Homogeneous polynomial of degree $n$ in $n^2$ indeterminates that cannot be the sum of the terms of degree $n$ in $P_1Q_1 + ... + P_k Q_k$ This is a "difficult" exercise taken from Bourbaki's "Commutative Algebra". I have no idea on how to solve it, nor tackle it.
Let $K$ be a finite field. Prove that for all $k \in \mathbb N$, there exists $n_k \in \mathbb N$ such that : for all $n \geq n_k$, there exists a homogeneous polynomial $F_n \in K[x_1,\ldots,x_{n^2}]$ ($n^2$ indeterminates) of degree $n$ that cannot be the sum of the terms of degree $n$ in any polynomial of the form $P_1Q_1 + \ldots + P_k Q_k$, where $P_i$ and $Q_i \in K[x_1,\ldots,x_{n^2}]$ are polynomials without constant term.
Does anyone have a hint or an idea on how to solve this ? Also, I really cannot see how the fact that $K$ is a finite field will help us.
 A: Hint. Compute the number of homogeneous polynomials of degree $n$ (this number is finite because $K$ is a finite field).
Solution (spoilerised, in case you would like to try the hint first).

 This problem is more combinatorics than algebra. Let's first compute the number of monomials of (total) degree $n$. It's given by the number of ways to choose $n$ variables from the $n^2$ variables $x_1,\dotsc,x_{n^2}$, with repetitions being allowed. Thus it equals $\binom{n^2+n-1}n$. This implies that the number of homogeneous polynomials of degree $n$ equals $$|K|^{\binom{n^2+n-1}n}\,,$$as there are $|K|$ choices of coefficient for each of the $\binom{n^2+n-1}n$ monomials.Next, observe that we may assume that all $P_i$ and $Q_i$ have degree $\leq n-1$. Indeed, all terms in $P_i$ of degree $\geq n$ won't contribute to the degree $n$ part of $P_iQ_i$,  since $Q_i$ has no constant term, and vice versa. So let's compute the number of polynomials of degree $\leq n-1$ (from now on we ignore the requirement that there be no constant coefficients, since it will be irrelevant for the estimate). The number of polynomials of degree $\leq n-1$ in $n^2$ variables $x_1,\dotsc,x_{n^2}$ is the same as the number of homogeneous polynomials of degree $n-1$ in $n^2+1$ variables $x_1,\dotsc,x_{n^2},t$, since we can use the additional variable $t$ to homogenise  in a unique way. Hence we can use a similar calculation as above to see that there are $$|K|^{\binom{n^2+1+(n-1)-1}{n-1}}$$polynomials of degree $\leq n-1$. Hence there are at most $$|K|^{2k\binom{n^2+n-1}{n-1}}$$choices for $P_1,\dotsc,P_k,Q_1,\dotsc,Q_k$ (under the assumption that all have degree $\leq n-1$).Finally, observe that $\binom{n^2+n-1}n=n\binom{n^2+n-1}{n-1}$. Hence, for $n>2k$ we see that there are more choices of homogeneous polynomials of degree $n$ than there are choices of $P_1,\dotsc,P_k,Q_1,\dotsc,Q_k$ (of degrees $\leq n-1$), which proves the desired claim. $\square$

