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Tag 00M5

Chapter 10: Commutative Algebra > Section 10.74: Tor groups and flatness

Lemma 10.74.8. Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent:

  1. The module $M$ is flat over $R$.
  2. For all $i>0$ the functor $\text{Tor}_i^R(M, -)$ is zero.
  3. The functor $\text{Tor}_1^R(M, -)$ is zero.
  4. For all ideals $I \subset R$ we have $\text{Tor}_1^R(M, R/I) = 0$.
  5. For all finitely generated ideals $I \subset R$ we have $\text{Tor}_1^R(M, R/I) = 0$.

Proof. Suppose $M$ is flat. Let $N$ be an $R$-module. Let $F_\bullet$ be a free resolution of $N$. Then $F_\bullet \otimes_R M$ is a resolution of $N \otimes_R M$, by flatness of $M$. Hence all higher Tor groups vanish.

It now suffices to show that the last condition implies that $M$ is flat. Let $I \subset R$ be an ideal. Consider the short exact sequence $0 \to I \to R \to R/I \to 0$. Apply Lemma 10.74.2. We get an exact sequence $$ \text{Tor}_1^R(M, R/I) \to M \otimes_R I \to M \otimes_R R \to M \otimes_R R/I \to 0 $$ Since obviously $M \otimes_R R = M$ we conclude that the last hypothesis implies that $M \otimes_R I \to M$ is injective for every finitely generated ideal $I$. Thus $M$ is flat by Lemma 10.38.5. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 17775–17787 (see updates for more information).

    \begin{lemma}
    \label{lemma-characterize-flat}
    Let $R$ be a ring. Let $M$ be an $R$-module.
    The following are equivalent:
    \begin{enumerate}
    \item The module $M$ is flat over $R$.
    \item For all $i>0$ the functor $\text{Tor}_i^R(M, -)$ is zero.
    \item The functor $\text{Tor}_1^R(M, -)$ is zero.
    \item For all ideals $I \subset R$ we have $\text{Tor}_1^R(M, R/I) = 0$.
    \item For all finitely generated ideals $I \subset R$ we have
    $\text{Tor}_1^R(M, R/I) = 0$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Suppose $M$ is flat. Let $N$ be an $R$-module.
    Let $F_\bullet$ be a free resolution of $N$.
    Then $F_\bullet \otimes_R M$ is a resolution of $N \otimes_R M$,
    by flatness of $M$. Hence all higher Tor groups vanish.
    
    \medskip\noindent
    It now suffices to show that the last condition implies that
    $M$ is flat. Let $I \subset R$ be an ideal.
    Consider the short exact sequence
    $0 \to I \to R \to R/I \to 0$. Apply
    Lemma \ref{lemma-long-exact-sequence-tor}. We get an
    exact sequence
    $$
    \text{Tor}_1^R(M, R/I) \to
    M \otimes_R I \to
    M \otimes_R R \to
    M \otimes_R R/I \to
    0
    $$
    Since obviously $M \otimes_R R = M$ we conclude that the
    last hypothesis implies that $M \otimes_R I \to M$ is
    injective for every finitely generated ideal $I$.
    Thus $M$ is flat by Lemma \ref{lemma-flat}.
    \end{proof}

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