K(a) is isomorphic to K[x] modulo the minimal polynomial

Statement

Let $L/K$ be a field extension and let $a \in L$ be algebraic over $K$ with minimal polynomial $m_a$. Then the evaluation map $K[x] \to L$, $f \mapsto f(a)$, induces a ring isomorphism $$K[x]/(m_a) \cong K(a).$$ In particular, $K(a) = K[a] = \{ f(a) : f \in K[x] \}$.

Remarks

The proof: the kernel of the evaluation map is $(m_a)$, by the divisibility property. The first isomorphism theorem gives $K[x]/(m_a) \cong K[a]$. Since $m_a$ is irreducible, $K[x]/(m_a)$ is a field. So $K[a]$ is a field, and $K[a] = K(a)$. If $a$ is transcendental over $K$, then the evaluation map is injective, and it extends to an isomorphism $K(x) \cong K(a)$.

Explanations

L1 human

An algebraic element adds nothing to the field beyond the relation given by its minimal polynomial. So $K(a)$ is the same as polynomials in a variable, computed modulo the minimal polynomial. For example, $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}[x]/(x^2 - 2)$.