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Let $K$ be a field and $K(x_1,\cdots,x_n)$ be the degree-$n$ purely transcendental extension of $K$. Given homogeneous polynomials $f_1,\cdots,f_n\in K[x_1,\cdots,x_n]\setminus K$ with $\deg f_i=d_i$, if $K(x_1,\cdots,x_n)$ is algebraic over $K(f_1,\cdots,f_n)$ and $(f_1,\cdots,f_n)$ is a regualar sequence in $K[x_1,\cdots,x_n]$, can we conclude that $[K(x_1,\cdots,x_n):K(f_1,\cdots,f_n)]=d_1\cdots d_n$?

Generally, how to determine the degree of a rational function field over a relatively algebraic subfield? Thanks.

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  • $\begingroup$ This is definitely not true as stated. Even the question of when $K(f_1,\ldots,f_n)$ is equal to $K(x_1,\ldots,x_n)$ is a subtle one; in characteristic $0$ there is for instance the Jacobian conjecture. For example, $f_1 = x_1+y_2^{100}$ and $f_2 = y_2$ generate $K(x_1,x_2)$, but $d_1d_2 > 1$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 21, 2022 at 18:24
  • $\begingroup$ This is the number of roots of $f_1=y_1,\dots, f_n=y_n$ in variables $x_1,\dots, x_n$ over the field $\overline{K(y_1,\dots,y_n)}$. As a computational problem, this is surely solvable - I'm going to take a wild guess that the answer involves Grobner bases. $\endgroup$
    – Will Sawin
    Commented Mar 21, 2022 at 20:29
  • $\begingroup$ Dear Sawin, can you give some references about your claim? Thanks. Besides, I think my problem is about Bezout's theorem. $\endgroup$
    – GiS
    Commented Mar 22, 2022 at 4:19
  • $\begingroup$ Dear van Dobben de Bruyn, I think you indeed give a counterexample when some $f_i$ are inhomogeneous, that's why I assume that $f_i$ are homogeous. Besides, I guess that my problem is perhaps about projective geometry, and your example is about affine geometry. $\endgroup$
    – GiS
    Commented Mar 22, 2022 at 4:22
  • $\begingroup$ I find a counterexample when $n=2$ and $[K(x_1,x_2):K(x_1,x_1x_2)]=1$. Hence I furthermore require $(f_1,\cdots,f_n)$ is a regular sequence. $\endgroup$
    – GiS
    Commented Mar 22, 2022 at 12:13

1 Answer 1

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To the first question, the answer is: "Yes."

The regularity of the sequence implies that the set $V(f_1, \ldots, f_n)$ of zeroes of $f_1, \ldots, f_n$ is zero dimensional. The homogeneity of the $f_j$ then implies that $V(f_1, \ldots, f_n)$ consists only of the origin of $K^n$, so Bézout's theorem, as stated for example in this recent question of mine, applies.

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