Prove that $\mathbb K[x] \times \mathbb K[y] \rightarrow \mathbb K[x,y] \,$ induces an isomorphism of $\mathbb K$-algebras. I'm having some trouble with the following problem.
Let $x, y$ be two commuting variables, and let $\mathbb K$ a field. We define the following function $$\mathbb K[x] \times \mathbb K[y] \rightarrow \mathbb K[x,y], \, (P(x),Q(y))\mapsto P(x)Q(y).$$
I have two questions.
1.) I don't see why this function is bilinear.
2.) Why can we say that our function induces an  isomorphic $\mathbb K[x] \otimes \mathbb K[y] \rightarrow \mathbb K[x,y]?$
(For question 2., can we use the universal property of the tensor product?)
 A: Your question makes little sense. You also need to specify what kind of isomorphism. Of vector spaces over $\mathbb{K}$? As $\mathbb{K}$-algebras? As $\mathbb{Z}$-algebras (= rings)?
You can show that
$$\mathbb{K}[x] \otimes \mathbb{K}[y]\cong \mathbb{K}[x,y]$$ as $\mathbb{K}$-algebras.
I'll give a brief sketch. You can use the comments to ask for clarification if necessary.
Define the mapping
$$\mathbb{K}[x]  \times \mathbb{K}[y]\to \mathbb{K}[x,y]: (P(x), Q(y)) \mapsto P(x)Q(y)$$
This is a bilinear map by distributivity in $\mathbb{K}[x,y]$. The universal property of the tensor product gives a unique linear map
$$\mathbb{K}[x] \otimes \mathbb{K}[y] \to \mathbb{K}[x,y]: P(x) \otimes Q(y) \mapsto P(x)Q(y)$$
You have to show this map is a bijection and you will be done. Surjectivity ought to be clear, but you need to work a bit to show injectivity.
One possible approach (see the top comment for another approach): You can try to define an explicit inverse. For example, define
$$\mathbb{K}[x,y] \to \mathbb{K}[x] \otimes \mathbb{K}[y]$$
by $x \mapsto x \otimes 1$ and $y \mapsto 1 \otimes y$ and extend multiplicatively and linearly.
