subset of a ring $R$, and let $M$ be an $R-$module. Then
$$
(m,s) \sim (m',s') : \Leftrightarrow \text{ there is an element } u \in S \text{ such that } u (s'm - sm') = 0
$$
is an equivalence relation on $M \times S$. We denote the equivalence class of $(m,s) \in M \times S$ by $\frac{m}{s}$. The set of all equivalence classes
$$
S^{-1} M := \left\{ \frac{m}{s} : m \in M , s \in S \right\}
$$
is then called the localization of $M$ at $S$. It is an $S^{-1} R-$module together with the addition and scalar multiplication
$$
\frac{m}{s} + \frac{m'}{s'} := \frac{s'm +sm'}{ss'} \quad \text{and} \quad \frac{a}{s} \cdot \frac{m'}{s'} = \frac{am'}{ss'}
$$
for all $a \in R$, $m,m' \in M$, and $s,s' \in S$.If $S = R / \mathfrak{p}$ is the complement of a prime ideal $\mathfrak{p}$, then $S^{-1} M$ is denoted $M_\mathfrak{p}$.

 

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Is $\mathfrak{p}$ a prime ideal? And what does $M_{\mathfrak{p}}$ mean? I studied some of this a while ago I am Aluffi and Jacobsen but have forgotten most of it.

Yes, $\mathfrak{p}$ is a prime ideal. The set $\operatorname{Spec}R$ is the spectrum of $R$, the set of all prime ideals of $R$.

 

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To solve this problem, consider showing that $\mathfrak{q}\notin (\phi^*)^{-1}(\operatorname{Supp} M) \iff \mathfrak{q}\notin\operatorname{Supp}(S \otimes_R M)$ using Nakayama's lemma.

 

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I hope you enjoyed the challenge! Here is my solution.

Spoiler

If $\mathfrak{q}\notin (\phi^*)^{-1}(\operatorname{Supp}M)$, the localization $M_{\phi^{-1}(\mathfrak{q})} = 0$. Thus $$(S\otimes_R M)_\mathfrak{q} \simeq S_\mathfrak{q}\otimes_S(S\otimes_R M) \simeq S_\mathfrak{q}\otimes_R M\simeq (S_\mathfrak{q}\otimes_{R_{\phi^{-1}(\mathfrak{q})}} R_{\phi^{-1}(\mathfrak{q})}) \otimes_R M\simeq S_\mathfrak{q}\otimes_{R_{\phi^{-1}(\mathfrak{q})}} M_{\phi^{-1}(\mathfrak{q})}=0$$ which shows $\mathfrak{q}\notin \operatorname{Supp}(S\otimes_R M)$.

Conversely, if $\mathfrak{q}\notin \operatorname{Supp}(S\otimes_R M)$, by the above calculation $S_\mathfrak{q}\otimes_{R_{\phi^{-1}(\mathfrak{q})}} M_{\phi^{-1}(\mathfrak{q})} = 0$. Let $k(\mathfrak{q})$ and $k(\phi^{-1}(\mathfrak{q}))$ be the residue fields of $S_\mathfrak{q}$ and $R_{\phi^{-1}(\mathfrak{q})}$, respectively. Suppose $\mathfrak{m}$ is the maximal ideal of $R_{\phi^{-1}(\mathfrak{q})}$. There are isomorphisms \begin{align*}0 &= k(\mathfrak{q})\otimes_{k(\mathfrak{q})} (S_\mathfrak{q} \otimes_{R_{\phi^{-1}(\mathfrak{q})}} M_{\phi^{-1}(\mathfrak{q})})\\ &\simeq k(\mathfrak{q}) \otimes_{R_{\phi^{-1}(\mathfrak{q})}} M_{\phi^{-1}(\mathfrak{q})}\\ &\simeq (k(\mathfrak{q}) \otimes_{k(\phi^{-1}(\mathfrak{q}))} k(\phi^{-1}(\mathfrak{q}))) \otimes_{R_{\phi^{-1}(\mathfrak{q})}} M_{\phi^{-1}(\mathfrak{q})}\\
&\simeq k(\mathfrak{q}) \otimes_{k(\phi^{-1}(\mathfrak{q}))} \frac{M_{\phi^{-1}(\mathfrak{q})}}{\mathfrak{m}M_{\phi^{-1}(\mathfrak{q})}} \end{align*} Since $M$ is a finitely generated $R$-module, $M_{\phi^{-1}(\mathfrak{q})}/\mathfrak{m}M_{\phi^{-1}(\mathfrak{q})}$ is a finite-dimensional vector space over $k(\phi^{-1}(\mathfrak{q}))$, say, of dimension $n$. Then $k(\mathfrak{q}) \otimes_{k(\phi^{-1}(\mathfrak{q}))} \frac{M_{\phi^{-1}(\mathfrak{q})}}{\mathfrak{m}M_{\phi^{-1}(\mathfrak{q})}} \simeq k(\mathfrak{q}) \otimes_{k(\phi^{-1}(\mathfrak{q}))} [k(\phi^{-1}(\mathfrak{q}))]^n \simeq k(\mathfrak{q})^n$. It follows that $k(\mathfrak{q})^n = 0$, so $n = 0$; this forces $M_{\phi^{-1}(\mathfrak{q})} = \mathfrak{m}M_{\phi^{-1}(\mathfrak{q})}$. Nakyama's lemma implies $M_{\phi^{-1}(\mathfrak{q})} = 0$, i.e., $\mathfrak{q}\notin (\phi^*)^{-1}(\operatorname{Supp} M)$.