---
tag: ND2TPP8Q2W
kind: proof
name: Proof of the existence of splitting fields
msc: [12F05]
proves: 01441KDGBY
uses: [0QA8CHH8MG, 02NZZ1J0FH, 0SHMC00RJV, 0MA5B744K3, 0VG02M1FEZ, 0ZN39QDTP9, 0A7KJ68XHT, 0K3MWA7DMC, 0T233AZ1GE, 0ASJVEFA9J]
methods: [0ASJVEFA9J]
sources:
  - url: https://www.jmilne.org/math/CourseNotes/FT.pdf
    locator: Proposition 2.7
  - url: https://en.wikipedia.org/wiki/Splitting_field
    locator: Constructing splitting fields
---

## Proof

We argue by [[induction|0ASJVEFA9J]] on $n$. The claim for $n$ is as follows. For every [[field|0QA8CHH8MG]] $K$ and every $f \in K[x]$ of degree $n$, there is a splitting field $L$ of $f$ over $K$ with $[L : K] \le n!$.

1. Case $n = 1$. Write $f = c x + d$ with $c \ne 0$, and let $a_1 = -d/c \in K$. Then $f = c (x - a_1)$, so $f$ [[splits|0SHMC00RJV]] over $K$.
2. Since $a_1 \in K$, the [[generated field|0MA5B744K3]] $K(a_1)$ is $K$. So $L = K$ is a [[splitting field|02NZZ1J0FH]] of $f$, and $[L : K] = 1 = 1!$.
3. Case $n \ge 2$. Assume the claim for $n - 1$, and let $f \in K[x]$ have degree $n$.
4. [[There is an extension|0VG02M1FEZ]] $K'/K$ with $[K' : K] \le n$ and a root $a_1 \in K'$ of $f$. Let $K_1 = K(a_1) \subseteq K'$.
5. By the [[tower law|0A7KJ68XHT]], $[K' : K] = [K' : K_1] \cdot [K_1 : K]$, so $[K_1 : K] \le [K' : K] \le n$.
6. By the [[factor theorem|0K3MWA7DMC]] in $K_1[x]$, $f = (x - a_1) g$ with $g \in K_1[x]$. By the [[degree of a product|0T233AZ1GE]], $\deg g = n - 1$.
7. By the induction hypothesis, $g$ has a splitting field $L$ over $K_1$ with $[L : K_1] \le (n-1)!$. So $g = c (x - a_2) \cdots (x - a_n)$ with $c, a_i \in L$, and $L = K_1(a_2, \dots, a_n)$.
8. Then $f = c (x - a_1)(x - a_2) \cdots (x - a_n)$ in $L[x]$, so $f$ splits over $L$, with roots $a_1, \dots, a_n$.
9. Let $F$ be the subfield of $L$ generated by $K$ and $a_1, \dots, a_n$. We show $F = L$. Here $F \subseteq L$ because $F$ is a subfield of $L$.
10. By the [[structure of simple extensions|0ZN39QDTP9]], $K_1 = \set{ h(a_1) : h \in K[x] }$. These elements lie in $F$, so $K_1 \subseteq F$.
11. So $F$ is a subfield of $L$ that contains $K_1$ and $a_2, \dots, a_n$. Since $L = K_1(a_2, \dots, a_n)$ is the smallest such subfield, $L \subseteq F$.
12. So $L = K(a_1, \dots, a_n)$, and with Step 8, $L$ is a splitting field of $f$ over $K$.
13. By the tower law and Steps 5 and 7, $[L : K] = [L : K_1] \cdot [K_1 : K] \le (n-1)! \cdot n = n!$.
14. This completes the induction.

## Idea

Adjoin one root of $f$, which costs a factor of at most $n$ in the degree. Over the bigger field, $f$ is $x - a_1$ times a polynomial of degree $n - 1$, and induction handles that polynomial. The degrees multiply by the tower law, giving at most $n \cdot (n-1)! = n!$.