---
tag: 7464TECR9S
kind: proof
name: Proof of the uniqueness of splitting fields
msc: [12F05]
proves: 069A3BC3EP
uses: [0QA8CHH8MG, 02NZZ1J0FH, 0SHMC00RJV, 0MA5B744K3, 0DQMDSVSXJ, 0GK1WB9R98, 06K7SARZKR, 0NFEG48RDR, 0ZN39QDTP9, 0N5QW9K7MH, 0A7KJ68XHT, 0ZMYX7GMH0, 0B6QC0C0G1, 0DQG39K2NH, 0QKVE1GD9A, 055HT85T3S, 0QRCFZKXH8, 0MNHXBSX3M, 0T233AZ1GE, 0ASJVEFA9J]
methods: [0ASJVEFA9J]
sources:
  - url: https://www.jmilne.org/math/CourseNotes/FT.pdf
    locator: Proposition 2.12 and Corollary 2.13
---

## Proof

We prove a stronger claim, which allows a change of base field, by strong [[induction|0ASJVEFA9J]] on the degree $[L : K]$.

1. For a ring isomorphism $\tau \colon K \to K'$ of fields and $h = \sum_i c_i x^i \in K[x]$, write $\tau h = \sum_i \tau(c_i) x^i$. By the [[universal property of the polynomial ring|0DQMDSVSXJ]], $h \mapsto \tau h$ is a ring isomorphism $K[x] \to K'[x]$, with inverse $h \mapsto \tau^{-1} h$.
2. Claim: let $\tau \colon K \to K'$ be a [[ring isomorphism|0B6QC0C0G1]] of [[fields|0QA8CHH8MG]], and $f \in K[x]$ nonzero. Let $L$ be a [[splitting field|02NZZ1J0FH]] of $f$ over $K$, and $L'$ one of $\tau f$ over $K'$. Then some ring isomorphism $\sigma \colon L \to L'$ extends $\tau$.
3. The statement is the case $K' = K$ and $\tau = \id$, so it suffices to prove the claim. First, $[L : K]$ is finite. Indeed, $L = K(a_1, \dots, a_n)$ for roots $a_i$ of $f$, which are algebraic over $K$.
4. Adjoining $a_1, \dots, a_n$ one at a time gives a chain of fields from $K$ to $L$. Each step is finite by the [[degree of a simple extension|0N5QW9K7MH]], so $[L : K]$ is finite by the [[tower law|0A7KJ68XHT]].
5. Write $f = c (x - a_1) \cdots (x - a_n)$ in $L[x]$ and $\tau f = c' (x - b_1) \cdots (x - b_n)$ in $L'[x]$, with $L = K(a_1, \dots, a_n)$ and $L' = K'(b_1, \dots, b_n)$, as in the definitions of [[splitting|0SHMC00RJV]] and [[generated field|0MA5B744K3]]. Let $d = [L : K]$.
6. Case $d = 1$. Then $L = K$, so all $a_i \in K$. Applying the ring homomorphism of Step 1 gives $\tau f = \tau(c) (x - \tau(a_1)) \cdots (x - \tau(a_n))$ in $K'[x]$.
7. Comparing with Step 5 by unique factorisation in $L'[x]$, which is a [[Euclidean domain|0DQG39K2NH]], hence a [[PID|0QKVE1GD9A]] and a [[UFD|055HT85T3S]], each $b_j$ is some $\tau(a_i) \in K'$. So $L' = K'$, and $\sigma = \tau$ works.
8. Case $d > 1$, assuming the claim whenever the degree is less than $d$. If all $a_i$ were in $K$, then $L = K$ and $d = 1$. So some root $a = a_i$ is not in $K$.
9. Let $m$ be the [[minimal polynomial|0GK1WB9R98]] of $a$ over $K$. It is [[irreducible|0NFEG48RDR]], it [[divides|06K7SARZKR]] $f$, and $\deg m = [K(a) : K] > 1$ since $a \notin K$.
10. By Step 1, $\tau m$ is monic, irreducible in $K'[x]$, of degree $\deg m$, and $\tau m$ divides $\tau f$. So $\tau m$ divides $\prod_j (x - b_j)$ in $L'[x]$, as $c' \ne 0$.
11. Let $\pi$ be a monic divisor of $\tau m$ in $L'[x]$ of least degree $\ge 1$. A factorisation of $\pi$ into two nonunits would give a monic divisor of smaller [[degree|0T233AZ1GE]] $\ge 1$.
12. So $\pi$ is irreducible. By Step 7, $L'[x]$ is a UFD, so $\pi$ is [[prime|0QRCFZKXH8]].
13. By the definition of a [[prime element|0MNHXBSX3M]], $\pi$ divides some factor $x - b_j$. Comparing degrees, $\pi = x - b_j$. So $b = b_j$ is a root of $\tau m$.
14. Since $\tau m$ is monic, irreducible and $\tau m(b) = 0$, divisibility gives that $\tau m$ is the minimal polynomial of $b$ over $K'$.
15. Let $\psi \colon K[x] \to K'(b)$, $h \mapsto (\tau h)(b)$. It is a ring homomorphism, and it is surjective because $K'(b) = \set{ u(b) : u \in K'[x] }$ by the [[structure of simple extensions|0ZN39QDTP9]].
16. Its kernel is the set of $h$ with $\tau m \mid \tau h$, by Step 14 and divisibility. By Step 1, this is the set of $h$ with $m \mid h$, that is, $(m)$.
17. By the [[first isomorphism theorem|0ZMYX7GMH0]], $K[x]/(m) \cong K'(b)$ through $h + (m) \mapsto (\tau h)(b)$. Also $K[x]/(m) \cong K(a)$ through $h + (m) \mapsto h(a)$, by the structure of simple extensions.
18. Composing, $\sigma_1 \colon K(a) \to K'(b)$, $h(a) \mapsto (\tau h)(b)$, is a ring isomorphism. On constants it equals $\tau$, so it extends $\tau$, and $\sigma_1 f = \tau f$.
19. The field $L$ is a splitting field of $f$ over $K(a)$: $f$ splits over $L$, and $L$ is generated over $K(a)$ by $a_1, \dots, a_n$, since $K(a) \subseteq L$.
20. In the same way, $L'$ is a splitting field of $\sigma_1 f = \tau f$ over $K'(b)$, because $b$ is one of the $b_j$.
21. By the tower law, $[L : K(a)] = d / [K(a) : K] < d$. The induction hypothesis, for $\sigma_1$, gives a ring isomorphism $\sigma \colon L \to L'$ extending $\sigma_1$.
22. Then $\sigma$ extends $\tau$, which proves the claim. With $\tau = \id_K$, we get $\sigma \colon L \to L'$ with $\sigma(c) = c$ for all $c \in K$.

## Idea

Build the isomorphism one root at a time. A root $a$ of $f$ outside $K$ has a minimal polynomial $m$, and its image $\tau m$ has a root $b$ among the roots of $\tau f$. Both $K(a)$ and $K'(b)$ are copies of $K[x]/(m)$, so $\tau$ extends to them. Over these larger fields the degree is smaller, and induction finishes.