---
tag: 9SB00E1N98
kind: proof
name: Proof that the degree of a simple extension is the degree of the minimal polynomial
msc: [12F05]
proves: 0N5QW9K7MH
uses: [0WQ1A8SHTZ, 0GK1WB9R98, 0ZN39QDTP9, 0MES3KS0ZC, 06K7SARZKR, 0T233AZ1GE, 0GS74FGFEK, 0VTQTPQGAV, 088RYKAWK9, 0EVHJFTG9Z, 05G2G0P224]
sources:
  - url: https://en.wikipedia.org/wiki/Simple_extension
    locator: Definition
  - url: https://www.jmilne.org/math/CourseNotes/FT.pdf
    locator: Chapter 1, Stem fields
---

## 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]], of degree $n$.
2. By the [[structure of simple extensions|0ZN39QDTP9]], $K(a) = \set{ f(a) : f \in K[x] }$.
3. Spanning. Let $z \in K(a)$, so $z = f(a)$ for some $f \in K[x]$ by Step 2.
4. Since $m_a$ is monic, [[polynomial division|0MES3KS0ZC]] gives $f = q m_a + r$ with $\deg r < n$. Write $r = \sum_{i=0}^{n-1} c_i x^i$ with $c_i \in K$.
5. Then $z = f(a) = q(a) m_a(a) + r(a) = r(a) = \sum_{i=0}^{n-1} c_i a^i$. So $1, a, \dots, a^{n-1}$ [[span|088RYKAWK9]] $K(a)$ over $K$.
6. [[Linear independence|0EVHJFTG9Z]]. Let $c_0, \dots, c_{n-1} \in K$ with $\sum_{i=0}^{n-1} c_i a^i = 0$, and let $r = \sum_{i=0}^{n-1} c_i x^i$. Then $r(a) = 0$.
7. By [[divisibility|06K7SARZKR]], $r = m_a s$ for some $s \in K[x]$. If $s \ne 0$, then $\deg r = n + \deg s \ge n$ by the [[degree of a product|0T233AZ1GE]], which is false.
8. So $s = 0$, hence $r = 0$, and all $c_i = 0$.
9. By Steps 5 and 8, $1, a, \dots, a^{n-1}$ is a [[basis|0VTQTPQGAV]] of $K(a)$ over $K$.
10. This basis has $n$ elements, so by the definition of [[dimension|05G2G0P224]], $[K(a) : K] = n$, as a [[degree|0GS74FGFEK]].

## Idea

Every element of $K(a)$ is a polynomial in $a$, and dividing by $m_a$ cuts its degree below $n$. So the powers $1, a, \dots, a^{n-1}$ span. A linear relation among them would be a nonzero polynomial of degree less than $n$ with root $a$, which the minimality of $m_a$ forbids.