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

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?

• $\operatorname{Im}(f\otimes Id_N)=\operatorname{Im}(f)\otimes N$ is not true in general. What you did in 2. is an actual counter-example to this equality. Jan 2, 2022 at 15:55
• Oh, and when is it true then? Jan 2, 2022 at 15:57
• It is certainly true when $N$ is flat or when $f$ is a pure morphism, but I don't know what kind of condition you are expecting... Jan 2, 2022 at 15:59

## 2 Answers

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...

• Thank you very much!! Jan 3, 2022 at 8:57

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,

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

2. 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

$$N\otimes_R \mathrm{Im}(f) \longrightarrow N\otimes_R M' \longrightarrow \mathrm{Coker}(1\otimes_R f) \longrightarrow 0.$$

• I see, in particular, the kernel of $N\otimes_RIm(f)\to Im(1\otimes f)$ is $Im\delta$, which is some quotient of $Tor^1(N,Coker(f))$. Jan 3, 2022 at 15:33