Discrete valuation on rational function field of multivariables

Question

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.

Answer 1

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 $L$.