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?

  • $\begingroup$ $\operatorname{Im}(f\otimes Id_N)=\operatorname{Im}(f)\otimes N$ is not true in general. What you did in 2. is an actual counter-example to this equality. $\endgroup$
    – Roland
    Jan 2, 2022 at 15:55
  • $\begingroup$ Oh, and when is it true then? $\endgroup$
    – MJane
    Jan 2, 2022 at 15:57
  • $\begingroup$ 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... $\endgroup$
    – Roland
    Jan 2, 2022 at 15:59

2 Answers 2


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...

  • $\begingroup$ Thank you very much!! $\endgroup$
    – MJane
    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

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

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

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