For two vector spaces, $V$ and $W$, and a map $f: V \to W$, it is clear that: $$ \ker(f) \otimes V + V \otimes \ker(f) \subseteq \ker(f \otimes f). $$ Does the opposite inclusion hold? If so, I'd like a proof, and if not, a counterexample.

Bascially, given an element an element $\sum_i a_i \otimes b_i \in V \otimes V$, for which it holds that $$ \sum_i f(a_i) \otimes f(b_i) = 0 $$ can we show that $\sum_i a_i \otimes b_i \in \ker(f) \otimes V + V \otimes \ker(f)$?

  • $\begingroup$ Its image is zero - that's where I got the inclusion in the question from. What I want to know is whether the opposite inclusion holds. $\endgroup$ Jun 1, 2012 at 19:01
  • $\begingroup$ This is basic linear algebra - I don't think you need to worry about anything so sophisticated as flatness, exactness, projectivity ... I'm sure the question has a simple basis argument answer - I just can't see it. $\endgroup$ Jun 1, 2012 at 19:09
  • $\begingroup$ I think that does it! Write up your comment as an answer. $\endgroup$ Jun 1, 2012 at 19:28

1 Answer 1


Take a basis $\{x_i\}_{i \in I_0}$ for the kernel of $f$, and extend it to a basis $\{x_i\}_{i \in I}$ for $V$, where $I_0 \subset I$. Thus $\{x_i \otimes x_j : i,j \in I\}$ is a basis for $V \otimes V$. Write out a general element of $\ker(f\otimes f)$ in terms of this basis, and note that the set $\{f(x_i) \otimes f(x_j) : i,j\in I\setminus I_0\}$ is linearly independent.

  • $\begingroup$ Great. Thanks a lot. $\endgroup$ Jun 1, 2012 at 19:35
  • $\begingroup$ @MikhailMatrix Only after some fumbling. I think it's true over a commutative ring if you just assume that $V$ is flat, but you were right—no need for that language here. Might add it later just for fun. Cheers, $\endgroup$ Jun 1, 2012 at 19:39

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