If $V=k[x]$ then $V \otimes_k V \simeq k[x,y]$

Question

Let $V=k[x]$. Show that $V \otimes_k V \simeq k[x,y]$.

I consider the function $\phi : V \otimes_k V \to k[x,y]$ given by $\phi(f(x) \otimes g(y)) = f(x)g(y)$. I could show that $\phi$ is injective, but in the surjective I don't understand why any $h(x,y) \in k[x,y]$ is of the form $f(x)g(y) = h(x,y)$?

Some help.

Thank you.

Answer 1

Hint: you are right that not all $h(x,y)$ are of that form. There are elements in $V\otimes V$ which cannot be written as $f\otimes g$.

Answer 2

In fact, the elements $f \otimes g$ generate the tensor product, but are not all the elements of it. As you note we have a function $\mu : (p,q) \in \mathbb{k}[X] \times \mathbb{k}[Y] \mapsto p(X)q(Y) \in \mathbb{k}[X,Y]$ which is bilinear. A possible approach is to show that $\mathbb{k}[X,Y]$ toghether with $\mu$ satisfy the universal property of the tensor product.

Suppose that you have any bilinear map $\nu : \mathbb{k}[X] \times \mathbb{k}[Y] \to W$. Since $\mathcal{B}' = \{X^iY^j\}_{i,j}$ is a basis for $\mathbb{k}[X,Y]$, an assignment $X^iY^j \mapsto x_iy_j =: \nu(X^i,Y^j)$ yields a linear functional $$ \nu* : f(X,Y) \in \mathbb{k}[X,Y] \mapsto \sum_{i,j}(f)_{ij}x_iy_j \in W $$

Now, we claim that $\nu$ uniquely factorizes through $\mu$ as $\nu = \nu^*\mu$. This will show that the universal property is in fact verified (and uniqueness of factorization comes from the injectivity of $\mu$, since $\nu^*\mu = \nu = g\mu$ would imply $\nu^* = g$). In effect, if $f = \sum_{i=0}^na_iX^i$ and $g = \sum_{j=0}^mb_jY^j$ then

$$ \begin{align} \nu^*\mu(f(X),g(Y)) & = \nu^*(f(X)g(Y)) = \nu^*\left(\sum_{k = 0}^{m+n}\sum_{l=0}^{k}a_lb_{k-l}X^lY^{k-l}\right) = \\ = & \sum_{k = 0}^{m+n}\sum_{l=0}^{k}a_lb_{k-l}x_ly_{k-l} = \sum_{k = 0}^{m+n}\sum_{l=0}^{k}a_lb_{k-l}x_ly_{k-l} = \\ = &\sum_{k = 0}^{m+n}\sum_{l=0}^{k}a_lb_{k-l}\nu(X^l,Y^{k-l}) = \sum_{k = 0}^{m+n}\sum_{l=0}^{k}\nu(a_lX^l,b_{k-l}Y^{k-l}) \stackrel{(*)}{=} \\ = &\sum_{i = 0}^{n}\sum_{j=0}^{m}\nu(a_iX^i,b_jY^j) = \nu\left(\sum_{i = 0}^{n}a_iX^i,\sum_{j=0}^{m}b_jY^j\right) = \nu(f(X),g(Y)). \end{align} $$

as desired. Here we only use (bi) linearity of the functions involved, the formula for polynomial multiplication and the prior definitions. The tricky step may be $(*)$, I will try to give a brief justification as to why the two sums have the same terms: pick any term $\nu(a_iX^i,b_jY^j)$. Then if $k = i+j$, we get $0 \leq k \leq n+m$ and so in the former sum this term only can belong to the sum $\sum_{l=0}^{k}\nu(a_lX^l,b_{k-l}Y^{k-l})$, and it is uniquely determined here by $l$. Moreover all terms $\nu(a_lX^l,b_{k-l}Y^{k-l})$ are of the form $\nu(a_iX^i,b_jY^j)$ for some $i,j$. This proves that there is a bijective correspondence between the terms of the sum, and so their sums coincide.