Discrete valuation on rational function field of multivariables For one variable rational function field, I know that we can write $\frac{f(X)}{g(X)}$ as $X^n\frac{s(X)}{t(X)}$ where $s(0)$ and $t(0)$ are nonzero. Then 
$v(\frac{f(X)}{g(X)})=n$ is a valuation. But how to construct a valuation for two-variable or multivariable rational function fields? I am not sure whether I can separate the rational function as in one variable situation. Also, checking whether the construction is a valuation  becomes more complicated. 
 A: It sometimes helps to state a problem as precisely as possible in order to see the solution! Here, you constructed a valuation on the rational function field $K(X)$ of one variable over any field $K$, right? In particular, you can apply it to the case where $K$ itself is a function field of one variable, say $L(Y)$, right? This gives you a valuation on the field
$$
(L(Y))(X),
$$
which is nothing but
$$
L(Y,X).
$$
Indeed, of course, one has the inclusion
$$
L(Y,X)\subseteq (L(Y))(X),
$$
as the former is the ring of fractions of polynomials in $X$ and $Y$, and the latter is the ring of fractions of polynomials in $X$ whose coefficients are fractions of polynomials in $Y$.
This inclusion is an equality since one can chase denominators in the numerator and denominator of a fraction of polynomials in $X$ whose coefficients are fractions of polynomials in $Y$. Hence, we have a valuation on $L(Y,X)$. As you said, it is defined by
$$
v(X^n\tfrac{s(X)}{t(X)})=n
$$
for any $s,t\in L(Y)[X]$ with $s(0)$ and $t(0)$ nonzero in $L(Y)$!
It is now an easy matter to generalize directly to the case of $K=L(Y_1,\ldots,Y_n)$, and get a valuation on the rational function field
$$
L(Y_1,\ldots,Y_n,X)
$$
in $n+1$ variables over the field $L$.
