Contents

# Contents

## Definition

Let $k$ be a field, and $p \in k[x]$ a monic polynomial of degree $n$. A splitting field for $p$ is a field extension $i: k \hookrightarrow E$ such that, regarding $p$ as a polynomial in $E[x]$ by applying the map

$k[x] \cong k \otimes_k k[x] \stackrel{i \otimes 1}{\to} E \otimes_k k[x] \cong E[x],$

the polynomial factors or “splits” as a product of linear factors (possibly repeated):

$p(x) = (x - r_1)(x - r_2)\ldots (x - r_n)$

with all the roots $r_i$ lying in $E$, and the smallest subfield of $E$ containing $k$ and these roots is $E$ itself (so that the roots generate $E$: $E = k(r_1, \ldots, r_n)$.

More generally, given a set $S \subseteq k[x]$ of monic polynomials, an extension field $E/k$ for which each $p \in S$ splits over $E$, and which as a field is generated over $k$ by the set of accompanying roots, is called a splitting field for $S$.

## Construction

The existence of a splitting field for a single polynomial $p \in k[x]$ can be proven by induction on $n = \deg(p)$. As $k[x]$ is a unique factorization domain, $p$ is a product of irreducible polynomials $p_1 p_2 \ldots p_k$. The ideal $(p_1)$ is maximal and so $E = k[x]/(p_1)$ is an finite field extension of $k$ where the coset $r_1 \coloneqq x + (p_1) \in E$ is a root of $p_1(x) \in E[x]$, and $E = k(r_1)$. Then $q(x) = p(x)/(x - r_1) \in E[x]$ has degree $n-1$, and thus has a splitting field $F$ by induction, whence $q(x)$ splits as $(x - r_2)\ldots (x-r_n)$ in $F[x]$, and then $p(x) = (x - r_1)(x - r_2) \ldots (x - r_n)$ in $F[x]$. Finally we have the generation condition: $F = E(r_2, \ldots, r_n) = k(r_1)(r_2, \ldots, r_n) = k(r_1, r_2, \ldots, r_n).$

What is not clear from this construction is that splitting fields are unique (up to isomorphism). This is the case however:

###### Theorem

Splitting fields of a polynomial are unique up to (non-unique) isomorphism.

###### Proof

(After Conrad.) Suppose $E = k(r_1, \ldots, r_n)$ and $E' = k(r_1', \ldots, r_n')$ are two splitting fields of a polynomial $p \in k[x]$. The tensor product $E \otimes_k E'$ is a coproduct of $E$ and $E'$ in the category of commutative algebras over $k$; let $i: E \to E \otimes_k E'$ and $i': E' \to E \otimes_k E'$ be the coproduct injections. Observe that as a $k$-algebra, $E \otimes_k E'$ is generated by the $i(r_j)$ together with the $i'(r_j')$.

Now let $m$ be any maximal ideal of $E \otimes_k E'$ (note that $E \otimes_k E'$ is finite-dimensional over $k$, so the existence of $m$ does not involve any choice principles), and consider the composite

$\phi = (E \stackrel{i}{\to} E \otimes_k E' \stackrel{quot}{\to} E \otimes_k E'/m).$

As is the case for any homomorphism between fields, this composite $\phi$ is an injective map. Moreover, letting $\rho_j$ denote the coset $i(r_j) + m$ and $\rho_j'$ the coset $i'(r_j') + m$, the polynomial $p(x)$ splits over the field $E \otimes_k E'/m$ as

$\prod_j (x - \rho_j) = \prod_k (x - \rho_k')$

and so by unique factorization of polynomials, for each $\rho_k'$ there is a $\rho_j = \phi(r_j)$ with $\rho_k' = \rho_j$. This means the $\phi(r_j)$ exhaust the generators $\rho_j, \rho_k'$ of $E \otimes_k E'/m$. In other words, $\phi$ is also surjective and provides an isomorphism $E \cong E \otimes_k E'/m$. By symmetry we also have an isomorphism $E' \cong E \otimes_k E'/m$, and thus $E \cong E'$.

### Splitting fields of sets of polynomials

A splitting field for a finite set of polynomials $\{p_1, \ldots, p_k\}$ is just a splitting field of the product $p_1 p_2 \ldots p_k$.

