# Elementary result: If $m>n$, then any $f_1,...,f_m$ (non-zero polynomials) in $K[X_1,...,X_n]$ are algebraically dependent over $K$

I am looking into this proof: If $m$$> n, then any f_1,...,f_m (non-zero polynomials) in K[X_1,...,X_n] are algebraically dependent over K (K is a field). The proof starts by assuming that f_1,...,f_m are algebraically independent over K. But I don't understand how we can now deduce that f_1,...,f_m forms a transcendence basis of K[X_1,...,X_n] over K. In other words, why is K[X_1,...,X_n] algebraic over K(f_1,...,f_m)? The proof then goes on to say that X_1,...,X_n forms a transcendence basis of K[X_1,...,X_n] over K. Could someone explain why this is true? The contradiction at the end is now straightforward since all transcendence bases must have the same cardinality. Would appreciate any sort of explanation as I am not very competent in this area of mathematics. • Have you a reference for the proof you are considering? Jan 8, 2018 at 21:17 • Krister Forsman, Two themes in commutative algebra. This is a standard result, but everywhere I have looked the (brief) explanations are all the same! – user519430 Jan 8, 2018 at 21:22 • No, K[X_1,\ldots, X_n] may not be apriori algebraic over K(f_1,\ldots, f_m). But, if not, we may assume say, X_1 is not algebraic. Then, take f_{m+1}=X_1. Then, these are still algebraically independent and proceed. Jan 9, 2018 at 1:25 • Mar 22, 2018 at 12:38 ## 1 Answer I never learned the basics of transcendence degree myself, but fortunately you can write down a proof that completely avoids it. It has the benefit of being much more concrete. A polynomial in the f_1, \dots f_m is a sum of monomials \prod f_i^{e_i}, which has degree \sum e_i d_i where d_i = \deg f_i (by which I mean the largest sum of exponents of a monomial occurring in f_i). So the number of such monomials of degree at most D is the number of solutions to$$\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. • 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!
– user519430
Jan 9, 2018 at 13:38