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Given a commutative ring $R$ with $1$, an ideal $I$ and a left $R$ module $A$, how can I show that the homomorphism

$$f: R/I \otimes_R A \to A/IA$$

given by $f((r +I) \otimes a) = ra + IA$ is injective? I can easily see that it's surjective. (We define $IA$ to be the submodule of $A$ generated by elements of the form $ia$, $i \in I, a \in A$.)

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You mean $R$-module homomorphism, don't you? –  Berci Feb 12 '13 at 22:05
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When working with tensor products it is often easier to define homomorphisms in both directions and show that they are inverse to each other, rather than showing surjectivity and injectivity for one of them. In this case, you can define \begin{align*} A &\longrightarrow R/I \otimes_R A,\\ a &\longmapsto \overline{1} \otimes a \end{align*} whose kernel contains $IA$ (since $\overline{1} \otimes ia = \overline{i} \otimes a = 0\otimes a = 0$), so induces a well-defined homomorphism \begin{align*} A/IA &\longrightarrow R/I \otimes_R A,\\ \overline{a} &\longmapsto \overline{1} \otimes a. \end{align*} Now show that this is inverse to the map $R/I \otimes_R A \to A/IA$ that you already defined. By the way, all homomorphisms here are homomorphisms of $R$-modules, not only of abelian groups.

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Assume that $f([r]\otimes a)=0$, that is, $ra\in IA$. Then there are elements $i_k\in I$ and $a_k\in A$ such that $ra=\sum_k i_ka_k$. With that, using the tensor property: $$r\otimes a = 1\otimes ra=1\otimes \sum_k i_ka_k=\sum_k i_k\otimes a_k$$ and that is $0$ in $R/I \,\otimes\, A$.

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I think this proves the case for simple tensors, no? –  user62051 Feb 12 '13 at 23:11
    
What do you mean? You can consider $r+I$ and $i_k+I$ terms on the left side of $\otimes$ in the last line. Well, $i_k+I=I$ obviously. –  Berci Feb 12 '13 at 23:39
    
$(r+I)\otimes a=(1+I) r\,\otimes\, a$ using the (right) $R$-module structure on $R/I$, then this -over $R$- equals to $(1+I)\otimes ra$, and so on,, –  Berci Feb 12 '13 at 23:46
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