Tensor product and exact sequences I'm having some problems with tensor product.
I know that, in general, for $M$ an $R-$module and $I$ an ideal it holds
$$M \otimes_R R/I = M/IM.$$
I also know that for $M_1$, $M_2$ and $N$ $R-$modules and $f: M_1 \longrightarrow M_2$ it holds $$Im(f \otimes Id_N) = Im(f) \otimes N$$

*

*If I consider the short exact sequence:
$$0 \longrightarrow I \longrightarrow R \longrightarrow R/I \longrightarrow 0$$
where the first map is the natural inclusion $\mathcal{i}$. Tensoring with $- \otimes_R R/J$, where J is an ideal of $R$, I obtain the exact sequence:
$$ I \otimes_R R/J \longrightarrow  R \otimes_R R/J \longrightarrow R/I\otimes_R R/J \longrightarrow 0$$
I have to prove that $$Ker(\mathcal{i} \otimes id_{A/J}) = (I \cap J)/IJ$$
Why is it wrong to say that $Im(\mathcal{i} \otimes id_{R/J}) =Im(\mathcal{i}) \otimes R/J =I \otimes R/J =I/IJ$ and conclude that $Ker(\mathcal{i} \otimes id_{A/J}) =0$?


*Similarly, I have another problem looking at a counterexample that proves the fact that tensoring is not left exact.
Put $R= K[X]$ and consider the exact sequence: $$0 \longrightarrow R \longrightarrow R \longrightarrow R/(X) \longrightarrow 0$$ where the first map is multiplication $\cdot X$ and the second is the natural projection.
Tensoring we have an exact sequence
$$ R \otimes_R R/(X)\longrightarrow  R \otimes_R R/(X) \longrightarrow R/(X)\otimes_R R/(X) \longrightarrow 0$$
The first map is not injective since it is the zero map. This means that $Im(\cdot X \otimes id_{R/(X)})=0$. But using the two remarks above I have:
$$Im(\cdot X \otimes id_{R/(X)}) = Im(\cdot X) \otimes R/(X) = (X) \otimes R/(X) = (X)/ (X)^2.$$
Where is my mistake? Can anyone help me?
 A: Good question! I got really confused about this as well.
I thought of this "proof" of the fact $\mathrm{Im}(f\otimes 1_N)=\mathrm{Im}(f)\otimes_R N$:
$\subseteq$: We have $(f\otimes 1_N)(x\otimes a)=f(x)\otimes a\in \mathrm{Im}(f)\otimes_R N$ for $x\in M_1$ and $a\in N$.
$\supseteq$: We have $f(x)\otimes a=(f\otimes 1_A)(x\otimes a)\in \mathrm{Im}(f\otimes 1_N)$ for $x\in M_1$ and $a\in N$.
However, the problem here is that $\mathrm{Im}(f\otimes 1_N)$ and $\mathrm{Im}(f)\otimes_R N$ don't live in the same space! While $\mathrm{Im}(f\otimes 1_N)\subseteq M_2\otimes_R N$, the space $\mathrm{Im}(f)\otimes_R N$ doesn't live in $M_2\otimes_R N$, since $-\otimes_RN$ doesn't necessarily preserve the injection $\mathrm{Im}(f)\hookrightarrow M_2$!
What is true is that there is a $R$-bilinear map
\begin{align*}\mathrm{Im}(f)\times N&\to \mathrm{Im}(f\otimes1_N)\\(f(x),a)&\mapsto f(x)\otimes a=(f\otimes 1_N)(x\otimes a),\end{align*}
which induces a surjective $R$-homomorphism $\mathrm{Im}(f)\otimes_RN\to \mathrm{Im}(f\otimes 1_N)$.
I wonder whether the kernel of this map can be expressed in terms of the Tor functor...
A: To answer the question raised in the other answer, you have a map of $R$-modules $f :M\to M'$ and wonder what the relation between $N\otimes_R \mathrm{Im}(f)$ and $\mathrm{Im}(1\otimes_R f)$ are.
What you have is a short exact sequence
$$0 \longrightarrow  \mathrm{Im}(f) \longrightarrow M' 
\longrightarrow \mathrm{Coker}(f)
\longrightarrow  0$$
and tensoring with $N$ gives you a right exact sequence
$$ N\otimes_R \mathrm{Im}(f) \longrightarrow N\otimes_R M' 
\longrightarrow N\otimes_R \mathrm{Coker}(f)
\longrightarrow  0$$
The failure for the map being an injection is of course measured by $\mathrm{Tor}_R^1(N,\mathrm{Coker}(f))$, in the sense there is a map
$$\delta : \mathrm{Tor}_R^1(N,\mathrm{Coker}(f)) \to N\otimes_R \mathrm{Im}(f)$$
with image the kernel of $1\otimes_R i$. In particular,

*

*The map is an injection if $N$ or $\mathrm{Coker}(f)$ is flat.


*The kernel of the map is $\mathrm{Tor}_R^1(N,\mathrm{Coker}(f))$ if and only if $\mathrm{Tor}_R^1(N,M') = 0$.
Note.
What is true is that cockernels commute with tensor products, and in fact you can check that the image of the first arrow $1\otimes_R i$ is equal to the image of $1\otimes_R f$, so you can in fact replace the above with
$$ N\otimes_R \mathrm{Im}(f) \longrightarrow N\otimes_R M' 
\longrightarrow \mathrm{Coker}(1\otimes_R f)
\longrightarrow  0.$$
