# Tensor product of two vector spaces

There is something which always intrigue me.

Let $U$ and $V$ be vector spaces over $k$ a field. Is it true that if $U\otimes_{k}V=0$, then $U=0$ or $V=0$.

Note that $U$ and $V$ are not necessarily finite dimensional.

I know that if $\{u\}$ and $\{v\}$ are basis vectors of $U$ and $V$, then $\{u\otimes v\}$ forms a basis for $U\otimes_{k}V$ but

1) where can I find the proof of this?

2) How does this implies conclusion?

3) What is the shortest way of proving this?

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1) That $u \otimes v$ span $U \otimes V$ is a direct consequence of the construction. The construction is given for example in Atiyah-MacDonald starting on page 24. –  Rudy the Reindeer Jan 12 '13 at 21:56
For point 2), are you ready to accept as obvious that the Cartesian product of two sets is empty only if one of those two sets is empty? (You don't need the Axiom of Choice for this ;-) –  Marc van Leeuwen Jan 12 '13 at 22:08

You can find a proof in any algebra textbook that treats tensor products abstractly (i.e. that doesn't "define" $U\otimes_kV$ as "the $k$-vector space with basis $u_i\otimes v_j$ where the $u_i$ form a basis of $U$ and the $v_j$ form a basis of $V$." For example, Hungerford's graduate textbook, or (I'm pretty sure) Keith Conrad's online notes on tensor products (I can't remember if this result is in the first or second set of notes, but I'm pretty sure it's in one of them).

Why does this imply your result? In fact it implies that for $u\in U$ and $v\in V$ both non-zero, $u\otimes v\neq 0$ in $U\otimes_kV$. The reason is that we can choose a basis $\{u_i:i\in I\}$ for $U$ such that $u=u_{i_0}$ for some $i_0\in I$ and a basis $\{v_j:j\in J\}$ for $V$ such that $v=v_{j_0}$ for some $j_0\in J$. According to the result above, $u_{i_0}\otimes v_{j_0}$ is an element of a basis for $U\otimes_kV$. Therefore it can't be zero. In particular $U\otimes_kV\neq 0$ if $U,V\neq 0$.

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I am convinced! –  enoughsaid05 Jan 12 '13 at 22:20
Keenan, it's in the second set of notes (Lemma 5.10 and Theorem 5.11 in the current version of them). –  KCd Jan 13 '13 at 0:27
Dear @KCd, Thank you! –  Keenan Kidwell Jan 13 '13 at 0:29
Here is another solution (from the notes that Keenan mentioned). Assume $U$ and $V$ are nonzero $k$-vector spaces. Pick nonzero vectors $u_0 \in U$ and $v_0 \in V$. From the existence of bases (hence Zorn's lemma in general), there are linear maps $f \colon U \rightarrow k$ and $g \colon V \rightarrow k$ such that $f(u_0) = 1$ and $g(v_0) = 1$. There is a linear map $f \otimes g \colon U \otimes_k V \rightarrow k$ with effect $u \otimes v \mapsto f(u)g(v)$ on all elementary tensors. Since $(f \otimes g)(u_0 \times v_0) = 1 \cdot 1 = 1$, $u_0 \otimes v_0 \not=0$, so $U \otimes_k V \not= O$. –  KCd Jan 13 '13 at 0:30
That is a nice proof! –  enoughsaid05 Jan 14 '13 at 10:21

I was looking at the flatness of the modules. So hope I am right here!

Here we use the fact that a vector space is a flat module.

So phrasing differently, if $U\otimes_{k}V=0$, and $U\neq 0$, then I must show that $V=0$.

Thus let $u$ be a non-zero element of $u$ and we look at the exact sequence

$0\rightarrow ku\rightarrow U$.

Then by flatness we have

$0\rightarrow ku\otimes_{k}V\rightarrow U\otimes_{k}V$.

By hypothesis, this means that $ku\otimes_{k}V=0$.

Now it remains to show that we have an injection of vector space $V\hookrightarrow ku\otimes_{k}V$ via $v\mapsto u\otimes v$. So suppose $u\otimes v=0$. Then using the universal property

\begin{array}{ccc} ku \times V & & & \\ \downarrow & \searrow & &\\ ku\otimes_{k}V &\rightarrow& V \end{array}

With $f:ku\times V\rightarrow V$ via $f(ku,v)=kv$. This can be checked that it is bilinear. Hence there exists a unique homomorphism $\phi:ku\otimes_{k}V\rightarrow V$ that makes the diagram commutes.

Now, we have $\phi(u\otimes v)=\phi(0)=0=v$. Hence we have an exact sequence

$0\rightarrow V\rightarrow ku\otimes_{k}V$.

But $ku\otimes_{k}V=0$. Hence so is $V$.

Hope I am right here!

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