The Stacks Project


Tag 00HL

Chapter 10: Commutative Algebra > Section 10.38: Flat modules and flat ring maps

Lemma 10.38.12. Suppose that $R$ is a ring, $0 \to M'' \to M' \to M \to 0$ a short exact sequence, and $N$ an $R$-module. If $M$ is flat then $N \otimes_R M'' \to N \otimes_R M'$ is injective, i.e., the sequence $$ 0 \to N \otimes_R M'' \to N \otimes_R M' \to N \otimes_R M \to 0 $$ is a short exact sequence.

Proof. Let $R^{(I)} \to N$ be a surjection from a free module onto $N$ with kernel $K$. The result follows from the snake lemma applied to the following diagram $$ \begin{matrix} & & 0 & & 0 & & 0 & & \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes_R N & \to & M' \otimes_R N & \to & M \otimes_R N & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ 0 & \to & (M'')^{(I)} & \to & (M')^{(I)} & \to & M^{(I)} & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes_R K & \to & M' \otimes_R K & \to & M \otimes_R K & \to & 0 \\ & & & & & & \uparrow & & \\ & & & & & & 0 & & \end{matrix} $$ with exact rows and columns. The middle row is exact because tensoring with the free module $R^{(I)}$ is exact. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 8716–8726 (see updates for more information).

    \begin{lemma}
    \label{lemma-flat-tor-zero}
    Suppose that $R$ is a ring, $0 \to M'' \to M' \to M \to 0$
    a short exact sequence, and $N$ an $R$-module. If $M$ is flat
    then $N \otimes_R M'' \to N \otimes_R M'$ is injective, i.e., the
    sequence
    $$
    0 \to N \otimes_R M'' \to N \otimes_R M' \to N \otimes_R M \to 0
    $$
    is a short exact sequence.
    \end{lemma}
    
    \begin{proof}
    Let $R^{(I)} \to N$ be a surjection from a free module
    onto $N$ with kernel $K$. The result follows
    from the snake lemma applied to the following diagram
    $$
    \begin{matrix}
     & & 0 & & 0 & & 0 & & \\
     & & \uparrow & & \uparrow & & \uparrow & & \\
     & & M''\otimes_R N & \to & M' \otimes_R N & \to & M \otimes_R N & \to & 0 \\
     & & \uparrow & & \uparrow & & \uparrow & & \\
    0 & \to & (M'')^{(I)} & \to & (M')^{(I)} & \to & M^{(I)} & \to & 0 \\
     & & \uparrow & & \uparrow & & \uparrow & & \\
     & & M''\otimes_R K & \to & M' \otimes_R K & \to & M \otimes_R K & \to & 0 \\
     & & & & & & \uparrow & & \\
     & & & & & & 0 & &
    \end{matrix}
    $$
    with exact rows and columns. The middle row is exact because tensoring
    with the free module $R^{(I)}$ is exact.
    \end{proof}

    Comments (3)

    Comment #470 by JuanPablo on March 18, 2014 a 4:38 pm UTC

    There is a typo, the bottom row should have $K$ instead of $N$.

    Comment #485 by Johan (site) on March 24, 2014 a 2:00 pm UTC

    Thanks! Fixed here.

    Comment #782 by Anfang Zhou on July 2, 2014 a 3:43 pm UTC

    Hi, I think it's better to say "snake lemma" here than "diagram chasing".

    There is also 1 comment on Section 10.38: Commutative Algebra.

    Add a comment on tag 00HL

    Your email address will not be published. Required fields are marked.

    In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?