The following is from A. Białynicki-Birula, "Algebra" (the translation is mine).
Let $K$ be a field. (...) Let $I_a$ denote ideal of these polynomials $f \in K[x]$ for which $f(a) = 0$.
(...)
Theorem 1.4 Let $L$ be an extension of the field $K$ and let $a \in L$. Then the following conditions are equivalent:
(a) $a$ is an algebraic element over $K$,
(b) $K[a]$ is a field, that is $K[a] = K(a)$
(...)
Proof. (a) => (b). If $a$ is an algebraic element over $K$, then $I_a \neq (0)$. Since beside this $I_a$ is a prime ideal, and $K[x]$ is a PID, then $I_a$ is a maximal ideal. Hence the ring $K[a]$ is isomorphic with $K[x]/I_a$ , hence it is a field.
How can one show that "the ring $K[a]$ is isomorphic with $K[x]/I_a$"? I searched the book for various results related to maximal ideals but I wasn't able to use any of them to prove this fact.
Consider the evaluation morphism $K[x] \to K$ which sends a polynomial $f$ to $f(a)$. Its image is $K[a]$ and its kernel is $f\in K[x]$ with $f(a) = 0$, that is $I_a$. The first isomorphism theorem for rings then gives $K[a] \cong K[x]/I_a$.