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
the polynomial factors or “splits” as a product of linear factors (possibly repeated):
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 $S$.
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:
Splitting fields of a polynomial are unique up to (non-unique) isomorphism.
(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
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
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 $E \cong E'$.
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
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.
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.
(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
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 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, 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
Last revised on October 25, 2023 at 12:15:31. See the history of this page for a list of all contributions to it.