2
$\begingroup$

I am looking into this proof: If $m$$>$ $n$, then any $f_1,...,f_m$ (non-zero polynomials) in $K[X_1,...,X_n]$ are algebraically dependent over $K$ ($K$ is a field).

The proof starts by assuming that $f_1,...,f_m$ are algebraically independent over $K$. But I don't understand how we can now deduce that $f_1,...,f_m$ forms a transcendence basis of $K[X_1,...,X_n]$ over $K$. In other words, why is $K[X_1,...,X_n]$ algebraic over $K(f_1,...,f_m)$?

The proof then goes on to say that $X_1,...,X_n$ forms a transcendence basis of $K[X_1,...,X_n]$ over $K$. Could someone explain why this is true?

The contradiction at the end is now straightforward since all transcendence bases must have the same cardinality.

Would appreciate any sort of explanation as I am not very competent in this area of mathematics.

$\endgroup$
  • $\begingroup$ Have you a reference for the proof you are considering? $\endgroup$ – Angina Seng Jan 8 '18 at 21:17
  • $\begingroup$ Krister Forsman, Two themes in commutative algebra. This is a standard result, but everywhere I have looked the (brief) explanations are all the same! $\endgroup$ – user519430 Jan 8 '18 at 21:22
  • 1
    $\begingroup$ No, $K[X_1,\ldots, X_n]$ may not be apriori algebraic over $K(f_1,\ldots, f_m)$. But, if not, we may assume say, $X_1$ is not algebraic. Then, take $f_{m+1}=X_1$. Then, these are still algebraically independent and proceed. $\endgroup$ – Mohan Jan 9 '18 at 1:25
  • $\begingroup$ Related: mathoverflow.net/questions/41535/… $\endgroup$ – Watson Mar 22 '18 at 12:38
2
$\begingroup$

I never learned the basics of transcendence degree myself, but fortunately you can write down a proof that completely avoids it. It has the benefit of being much more concrete.

A polynomial in the $f_1, \dots f_m$ is a sum of monomials $\prod f_i^{e_i}$, which has degree $\sum e_i d_i$ where $d_i = \deg f_i$ (by which I mean the largest sum of exponents of a monomial occurring in $f_i$). So the number of such monomials of degree at most $D$ is the number of solutions to

$$\sum_{i=1}^m e_i d_i \le D, e_i \ge 0.$$

As a function of $D$ this looks like the volume of a generalized right triangle; the leading term asymptotically is $\frac{D^m}{d_1 \dots d_m m!}$, and in particular it grows like $D^m$. On the other hand, the dimension of the space of polynomials in $K[x_1, \dots x_n]$ of degree at most $D$ is the number of solutions to

$$\sum_{i=1}^n e_i \le D, e_i \ge 0$$

and so it grows like $\frac{D^n}{n!}$; since $m > n$ the former eventually overtakes the latter, and so the monomials $\prod f_i^{e_i}$ eventually cannot be linearly independent.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How did you deduce that the leading term asymptotically is $\frac{D^m}{d_1 \dots d_m m!}$, and in particular grows like $D^m$? Is this a topic of mathematics? Because if so then I haven't covered it! $\endgroup$ – user519430 Jan 9 '18 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy