---
tag: MKRHF6333F
kind: proof
name: Proof that K(a) is isomorphic to K[x] modulo the minimal polynomial
msc: [12F05]
proves: 0ZN39QDTP9
uses: [0WQ1A8SHTZ, 0GK1WB9R98, 0125QVNCFQ, 0DQMDSVSXJ, 0AAHE7GEW6, 06K7SARZKR, 0ZMYX7GMH0, 0NFEG48RDR, 0AWMPX35CW, 0MA5B744K3, 07QY5RPFBR, 0B6QC0C0G1, 0VFW4Y69BT]
sources:
  - url: https://en.wikipedia.org/wiki/Simple_extension
    locator: Structure of simple extensions
  - url: https://www.jmilne.org/math/CourseNotes/FT.pdf
    locator: Chapter 1, Algebraic and transcendental elements
---

## Proof

1. Let $L/K$ be a [[field extension|0WQ1A8SHTZ]], let $a \in L$ be algebraic over $K$, and let $m_a$ be its [[minimal polynomial|0GK1WB9R98]].
2. Let $\operatorname{ev} \colon K[x] \to L$, $f \mapsto f(a)$. By the [[universal property of the polynomial ring|0DQMDSVSXJ]], applied to the inclusion $K \to L$ and $x \mapsto a$, it is a ring homomorphism. It is the restriction of the [[evaluation homomorphism|0125QVNCFQ]] of $L[x]$ at $a$.
3. Let $K[a] = \set{ f(a) : f \in K[x] }$, the image of $\operatorname{ev}$.
4. The [[kernel|0AAHE7GEW6]] 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 $f$|06K7SARZKR]], so $f \in (m_a)$.
6. By the [[first isomorphism theorem for rings|0ZMYX7GMH0]], the map $f + (m_a) \mapsto f(a)$ is a [[ring isomorphism|0B6QC0C0G1]] from the [[quotient ring|0VFW4Y69BT]] $K[x]/(m_a)$ onto $K[a]$.
7. The polynomial $m_a$ is [[irreducible|0NFEG48RDR]]. 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|07QY5RPFBR]] 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 [[$K(a)$|0MA5B744K3]] 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 $K[x]/(m_a) \cong K(a)$.

## Idea

Evaluation at $a$ is a ring homomorphism whose kernel is generated by the minimal polynomial. The first isomorphism theorem identifies its image $K[a]$ with $K[x]/(m_a)$. Since $m_a$ is irreducible, this quotient is a field, so the ring $K[a]$ is already the field $K(a)$.