Assume that $f_1\colon V_1\to W_1, f_2\colon V_2\to W_2$ are $k$-linear maps between $k$-vector spaces (over the same field $k$, but the dimension may be infinity). Then the tensor product $f_1\otimes f_2\colon V_1\otimes V_2\to W_1\otimes W_2$ is defined, and it's obvious that $\ker f_1\otimes V_2+ V_1\otimes \ker f_2 \subseteq \ker (f_1\otimes f_2)$. My question is whether the relation $\subseteq$ is in fact $=$.

If this does not hold, how about assuming all these vector spaces are commutative associated $k$-algebras with identity and that all the maps are $k$-algebra homomorphisms? Or can you give a "right" form of the kernel $\ker (f_1\otimes f_2)$?

  • $\begingroup$ You cannot form the tensor product of two vector spaces unless they are vector spaces over the same field. Also, a linear map (or a $K$-homomorphism) of two vector spaces has to commute with the scalars, so it is necessary to assume that $W_i, V_i$ are in fact vector spaces over the same field $K$. $\endgroup$
    – BIS HD
    Oct 27 '13 at 14:56
  • $\begingroup$ Of course, I will make it clear now! $\endgroup$
    – Lao-tzu
    Oct 27 '13 at 14:58

Yes, that's true. Let $f_i : V_i \to W_i$ be two linear maps. Since $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $W_1 \otimes W_2$, we may assume that $f_1,f_2$ are surjective. But then they are split, so that we can assume that $V_i = W_i \oplus U_i$ and that $f_i$ equals the projection $V_i \to W_i$, with kernel $U_i$. Then $V_1 \otimes V_2 = W_1 \otimes W_2 \oplus W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2$ and $f_1 \otimes f_2$ equals the projection of $V_1 \otimes V_2$ onto $W_1 \otimes W_2$. Hence the kernel is $W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2 = U_1 \otimes V_2 + V_1 \otimes U_2$.

This shows even more: The kernel is the pushout $(\ker(f_1) \otimes V_2) \cup_{\ker(f_1) \otimes \ker(f_2)} (V_1 \otimes \ker(f_2))$.

By the way, this argument is purely formal and works in every semisimple abelian $\otimes$-category. What happens when we drop semisimplicity, for example when we consider modules over some commutative ring $R$? Then we only need some flatness assumptions:

Let $f_1 : V_1 \to W_1$ and $f_2 : V_2 \to W_2$ be two morphisms in an abelian $\otimes$-category (for example $R$-linear maps between $R$-modules). If $f_1,f_2$ are epimorphisms, then we have exact sequences $\ker(f_1) \to V_1 \to W_1 \to 0$ and $\ker(f_2) \to V_2 \to W_2 \to 0$. Applying the right exactness of the tensor product twice(!), we get that then also

$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2 \to 0$

is exact. If $f_1,f_2$ are not epi, we can still apply the above to their images and get the exactness of

$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to \mathrm{im}(f_1) \otimes \mathrm{im}(f_2) \to 0.$

Now assume that $\mathrm{im}(f_1)$ and $W_2$ are flat. Then $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $\mathrm{im}(f_1) \otimes W_2$ which embeds into $W_1 \otimes W_2$. Hence, we have still that the sequence

$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2$

is exact. In other words, we have a sum decomposition $$\ker(f_1 \otimes f_2) = \alpha(\ker(f_1) \otimes V_2) + \beta(V_1 \otimes \ker(f_2)),$$ where $\alpha : \ker(f_1) \otimes V_2 \to V_1 \otimes V_2$ and $\beta : V_1 \otimes \ker(f_2) \to V_1 \otimes V_2$ are the canonical morphisms. In general, these are not monic! However, this is the case, by definition, when $V_1$ and $V_2$ are flat. So in this case we can safely treat $\alpha$ and $\beta$ as inclusions and write $$\ker(f_1 \otimes f_2) = V_1 \otimes \ker(f_2) + \ker(f_1) \otimes V_2.$$

  • 1
    $\begingroup$ Thank you very very much! You are so great! $\endgroup$
    – Lao-tzu
    Oct 28 '13 at 1:55
  • $\begingroup$ @Martin Brandenburg: First of all, a great answer. However, isn't it enough to assume flatness of $W_1$,$W_2$,$im(f_1)$ and $im(f_2)$? The step that $\alpha:ker(f_1) \otimes V_2 \rightarrow V_1 \otimes V_2$ is injective seems to always hold if $W_1$ and $W_2$ are flat (by using the tor-functors). Maybe I've missed something here, but it seems to be the case. $\endgroup$
    – user101036
    Oct 12 '14 at 15:48
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    $\begingroup$ Can you flesh out a bit what you mean by "applying the right exactness of the tensor product (twice!)"? $\endgroup$
    – Eric Auld
    Jan 21 '17 at 6:27
  • $\begingroup$ @EricAuld: See Lemma 3.1.20 in arxiv.org/pdf/1410.1716.pdf for instance (applied to $\mathcal{C}=\mathrm{Vect}(k)$). $\endgroup$ Apr 12 '17 at 11:39
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    $\begingroup$ @EricAuld - by right exactness of $-\otimes V_2$, we have $\ker f_1\otimes V_2\rightarrow V_1\otimes V_2\rightarrow W_1\otimes V_2\rightarrow 0$ exact, in particular the kernel of $V_1\otimes V_2\rightarrow W_1\otimes V_2$ is the image of $\ker f_1\otimes V_2$ in $V_1\otimes V_2$. Meanwhile (by similar logic), right exactness of $W_1\otimes -$ implies the kernel of $W_1\otimes V_2\rightarrow W_1\otimes W_2$ is the image of $W_1\otimes \ker f_2$ in $W_1\otimes V_2$. Now $f_1\otimes f_2$ is the composed map $V_1\otimes V_2 \rightarrow W_1\otimes V_2 \rightarrow W_1\otimes W_2$. Its kernel is the $\endgroup$ Jan 26 '21 at 3:47

The isomorphism $$ \ker (f_1\otimes f_2)=V_1\otimes \ker f_2 + \ker f_1 \otimes V_2 $$ is equivalent to the isomorphism $$ \operatorname{coker} (f_1\otimes f_2)=\operatorname{coker} f_1\otimes \operatorname{coker} f_2, $$ since for any ideals $I,J\subset R$ we have $$ R/(I+J)=R/I\otimes R/J $$

  • $\begingroup$ This is not a correct argument. $\endgroup$ Apr 12 '17 at 11:37

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