I am trying to understand what tensor products are, but one particular thing confuses me slightly. For $V$ a vector space over a field $\mathbb{k}$, why do we have that $V \otimes \mathbb{k} \cong V$?

The explanation the book I'm using gives is the following: "There is also a special tensor product with zero factors. By definition this is the universal linear function in no variables. This is $\mathbb{k}$ itself, so by the rules above we get canonical identifications $V \otimes \mathbb{k} \cong V \cong \mathbb{k} \otimes V$". I don't quite understand the "by definition" part and everything that follows.

Instead, I have tried to prove this directly by constructing an isomorphism $V \rightarrow V \otimes \mathbb{k}$, but nothing came out of it. Is there a better explanation?

  • $\begingroup$ One possible definition of a tensor product is to define $V_1\otimes\ldots\otimes V_k$ as the space of multilinear functions from $V_1^*\times \ldots V_k^*$ to the base field. Then $0$-tensors are, by definition, "constants", in other words, "linear function in no variables" -- that is just the field $k$ itself. $\endgroup$ – Peter Franek Jun 2 '15 at 11:55
  • $\begingroup$ Thanks! I have not yet studied dual spaces so I think of tensor product mainly in terms of their universal property, but I'll come back to your comment. $\endgroup$ – Vitaly B Jun 2 '15 at 12:12
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    $\begingroup$ @PeterFranek, that only works if $V$ is assumed finite-dimensional. $\endgroup$ – goblin Jun 2 '15 at 13:17

One isomorphism is $v \mapsto v \otimes 1$. I think you'll find it's not too hard to prove that map is surjective. Injective might be a little trickier.

I, too, don't understand the notion of a "universal linear function in no variables", but I can think of $\mathbb{k}$ as a 1-dim'l vector space over $\mathbb{k}$ (with basis "1"), and then you have a tensor product of two spaces, for which a basis is given by $e_i \otimes f_j$, where the $e$s and $f$s form bases for the factor spaces. So a basis for the right hand side is $\{b_i \otimes 1 \mid i = 1, \ldots, n\}$, where the $b$s form a basis of $V$.

Added as clarification

OP asks how to prove $v \mapsto v \otimes 1$ is injective.

One of the fundamental properties of the tensor product is that for any bilinear map $F: V \times W \to Y$ (where $V, W, Y$ are vector spaces), there's a linear map $G: V \otimes W \to Y$ with the property that $$ G(x \otimes y) = F(x, y). $$ If we define $i: V \times W \to V \otimes W$, then this can be stated as $$ G \circ i = F, $$ which may or may not be clearer. And this correspondence is an isomorphism: for every linear map on $V \otimes W$, there's a corresponding bilinear map on $V \times W$.

Anyhow, assuming you've seen that theorem, let's look at $V \otimes \mathbb{k}$. I'm going to define a linear map on this space, namely $$ P : V \otimes \mathbb{k} \to V : v \otimes c \mapsto cv. $$ (I've chosen "$P$" here to denote "product" :).)

You might say "But hoe do I know that every element of $V V \otimes \mathbb{k}$ has the form $v \otimes c$ rather than being some sum of things like this? Answer: anything more complex can be simplified, like this: \begin{align} v_1 \otimes 3 + v_2 \otimes 2 &= 3 (v_1 \otimes 1) + 2(v_2 \otimes 1)\\ &= (3 v_1) \otimes 1 + (2v_2) \otimes 1\\ &= (3 v_1 + 2v_2) \otimes 1. \end{align}

OK. Let's call the original alleged isomorphism $v \mapsto v \otimes 1$ by the name $f$. Now look at $P \circ f$. I claim it's the identity, which is pretty clear by inspection.

But that means that $f$ must be injective! (If not, then $P \circ f$ could not be injective, and hence would not be an isomorphism.)

  • $\begingroup$ This is very helpful, thanks! A follow-up question is how would one show injectivity of the $v \mapsto v \otimes 1$? As far as I understand, one would need to show that the kernel consists of just ${0}$ so the only case in which $v \otimes 1 = 0 \otimes 1$ is if $v = 0$. This seems true, but tensor product feel unintuitive for me. $\endgroup$ – Vitaly B Jun 2 '15 at 12:09
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    $\begingroup$ See "added clarification." Sorry it's so long. Like Pascal, I lacked the time to make it shorter. $\endgroup$ – John Hughes Jun 2 '15 at 13:15

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