Tensor product and exact sequences

commutative-algebraexact-sequencemodulestensor-products

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 $$Im(f \otimes Id_N) = Im(f) \otimes N$$

  1. 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 $Ker(\mathcal{i} \otimes id_{A/J}) =0$?

  2. 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?

Best Answer

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...

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