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Suppose $V_1$ and $V_2$ are $k$-vector spaces with bases $(e_{i1})$ and $(e_{i2})$, respectively. I've seen the claim that the collection of elements of the form $e_{i1} \otimes e_{i2}$ forms a basis for $V_{1} \otimes V_{2}$, but I seem to get stuck with the proof.

My question: What's the easiest way to see that the above set is indeed linearly independent?

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They should be $e_{i1}\otimes e_{j2}$. –  user18119 Nov 3 '11 at 13:50

2 Answers 2

Construct a set $\{\phi_{i,j}\}$ of linear forms on the tensor product which is a dual basis. That immediaely implies linear independence.

If $\{\psi_i\}$ is a dual basis to your basis of $V_1$ and $\{\rho_j\}$ is a dual basis to your basis of $V_2$, then you can consider the map $\phi_{i,j}=\psi_i\otimes \rho_j:V_1\otimes V_2\to k\otimes k\cong k$.

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Just making what Mariano Suárez-Alvarez wrote explicit:

Let $V_1$ be a vector space with basis $\{e_i\}$ and $V_2$ a vector space with basis $\{f_j\}$. It's easy to see that $\{(e_i,f_j)\}$ is a basis for $V_1\times V_2$. Since $V_1 \otimes V_2$ is a quotient of $V_1 \times V_2$, the projection of the basis elements of $V_1 \times V_2$ must span $V_1 \otimes V_2$. (Edit: this is false! Please see Mariano and Alex's comments below.)

Let $\phi_{i,j}: V_1 \times V_2 \rightarrow k \;$ be the dual basis for $V_1 \times V_2$. That is, $\phi_{i,j}((e_i,f_j)) = 1$ and $f_{i,j} = 0$ on any other basis element. Note that each $\phi_{i,j}$ is bilinear.

Since $F$ is also an $F$-vector space, the universal property of tensor products implies that for each $\phi_{i,j}$ there is a unique homomorphism $\Phi_{i,j}: V_1 \otimes V_2 \rightarrow F$ such that $\Phi_{i,j}(e_i\otimes f_j) = \phi_{i,j} ((e_i,f_j))$.

Suppose $\sum_{i,j} c_{i,j} e_i \otimes f_j = 0$ for some scalers $c_{i,j}$ not all equal to 0. Then $\Phi_{k,l}(\sum_{i,j} c_{i,j} e_i \otimes f_j ) = c_{k,l}$ is not equal to 0 for some $(k,l)$. But $\Phi_{k,l}(\sum_{i,j} c_{i,j} e_i \otimes f_j ) = \Phi_{k,l}(0) = 0$. So no such $c_{i,j}$ exist, and the set of $e_i\otimes f_j$ is linearly independent.

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Can you be a little more explicit in your argument where you conclude that the basis of $V_1 \times V_2$ spans the tensor product of these spaces? Does this follow from some property of quotient maps in general? –  ItsNotObvious Nov 3 '11 at 14:25
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$V_1\otimes V_2$ is not a quotient of $V_1\times V_2$. For example, if both $V_1$ and $V_2$ have dimension $3$, the former has dimension $9$ while the latter only has dimension $6$. –  Mariano Suárez-Alvarez Nov 3 '11 at 14:52
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Agreed with Mariano. What perhaps IAmBrianDawkins meant is that $V_1\otimes V_2$ is a quotient of the free $\mathbb{Z}$-module on $\left\{m\otimes n:m,n\in M,N\right\}$. The ideal we mod has the nice quality that it precisely makes the dual basis...a basis (i.e. it adds in bilinearity). –  Alex Youcis Nov 3 '11 at 17:04
    
Thank you, Mariano and Alex. Much appreciated. –  Adam Saltz Nov 3 '11 at 18:04

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