I have a few problems with this exercise.
Let $m > n$ be positive integers. Let $K$ be a field, and let $u : K^n \to K^m$ be a morphism defined with polynomials $f_1, \dots, f_m \in K[X_1, \dots, X_n].$ Show that there exists a polynomial $g \in K[X_1, \dots, X_m]$ with $g(f_1, \dots, f_m) = 0.$
I just don't know where to start. I know that I can somehow use that the subspace of all polynomials of degree at most $d$ has dimension $d+m \choose m.$
Thanks for all hints in advance!
I got the answer, so if anyone ever needs it:
Let $k$ be infinite. We know $u: k^n \to k^m, \, p \mapsto \left(f_1(p), \dots , f_m(p)\right)$ and also $k[k^m]=k[X_1, \dots ,X_m]/I(k^m)=k[X_1, \dots ,X_m]$, because for a infinite $k$ it holds that $I(k^m)=0$.
Now take $u^*: k[k^m] \to k[k^m], \, g \mapsto g\left( f_1, \dots ,f_m \right)$. Restrict $u^*$ to the space $V_m^d \to V_n^{dl}$, where $V_m^d$ is subspace of $k[X_1, \dots ,X_m]$ of al polynomials of degree at most $d$ and $l=max_i\{deg(f_i)\}$. Now $u^*(cg+dh)=c\cdot g\left( f_1, \dots ,f_m \right) + d \cdot h\left( f_1, \dots ,f_m \right)$ for $g,h \in k[X_1, \dots ,X_m]$ and $c,d \in k$.
We now take a look at the kernal of $u^*|_{V_m^d}$. $dim(V_m^d)= \binom{d+m}{m}$ and $dim(V_n^{ld})= \binom{ld+n}{n}$.
$\binom{d+m}{m}\sim d^m$ and $\binom{ld+n}{n} \sim d^n$. So for $d$ big enough it is $dim(V_m^d) > dim(V_n^{ld})$ and this gives us , that $u^*$ can't be injective. So there is a $g \in ker(u^*)$ with $u*g=g\left( f_1, \dots ,f_m \right)=0$.
For the case, where $k$ is finite we can't assume, that $I(k^m)=0$. Take $g \in I(k^m)$, then $g(q)=0$ with $q=\left(f_1(p), \dots , f_m(p)\right)$.