Let $X, Y$ be affine varieties, we know that the coordinate ring of the product variety $X\times Y$ satisfies $k[X\times Y]\cong k[X]\otimes_k k[Y]$.
My question is is it true that for rational function field, we also have $k(X\times Y)\cong k(X)\otimes_k k(Y)$? If not, is there a way to relate $k(X\times Y)$ with $k(X)$, and $k(Y)$ ?
I just thought I would expand on Alex's answer a bit. First I want to make sure we're assuming that $k = \bar k$. Then we can always think of $k(X)\otimes_k k(Y)$ as a subset of $k(X\times Y)$ by the map extending $\phi\otimes\psi \mapsto \phi\psi$ (see note below), but you can see that in this way, $k(X)\otimes_k k(Y)$ only contains rational functions $f(\mathbf x, \mathbf y) / g(\mathbf x, \mathbf y) \in k(X\times Y)$ where $g(\mathbf x, \mathbf y)$ is a product $g_1(\mathbf x)g_2(\mathbf y)$ of polynomials in just $\mathbf x$ and $\mathbf y$ separately, and this is not all possible rational functions -- e.g. $\frac{1}{xy-1}$ for $X=Y=\mathbb{A}^1_k$. Of course, you may write $k(X\times Y) \cong \mbox{frac}(k(X)\otimes_k k(Y)) \cong \mbox{frac}(k[X]\otimes_k k[Y])$, or you may just localize $k(X)\otimes_k k(Y)$ at the set of polynomials which are not of the above form, but these are probably not super helpful.
Note. It is non-trivial that, when $k = \bar k$ and $A,B$ are $k$-algebras and also domains, then $A\otimes_k B$ is a domain. You can find a discussion of this here. It is not always true when $k\ne \bar k$.
It is not true in general that $k(X\times Y)\cong k(X)\otimes_k k(Y)$.
For example, let $k=\mathbb{Q}$, $k[X]=k[Y]=\mathbb{Q}(i)$ then $k(X)=k(Y) = \mathbb{Q}(i)$, but $\mathbb{Q}(i) \otimes_{\mathbb{Q}} \mathbb{Q}(i)$ is not even a field.
It is also not true for product of two affine lines: $k(x) \otimes_k k(y) \subsetneq k(x,y)$.
