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

Proof

1. Let $L/K$ be a field extension, let $a \in L$ be algebraic over $K$, and let $m_a$ be its minimal polynomial.
2. Let $\operatorname{ev} \colon K[x] \to L$, $f \mapsto f(a)$. By the universal property of the polynomial ring, applied to the inclusion $K \to L$ and $x \mapsto a$, it is a ring homomorphism. It is the restriction of the evaluation homomorphism of $L[x]$ at $a$.
3. Let $K[a] = \set{ f(a) : f \in K[x] }$, the image of $\operatorname{ev}$.
4. The kernel of $\operatorname{ev}$ is $(m_a)$. Indeed, if $f = q m_a$, then $f(a) = q(a) m_a(a) = 0$.
5. Conversely, if $f(a) = 0$, then $m_a$ divides $f$, so $f \in (m_a)$.
6. By the first isomorphism theorem for rings, the map $f + (m_a) \mapsto f(a)$ is a ring isomorphism from the quotient ring $K[x]/(m_a)$ onto $K[a]$.
7. The polynomial $m_a$ is irreducible. So [[$K[x]/(m_a)$ is a field|0AWMPX35CW]].
8. By Steps 6 and 7, $K[a]$ is a subring of $L$ that is a field. So every nonzero element of $K[a]$ has its inverse in $K[a]$, and $K[a]$ is a subfield of $L$.
9. The subfield $K[a]$ contains $K$, as the values of the constant polynomials, and contains $a = \operatorname{ev}(x)$.
10. The field $K(a)$ is the intersection of all subfields of $L$ that contain $K$ and $a$. By Step 9, $K(a) \subseteq K[a]$.
11. Conversely, $K(a)$ is a subfield containing $K$ and $a$. It is closed under sums and products, so it contains every $f(a) = \sum_i c_i a^i$ with $c_i \in K$. So $K[a] \subseteq K(a)$.
12. Hence $K(a) = K[a] = \set{ f(a) : f \in K[x] }$. With Step 6, the map induced by $\operatorname{ev}$ is a ring isomorphism $K[x]/(m_a) \cong K(a)$.