Polynomial takes distinct values at set of points

Question

Let $f = a_1 x_1 + \cdots + a_n x_n$ be a polynomial in $n$ variables $x_j$ over an infinite field $F$ with coefficients $a_j\in F$.

Let $p_1, \dots, p_m$ be $m$ distinct points in $F^n$.

Now consider the following statement:

One can choose $a_1, \dots, a_n$ such that $f$ takes distinct values at the $p_j$.

This statement is clear intuitively, as a "random" choice of the $a_j$ should guarantee this.

However, is there a rigerous proof of this fact?

To make sure: is $f$ necessarily homogeneous of degree $1$? This is what's written now, but it seems possible that it's not what you meant. — diracdeltafunk, Mar 12 '23 at 23:23

score 2 · Accepted Answer · answered Mar 12 '23 at 23:41

The set of all possible $(a_1,\dots,a_n)$ is itself an $n$-dimensional vector space over $F$. The subset of $(a_1,\dots,a_n)$ for which the undesired $f(p_i)=f(p_j)$ occurs is a subspace of that vector space of dimension $n-1$. There are finitely many such subspaces, and it's not hard to check that the union of finitely many subspaces of a vector space over a field with infinitely many elements cannot equal the entire space (indeed has measure $0$, is a proper subvariety, etc.—there are lots of ways to phrase the intuition that a random element of the vector space is not in this subspace).

Here is a proof of the claim that a vector space over an infinite field is not a finite union of proper subspaces. — diracdeltafunk, Mar 12 '23 at 23:44

Polynomial takes distinct values at set of points

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