Polynomial over Field has Finitely Many Roots

Theorem
Let $F$ be a field.

Let $F \left[{x}\right]$ be the ring of polynomial functions in the indeterminate $x$.

If $p \in F \left[{x}\right]$ be non-null, then $p$ has finitely many roots in $F$.

Proof
Let $n \ge 1$ be the degree of $p$.

We argue that $p$ has at most $n$ roots in $F$.

Let $A$ be the set of roots of $p$.

Let $a \in A$.

By the Polynomial Factor Theorem:
 * $p \left({x}\right) = q_1 \left({x}\right) \cdot \left({x - a}\right)$
 * where $\deg q_1 = n - 1$.

Let $a' \in A$ such that $a' \ne a$.

Then since:
 * $p \left({a'}\right) = 0$

but:
 * $\left({a' - a}\right) \ne 0$

it follows that:
 * $q_1 \left({a'}\right) = 0$

Again by the Polynomial Factor Theorem:
 * $q_1 \left({x}\right) = q_2 \left({x}\right) \cdot \left({x - a'}\right)$

Therefore:
 * $p \left({x}\right) = q_2 \left({x}\right) \cdot \left({x - a'}\right) \cdot \left({x - a}\right)$

where $\deg q_2 = n-2$.

We can repeat this procedure as long as there are still distinct roots in $A$.

After the $i$th iteration we obtain a polynomial $q_i$ of degree $n-i$.

Now the degree of $q_i$ decreases each time we factor out a root of $p$, so $q_n$ is necessarily a constant term.

Therefore $q_n$ can share no roots with $p$.

So this procedure must stop by at most the $n$th step.

That is, we can pick at most $n$ distinct elements from the set $A$, and:
 * $\left\vert{A}\right\vert \le n$