Now let $S \subseteq k[x]$ be an arbitrary set of monic polynomials. For each $p \in S$, let $n_p$ denote the degree of $p$, and let us introduce a set of indeterminates $X_p = \{x_{p, 1}, \ldots, x_{p, n_p}\}$. Let $X$ be the coproduct $\sum_{p \in S} X_p$, and form the polynomial algebra $R = k[X]$. For each $p \in S$ let us write

$p(t) - \prod_{i = 1}^{n_p} (t - x_{p, i}) = \sum_{j = 0}^{n_p - 1} s_{p, j} t^j$

for some elements $s_{p, j} \in R$. We claim that the set $\Sigma$ of all $s_{p, j}$ with $p$ ranging over $S$ generates a proper ideal in $R$. For given an $R$-linear combination $g = g_1 s_1 + \ldots + g_k s_k$ with $s_i \in \Sigma$, where $s_i$ is associated with a polynomial $p_i$, let $E$ be the splitting field of $p = p_1 \ldots p_k$, and define a homomorphism $k[X] \to E$ that sends distinct elements of the form $x = x_{p_i, j}$ to distinct roots $r_j$ of $p_i$, and otherwise sends $x = x_{f, j}$ with $f \notin \{p_1, \ldots, p_k\}$ to $0$. Then $g$ is sent to $0$ in $E$, hence $g$ cannot be $1$.

Let $\mathfrak{p}$ be a prime ideal containing the ideal generated by $\Sigma$; the existence of such $\mathfrak{p}$ is guaranteed under the ultrafilter principle. The ring $R/\mathfrak{p}$ is an integral domain; by construction every $p \in S$ splits over $R/\mathfrak{p}$ and the roots of such $p$ generate $R/\mathfrak{p}$ as a $k$-algebra. These roots thus generate its field of fractions $F$ as a field and $F$ is therefore a splitting field for $S$.

In particular, taking $S$ to be the set of all monic polynomials in $k[x]$, this gives an efficient construction of the algebraic closure of $k$ that invokes not the full strength of the axiom of choice (in the form of Zorn's lemma), but only of the weaker ultrafilter principle.

###### Theorem

Any two splitting fields of a given set $S$ of polynomials are isomorphic.

Again we prove this only under the assumption of the ultrafilter principle, and not under the assumption of full axiom of choice.

###### Proof

(After Caicedo.) Let $E, F$ be splitting fields for $S$; without loss of generality we may suppose $S$ is closed under multiplication of polynomials. For $p \in S$ let $E_p, F_p$ denote the corresponding subfields obtained by adjoining all roots of $p$ that occur in $E, F$ to the ground field $k$. Any isomorphism $E \to F$ restricts to an isomorphism $E_p \to F_p$. Notice that the $E_p$ form a directed system of fields, with an inclusion morphism $E_p \to E_q$ if $p$ divides $q$, and $E$ is the colimit of the $E_p$ over this directed system. We are required to show that the set of isomorphisms $Iso(E, F)$ is nonempty; we have restriction maps $Iso(E, F) \to Iso(E_p, F_p)$ which exhibit $Iso(E, F)$ as the inverse limit of $Iso(E_p, F_p)$ over the corresponding system, with transition maps $Res_{p q}: Iso(E_q, F_q) \to Iso(E_p, F_p)$ for $p|q$ also given by restriction. So we are required to show that this inverse limit is inhabited.

By Theorem , the set of isomorphisms $Iso(E_p, F_p)$ is inhabited, and it is also finite (being a torsor of the group $Aut(E_p)$ which has cardinality at most $n!$ where $n$ is the number of roots). It can then be proven from the ultrafilter theorem that $\Phi = \prod_{p \in S} Iso(E_p, F_p)$ is nonempty; see here for a proof. Similarly, the subset

$\Phi_{p q} \coloneqq \{\phi = (\phi_p)_{p \in S} \in \Phi: Res_{p q}(\phi_q) = \phi_p\}$

is also nonempty, as is any finite intersection of such subsets $\Phi_{p q}$ (which actually is another $\Phi_{p' q'}$, by considering the product of $p$‘s and $q$’s).

Now $\Phi$ topologized as a product of finite discrete spaces is a compact Hausdorff space; this version of the Tychonoff theorem also follows from the ultrafilter theorem. We conclude from compactness that the full intersection $\bigcap_{p|q} \Phi_{p q}$ is inhabited. But this intersection is just the inverse limit which is $Iso(E, F)$, which concludes the proof.

An alternative proof based on a similar pattern of reasoning, but more explicitly in the language of model theory and the compactness theorem, was given by Joel David Hamkins, in an answer following Caicedo’s at MathOverflow.

• Bernhard Banaschewski, Algebraic closure without choice, Mathematical Logic Quarterly, Vol. 38 Issue 1 (1992), 383–385. (doi: 10.1002/malq.19920380136, paywall)
• Keith Conrad, Isomorphisms of Splitting Fields, Galois theory notes (pdf).
• Andres Caicedo (http://mathoverflow.net/users/6085/andres-caicedo), Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?, URL (version: 2010-11-19): http://mathoverflow.net/q/46568
• Joel David Hamkins (http://mathoverflow.net/users/1946/joel-david-hamkins), Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?, URL (version: 2010-11-20): http://mathoverflow.net/q/46729