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Tag 0AV7

Chapter 15: More on Algebra > Section 15.21: Reflexive modules

Lemma 15.21.10. Let $R$ be a Noetherian ring. Let $\varphi : M \to N$ be a map of $R$-modules. Assume that for every prime $\mathfrak p$ of $R$ at least one of the following happens

  1. $M_\mathfrak p \to N_\mathfrak p$ is injective, or
  2. $\mathfrak p \not \in \text{Ass}(M)$.

Then $\varphi$ is injective.

Proof. Let $\mathfrak p$ be an associated prime of $\mathop{\rm Ker}(\varphi)$. Then there exists an element $x \in M_\mathfrak p$ which is in the kernel of $M_\mathfrak p \to N_\mathfrak p$ and is annihilated by $\mathfrak pR_\mathfrak p$ (Algebra, Lemma 10.62.15). This is impossible in both cases. Hence $\text{Ass}(\mathop{\rm Ker}(\varphi)) = \emptyset$ and we conclude $\mathop{\rm Ker}(\varphi) = 0$ by Algebra, Lemma 10.62.7. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 4419–4429 (see updates for more information).

    \begin{lemma}
    \label{lemma-check-injective-on-ass}
    Let $R$ be a Noetherian ring. Let $\varphi : M \to N$ be a map of
    $R$-modules. Assume that for every prime $\mathfrak p$
    of $R$ at least one of the following happens
    \begin{enumerate}
    \item $M_\mathfrak p \to N_\mathfrak p$ is injective, or
    \item $\mathfrak p \not \in \text{Ass}(M)$.
    \end{enumerate}
    Then $\varphi$ is injective.
    \end{lemma}
    
    \begin{proof}
    Let $\mathfrak p$ be an associated prime of $\Ker(\varphi)$.
    Then there exists an element $x \in M_\mathfrak p$ which is
    in the kernel of $M_\mathfrak p \to N_\mathfrak p$ and is
    annihilated by $\mathfrak pR_\mathfrak p$
    (Algebra, Lemma \ref{algebra-lemma-associated-primes-localize}).
    This is impossible in both cases. Hence
    $\text{Ass}(\Ker(\varphi)) = \emptyset$ and we conclude $\Ker(\varphi) = 0$ by
    Algebra, Lemma \ref{algebra-lemma-ass-zero}.
    \end{proof}

    Comments (2)

    Comment #2471 by Raymond Cheng on April 1, 2017 a 3:22 pm UTC

    Third sentence: "This is impossible in all three cases." There are only two cases!

    Comment #2504 by Johan (site) on April 14, 2017 a 12:03 am UTC

    Thanks, fixed here.

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