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Tag 080T

Chapter 10: Commutative Algebra > Section 10.40: Going up and going down

Lemma 10.40.12. Let $R \to S$ be a ring map. Let $N$ be a finite $S$-module flat over $R$. Endow $\text{Supp}(N) \subset \mathop{\rm Spec}(S)$ with the induced topology. Then generalizations lift along $\text{Supp}(N) \to \mathop{\rm Spec}(R)$.

Proof. The meaning of the statement is as follows. Let $\mathfrak p \subset \mathfrak p' \subset R$ be primes. Let $\mathfrak q' \subset S$ be a prime $\mathfrak q' \in \text{Supp}(N)$ Then there exists a prime $\mathfrak q \subset \mathfrak q'$, $\mathfrak q \in \text{Supp}(N)$ lying over $\mathfrak p$. As $N$ is flat over $R$ we see that $N_{\mathfrak q'}$ is flat over $R_{\mathfrak p'}$, see Lemma 10.38.19. As $N_{\mathfrak q'}$ is finite over $S_{\mathfrak q'}$ and not zero since $\mathfrak q' \in \text{Supp}(N)$ we see that $N_{\mathfrak q'} \otimes_{S_{\mathfrak q'}} \kappa(\mathfrak q')$ is nonzero by Nakayama's Lemma 10.19.1. Thus $N_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p')$ is also not zero. We conclude from Lemma 10.38.15 that $N_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p)$ is nonzero. Let $J \subset S_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p)$ be the annihilator of the finite nonzero module $N_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p)$. Since $J$ is a proper ideal we can choose a prime $\mathfrak q \subset S$ which corresponds to a prime of $S_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p)/J$. This prime is in the support of $N$, lies over $\mathfrak p$, and is contained in $\mathfrak q'$ as desired. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 9476–9481 (see updates for more information).

    \begin{lemma}
    \label{lemma-going-down-flat-module}
    Let $R \to S$ be a ring map. Let $N$ be a finite $S$-module flat over $R$.
    Endow $\text{Supp}(N) \subset \Spec(S)$ with the induced topology.
    Then generalizations lift along $\text{Supp}(N) \to \Spec(R)$.
    \end{lemma}
    
    \begin{proof}
    The meaning of the statement is as follows. Let
    $\mathfrak p \subset \mathfrak p' \subset R$ be primes. Let
    $\mathfrak q' \subset S$ be a prime $\mathfrak q' \in \text{Supp}(N)$
    Then there exists a prime $\mathfrak q \subset \mathfrak q'$,
    $\mathfrak q \in \text{Supp}(N)$ lying over $\mathfrak p$.
    As $N$ is flat over $R$ we see that $N_{\mathfrak q'}$ is flat
    over $R_{\mathfrak p'}$, see Lemma \ref{lemma-flat-localization}.
    As $N_{\mathfrak q'}$ is finite over $S_{\mathfrak q'}$
    and not zero since $\mathfrak q' \in \text{Supp}(N)$ we see
    that $N_{\mathfrak q'} \otimes_{S_{\mathfrak q'}} \kappa(\mathfrak q')$
    is nonzero by Nakayama's Lemma \ref{lemma-NAK}.
    Thus $N_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p')$
    is also not zero. We conclude from Lemma \ref{lemma-ff}
    that $N_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p)$
    is nonzero. Let
    $J \subset S_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p)$
    be the annihilator of the finite nonzero module
    $N_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p)$.
    Since $J$ is a proper ideal we can choose a prime $\mathfrak q \subset S$
    which corresponds to a prime of
    $S_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p)/J$.
    This prime is in the support of $N$, lies over $\mathfrak p$, and
    is contained in $\mathfrak q'$ as desired.
    \end{proof}

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