From Herstein's "Topics in Algebra" ($2^{\text{nd}}$ edition):
LEMMA $5.3.2$ $\;$ A polynomial of degree $n$ over a field can have at most $n$ roots in any extension field.
$\;$ Proof. $\;$ We proceed by induction on $n$, the degree of the polynomial $p(x)$. If $p(x)$ is of degree $1$, the it must be of the form $\alpha x + \beta$ where $\alpha , \beta$ are in a field $F$ and where $\alpha \neq 0$. Any $a$ such that $p(a) = 0$ must then imply that $\alpha a + \beta = 0$, from which we conclude that $a = (-\beta/\alpha)$. That is, $p(x)$ has a unique root $-\beta/\alpha$, whence the conclusion of the lemma certainly holds in this case.
$\;$ Assuming the result to be true in any field for all polynomials of degree less than $n$, let us suppose that $p(x)$ is of degree $n$ over $F$. Let $K$ be any extension of $F$. If $p(x)$ has no roots in $K$, then we are certainly done, for the number of roots in $K$, namely zero, is definitely at most $n$. So, suppose that $p(x)$ has at least one root $a \in K$ and that $a$ is a root of multiplicity $m$. Since $(x-a)^{m} \mid p(x)$, $m \leq n$ follows. Now $p(x) = (x-a)^{m} q(x)$, where $q(x) \in K[x]$ is of degree $n - m$. From the fact that $(x-a)^{m+1} \not\mid p(x)$, we get that $(x-a) \not\mid q(x)$, whence, by the corollary to Lemma $5.3.1$, $a$ is not a root of $q(x)$. If $b \neq a$ is a root, in $K$, of $p(x)$, then $0 = p(b) = (b-a)^m q(b)$; however, since $b - a \neq 0$ and since we are in a field, we conclude that $q(b) = 0$. That is, any root of $p(x)$, in $K$, other than $a$, must be a root of $q(x)$. Since $q(x)$ is of degree $n-m \lt n$, by our induction hypothesis, $q(x)$ has at most $n-m$ roots in $K$, which, together with the other root $a$, counted $m$ times, tell us that $p(x)$ has at most $m + (n-m)$ roots in $K$. This completes the induction and proves the lemma.
The corollary to Lemma $5.3.1$:
*If $a \in K$ is a root of $p(x) \in F[x]$, where $F \subset K$, then in $K[x]$, $(x-a) \mid p(x)$.
After the proof of Lemma $5.3.2$, Herstein points out that commutativity is necessary for the lemma to be true: In the ring of real quaternions, the polynomial $x^{2} + 1$ has an infinite number of roots. But where exactly is commutativity used in the proof of Lemma $5.3.2$?
