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Let $S$ be a commutative ring with unity, and let $A,B,A',B'$ be $S$-modules. If $\phi:A\rightarrow A'$ and $\psi:B\rightarrow B'$ are $S$-module homomorphisms, is it true that

$$\operatorname{im}(\phi\otimes\psi)=\operatorname{im}(\phi)\otimes_S \operatorname{im} (\psi)?$$

Bart Patzer
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  • There is also an explanation in Keith Conrad's notes on tensor products (Example 2.14)-http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod2.pdf – Artus Dec 28 '16 at 12:11

1 Answers1

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No, because the map $im(\phi)\otimes_S im(\psi) \to A'\otimes_S B'$ may not be injective.

E.g. consider the case $S = \mathbb Z$, $A = A' = B =\mathbb Z/2\mathbb Z,$ $B' = \mathbb Q/\mathbb Z$, $\phi = id_A$, and $\psi: \mathbb Z/2\mathbb Z \hookrightarrow \mathbb Q/\mathbb Z$ is the unique injection.

Then the map $A \otimes_S B \to im(\phi) \otimes_S im(\psi)$ is an isomorphism, while the map $\phi \otimes_S \psi$ is the zero map, since $A'\otimes_S B' = 0$.

More generally, if you restrict to the case when $\phi$ is the identity, and $\psi$ is injective, you are asking whether the injection $\psi: B \hookrightarrow B'$ induces an injection $A\otimes_S B \hookrightarrow A \otimes_S B'$. The answer is no in general because tensoring is not left exact. (The preceding example is one illustration of this.)

Matt E
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  • Thanks a lot for great answer. I don't quite see what the unique injection $\mathbb{Z}/2\mathbb{Z}\rightarrow\mathbb{Q}/\mathbb{Z}$ is though...could you please tell me? And what if all the modules $A,A',B,B'$ are free $S$- modules (even finitely generated), do we then have $$\operatorname{im}(\phi\otimes\psi)=\operatorname{im}(\phi)\otimes_{S}\operatorname{im}(\psi)?$$ – Bart Patzer May 16 '12 at 13:16
  • $\psi : A \rightarrow \psi(A)$ and $\phi: B \rightarrow \phi(B)$ are surjective. Thus $\psi\otimes 1_B:A \otimes B \rightarrow \psi(A) \otimes B$ and $1_{\psi(A)} \otimes \phi : \psi(A) \otimes B \rightarrow \psi(A) \otimes \phi(B)$ are surjective. Thus $\psi \otimes \phi = (1_{\psi(A)} \otimes g) \circ (\psi \otimes 1_B)$ is surjective. What am i missing? – Manos May 16 '12 at 13:51
  • Thus $im(\psi \otimes \phi) = \psi(A) \otimes \phi(B)=im(\psi) \otimes im(\phi)$. – Manos May 16 '12 at 13:59
  • @Manos: Dear Manos, The map from the tensor product of the images to the image of the tensor product is surjective, but not injective in general (e.g. as in the example I give). Regards, – Matt E May 16 '12 at 14:42
  • @BartPatzer: Dear Bart, The group $\mathbb Q/Z$ has a unique subgroup of order two; let $\psi$ be the unique isomorphism of $\mathbb Z/2\mathbb Z$ with this subgroup. Regards, – Matt E May 16 '12 at 14:43
  • @MattE: Dear Matt, thanks for your answer. I understand the argument that you gave in your answer. What i can't see, is where i am wrong in my argument in my comment. – Manos May 16 '12 at 15:23
  • @Manos: Dear Manos, You are trying to identify $\phi(A)\otimes \phi(B)$ with its image in $A'\otimes B'$, which is not possible in general since tensor products are not left-exact: there is a surjection from the first to the second, but it is not injective in general. Regards, – Matt E May 16 '12 at 20:20
  • What should be the case for linear maps between linear spaces? – Wfpiggie Nov 06 '12 at 12:09
  • @Wfpiggie For linear maps between linear spaces they are equal. This follows (and it is equivalent) from the property that the rank of Kronecker product of two matrices (over an arbitrary field) equals the product of ranks. (Btw, this property deserves a proof as long as many classical textbooks in linear algebra, e.g. Horn & Johnson, give proofs only for real matrices via SVD.) –  Nov 12 '12 at 16:46