Polynomial over Field has Finitely Many Roots

Theorem
Let $F$ be a field, and let $F[x]$ be the ring of polynomial functions in the indeterminate $x$.

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

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

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

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

By the Polynomial Factor Theorem, if $a\in A$, then we can write $p(x) = q_1 (x) \cdot (x-a)$ where $\operatorname{deg} q_1 = n-1$.

Now let $a' \ne a$ be another element of $A$.

Then since $p(a')=0$ but $(a'-a) \ne 0$, we must have $q_1 (a')=0$.

Again by the Polynomial Factor Theorem, we can write $q_1 (x) = q_2 (x) \cdot (x-a')$.

Therefore $p(x) = q_2 (x) \cdot (x-a') \cdot (x-a)$, where $\operatorname{deg} q_2 = n-2$.

We can repeat this procedure as long as there are still distinct roots in $A$, obtaining after the $i$th iteration 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 $p$, and 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 $|A| \leq n$