Homogeneous polynomial over algebraically closed field has non-trivial zero Let $k$ be an algebraically closed field and let $f\in k[x_1,\cdots,x_n]$ homogeneous of positive degree $d>0$ in $n>1$ variables. Why does $f$ have a nontrivial zero in $k^n$?
We could prove this using projective dimension theory, but surely this is an overkill. I would like an easy proof that does not involve algebraic geometry or (advanced results in) commutative algebra.
My first attempt was to simply fill in $1$ for every variable except for $x_1$, but $f(x_1,1,\cdots,1)$ need not have a zero in $k$ since it might be constant, e.g. when $f(x_1,x_2,x_3,x_4)=x_1x_2-x_1x_3+x_4^2$. Surely I am overlooking a simple argument here.
 A: I will prove the statement via induction on $n$, the number of variables. For the base case $n=2$, write $f(x_1,x_2) = \sum\limits_{i=0}^{d}{a_ix_1^ix_2^{d-i}}$. Notice that if $a_d = 0$, then $f(x_1,0) = 0$ since in each nonzero term of the sum we would have $x_2$ raised to a positive power, and hence $(x,0)$ for any arbitrary value of $x\ne 0$ is a nontrivial root of $f$. Hence, assume that $a_d\ne 0$. Notice that for $c\in k$ we have
$$f(cx_1,x_1) = \sum\limits_{i=0}^{d}{a_i(cx_1)^i(x_1)^{d-i}} = x_1^d\sum\limits_{i=0}^{d}{a_ic^i}$$
and $\sum\limits_{i=0}^{d}{a_ic^i}$ is a nonconstant polynomial in $c$ (as $a_d\ne 0$), so it has a nonzero root (call it $r$) since $k$ is algebraically closed. Then $f(r,1) = \sum\limits_{i=0}^{d}{a_ir^i} = 0$, so $(r,1)$ is a nontrivial root of $f$.
Now, suppose the statement is true for $n$; i.e. every homogeneous polynomial of degree $d$ in $n$ variables has a nontrivial root in $k^n$ for every $d>1$. Let $f\in k[x_1,\dots,x_n,x_{n+1}]$, and write $f(x_1,\dots,x_n,x_{n+1}) = \sum\limits_{i=0}^{d}{f_i(x_1,\dots,x_n)x_{n+1}^i}$, where $f_i\in k[x_1,\dots,x_n]$. By homogeneity, we see that each $f_i$ is homogeneous of degree $d-i$. If $f_0(x_1,\dots,x_n) = 0$ for all $x_1,\dots,x_n$ (which is equivalent to saying that $f_0 = 0$ since $k$ is infinite), then $f(x_1,\dots,x_n,0) = 0$ since in each nonzero term of the sum we would have $x_{n+1}$ raised to a positive power, and hence $f$ would have a nontrivial root. If $f_j = 0$ for all $j>0$, then $f(x_1,\dots,x_n,x_{n+1}) = f_0(x_1,\dots,x_n)$, and $f_0$ is homogeneous of degree $d$, so by the induction hypothesis it has a nontrivial root in $k^n$, and hence $f$ would also have a nontrivial root in $k^{n+1}$. Hence, assume $f_0\ne 0$ and that $f_j\ne 0$ for some $j>0$.
Notice that there exists $(y_1,\dots,y_n)\in k^n$ such that $f_0(y_1,\dots,y_n)\ne 0$ and $f_j(y_1,\dots,y_n)\ne 0$: otherwise, we would have $(f_0f_j)(y_1,\dots,y_n) = 0$ for all $(y_1,\dots,y_n)\in k^n$, which forces $f_0f_j = 0$ since $k$ is infinite, contradicting the assumption that $f_0\ne 0$ and $f_j\ne 0$. If we let $c_i = f_i(y_1,\dots,y_n)\in k$, then
$$ f(y_1,\dots,y_n,x_{n+1}) = \sum\limits_{i=0}^{d}{f_i(y_1,\dots,y_n)x_{n+1}^i} = \sum\limits_{i=0}^{d}{c_ix_{n+1}^i}. $$
Notice that the above polynomial in $x_{n+1}$ is not constant, since $c_j = f_j(y_1,\dots,y_n)\ne 0$, and it does not have zero as a root, since the constant term $c_0 = f_0(y_1,\dots,y_n)$ is not $0$. It follows that there exists a nonzero root $x_{n+1} = r$ of this polynomial, from which we have that $f(y_1,\dots,y_n,r) = 0$, i.e. $(y_1,\dots,y_n,r)$ is a nontrivial root of $f$ in $k^{n+1}$.
