I hope you can help me with these questions, I can't really come up with a solution!
Let $V_k$ be a vector space of dimension $n$ over a field $k$. Let $K=\bar k$ be the algebraic closure of $k$. A map $f:V_k \rightarrow k$ is called polynomial if it can be written as a polynomial expression (with coefficients in $k$) in terms of elements of the dual $V_k^*$. Let $k[V]$ be the set of polynomial maps on $V_k$.
1) How can we see that any polynomial map $f : V_k \rightarrow k$ can be extended to $V_K=K \bigotimes V_k$?
2) If I know that $V_k$ has dimension $n$, what do I know about the dimension of $V_K$?
Thanks!
The dimension of $V_K$ over $k$ is equal to $\dim_k(K)\cdot\dim_k(V_k)$ by general properties of the tensor product, simply viewing $K$ as a $k$-vector space. However, the dimension of $V_K$ as a $K$-vector space is the same as $\dim_k(V_k)$. You can easily check that if $B$ is a basis for $V_k$, then the set $\{ 1\otimes b \mid b\in B\}$ is a basis for $V_K$ over $K$.
There is nothing really scary going on here. I like to think of $k=\mathbb Q$ and $K=\mathbb R$, even though $K$ is not the algebraic closure of $k$ in this example. Now think of $V_k$ as $\mathbb Q^3$. The space $V_K$ will just be $\mathbb R^3$. This construction is sometimes called "extension of scalars" because that's all you really do, you leave the general structure of the vector space intact and allow more scalars.
Given a function $f:V_k\to k$, you get a function \begin{align*} f_K = K\otimes V_k &\longrightarrow K \\ x\otimes v &\longmapsto x\cdot f(v) \end{align*}
This is just the general construction $\operatorname{id}_K\otimes f: K\otimes_k V_k \to K\otimes_k k = K$.
However, you may want to think of this more concretely: Given a polynomial with coefficients in $k$, you can just as well interpret this polynomial as one with coefficients in $K$.