Tensorial product vector space over the algebraic closure of a field Let $V$ be a vector space over a field $K$, of finite dimensión $n$. Let $L$ be the algebraic closure of $K$. I must to show that $V\otimes_{K}L$ is a $n-$dimensional vector space over $L$.
My Aproach: If $\beta=\{x_{1},...,x_{n}\}$ is a basis of $V$ over $K$ then $\beta'=\{x_{1}\otimes 1,...,x_{n}\otimes 1\}$ is a basis of $V\otimes_{\mathbb{Z}}L$ over $L$. It is easy to prove that the vectors $x_{j}\otimes 1$ generates $V\otimes_{K}L$ over $L$ but I don't know why these vectors are linearlly independent.
 A: Here is one neat, purely formal way: if $W$ is a $K$-vectorspace, then for any other $K$-vectorspace $T$, we have
$$\text{Hom}_K(W,T)\cong \text{Hom}_K(K^{\dim W},T)\cong T^{\dim W}.$$
So we have
$$\text{Hom}_{L}(V\otimes_K L,T)=\text{Hom}_K(V,\text{Hom}_L(L,T))\cong \text{Hom}_K(V,T)\cong T^{\dim T}$$
and thus $V\otimes_K L$ is $\dim V$-dimensional as an $L$-vectorspace
A: Let $a_1,\dots,a_n$ be scalars in $L$ such that $\sum_i a_i(x_i \otimes 1) = 0$. Suppose by contradiction that $a_j \neq 0$ for some $j$. Then there exists a $K$-linear map $f_j : L \to K$ such that $f_j(a_j) \neq 0$. Thus, for the $K$-bilinear map $$b : V \times L \to K; \quad (v_1x_1+\cdots+v_nx_n,y) \mapsto v_j f_j(y)$$ (here we are writting every element of $V$ as a linear combination of $x_1,\dots,x_n$) there exists a $K$-linear map $\tilde b : V \otimes_K L \to K$ such that $\tilde b \circ \otimes = b$. Now, since $0 = \sum_i a_i(x_i \otimes 1) = \sum_i x_i \otimes (a_i1) = \sum_i x_i \otimes a_i$, if we apply $\tilde b$ on both sides of this equation we end up with $$0 = \tilde b \bigg( \sum_i x_i \otimes a_i \bigg) = \sum_i \tilde b(x_i \otimes a_i) = \sum_i b(x_i,a_i) = \sum_i \delta_{ij}f_j(a_i) = f_j(a_j)$$ that is, a contradiction.
