This question concerns teaching a proof of the theorem that if a polynomial $f \in k[x]$ over an infinite field $k$ is the zero function (i.e. $f(a) = 0$ for all $a \in k$) then it is also the zero polynomial (i.e. the coefficients of $f$ are all zero). (Note that over a finite field, one can have nonzero polynomials that are zero functions.)

Proof. We will use the fact that a nonzero polynomial in $k[x]$ of degree $n$ has at most $n$ distinct roots, that is, a finite number of roots. Since $f$ is the zero function (condition), $f(a) = 0$ for all $a \in k$ (definition). However, since $k$ is infinite (condition), $f$ has infinitely many roots and thus is the zero polynomial. Were any of the polynomial coefficients nonzero, this would lead to $f$ having a finite number of roots. However, this would contradict $f$ having infinitely many roots and hence there cannot be any nonzero coefficients.

What could be done to improve the clarity of the exposition so that the logic of the proof is easier to follow?


How about this:

Let $k$ be an infinite field, and let $f \in k[x]$. Assume $f(t) = 0$ for all $t \in k$.

Assume to the contrary that $f$ is not the zero polynomial. Then $f$ is a polynomial of degree $n \geq 1$.

Choose distinct elements $z_1, z_2, z_3, \dots, z_{n+1}$ of $k$.

By repeated use of the linear factor theorem, we know that $f(x) = (x-z_1)(x-z_2)(x-z_3) \dots (x-z_{n+1})g(x)$ for some nonzero polynomial $g \in k[x]$.

This show that $f$ has degree at least $n+1$, which contradicts the supposition that $f$ had degree $n$.

Thus our supposition was false, and $f$ must be the zero polynomial.

A few subtle points:

  1. There is actually a small inductive argument hidden in the "repeated use of linear factors theorem", which you may or may not want to glide over.
  2. One must be careful to explain why $g$ cannot be the zero polynomial. If it were, then $f$ would be the zero polynomial contradicting the hypothesis. Understanding the distinction between formal polynomials and polynomial functions is a likely stumbling block in this proof.
  3. It is worth pointing out that the linear factor theorem does not fail for the zero polynomial, since $(x-z)$ divides the zero polynomial for any $z \in k$.
  4. It is also worth showing that this theorem has content by displaying a polynomial over a finite field which is nonzero as a polynomial, but whose associated function is the constant zero function.

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