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Say we have a commutative ring $A$, an $A$-module $M$, and an ideal $\mathfrak{a}$ in $A$.

Then we have the short exact sequence:

$$ 0 \to \mathfrak{a} \hookrightarrow A \twoheadrightarrow A/\mathfrak{a} \to 0 $$

After tensoring with $M$ we're left with the exact sequence:

$\mathfrak{a} \otimes_A M \rightarrow A \otimes_A M \twoheadrightarrow A/\mathfrak{a} \otimes_A M \to 0$.

We know that $A \otimes_A M \simeq M$ and so $A \otimes_A M /\ker \pi \simeq A/\mathfrak{a} \otimes_A M$. Where $\pi$ is the surjection in the last sequence.

$\ker \pi = \{ \sum a_i \otimes m_i : \sum \pi(a_i) \otimes m_i = 0 \}$, so how do I conclude that $\ker \pi \simeq \mathfrak{a}M$?

Daniel Donnelly
  • 22,288

1 Answers1

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The tensor product is right exact, so the kernel of $\pi$ is the image of $\mathfrak{a} \otimes_A M \rightarrow A \otimes_A M=M$ which is $\mathfrak{a}M$.

Proving that the tensor product is right exact

Tsemo Aristide
  • 89,587