0

Let $A$ be ring and $\mathfrak{a}$ be an ideal of $A$. Then I was wondering whether $$\mathfrak{a}\otimes_AM \cong \mathfrak{a}M $$ where $M$ is an $A$-module. Clearly the map $f: \mathfrak{a} \times M \to M$ defined by $f(x,y) = xy$ is surjective and bilinear so it extends to a surjective homomophism $f^{*}:\mathfrak{a}\otimes_AM \to \mathfrak{a}M$. But there is no natural inverse of $f^{*}$. It seems intuitive because $A\otimes_AM \cong M$.

General Grievous
  • 2,121

1 Answers1

1

The sufficient (but not neccesary) condition for the canonical morphism $$\mathfrak{a}\otimes_A M\twoheadrightarrow \mathfrak{a}M$$ to be an isomorphism is $\mathrm{Tor}^A_1(A/\mathfrak{a},M) = 0$. This happens for instance when $M$ is flat. On the other hand if $\mathfrak{a} = 2\mathbb{Z}$ and $M = \mathbb{Z}/2\mathbb{Z}$, then $$\mathbb{Z}/2\mathbb{Z} \cong 2\mathbb{Z}\otimes_{\mathbb{Z}} \mathbb{Z}/2\mathbb{Z}\twoheadrightarrow 2\mathbb{Z}\cdot \mathbb{Z}/2\mathbb{Z} = 0$$

Slup
  • 4,673