nLab splitting field




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

k[x]k kk[x]i1E kk[x]E[x],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)=(xr 1)(xr 2)(xr n)p(x) = (x - r_1)(x - r_2)\ldots (x - r_n)

with all the roots r ir_i lying in EE, and the smallest subfield of EE containing kk and these roots is EE itself (so that the roots generate EE: E=k(r 1,,r n)E = k(r_1, \ldots, r_n).

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


The existence of a splitting field for a single polynomial pk[x]p \in k[x] can be proven by induction on n=deg(p)n = \deg(p). As k[x]k[x] is a unique factorization domain, pp is a product of irreducible polynomials p 1p 2p kp_1 p_2 \ldots p_k. The ideal (p 1)(p_1) is maximal and so E=k[x]/(p 1)E = k[x]/(p_1) is an finite field extension of kk where the coset r 1x+(p 1)Er_1 \coloneqq x + (p_1) \in E is a root of p 1(x)E[x]p_1(x) \in E[x], and E=k(r 1)E = k(r_1). Then q(x)=p(x)/(xr 1)E[x]q(x) = p(x)/(x - r_1) \in E[x] has degree n1n-1, and thus has a splitting field FF by induction, whence q(x)q(x) splits as (xr 2)(xr n)(x - r_2)\ldots (x-r_n) in F[x]F[x], and then p(x)=(xr 1)(xr 2)(xr n)p(x) = (x - r_1)(x - r_2) \ldots (x - r_n) in F[x]F[x]. Finally we have the generation condition: F=E(r 2,,r n)=k(r 1)(r 2,,r n)=k(r 1,r 2,,r n).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,,r n)E = k(r_1, \ldots, r_n) and E=k(r 1,,r n)E' = k(r_1', \ldots, r_n') are two splitting fields of a polynomial pk[x]p \in k[x]. The tensor product E kEE \otimes_k E' is a coproduct of EE and EE' in the category of commutative algebras over kk; let i:EE kEi: E \to E \otimes_k E' and i:EE kEi': E' \to E \otimes_k E' be the coproduct injections. Observe that as a kk-algebra, E kEE \otimes_k E' is generated by the i(r j)i(r_j) together with the i(r j)i'(r_j').

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

ϕ=(EiE kEquotE kE/m).\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 ϕ\phi is an injective map. Moreover, letting ρ j\rho_j denote the coset i(r j)+mi(r_j) + m and ρ j\rho_j' the coset i(r j)+mi'(r_j') + m, the polynomial p(x)p(x) splits over the field E kE/mE \otimes_k E'/m as

j(xρ j)= k(xρ k)\prod_j (x - \rho_j) = \prod_k (x - \rho_k')

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

Splitting fields of sets of polynomials

A splitting field for a finite set of polynomials {p 1,,p k}\{p_1, \ldots, p_k\} is just a splitting field of the product p 1p 2p kp_1 p_2 \ldots p_k.

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

p(t) i=1 n p(tx p,i)= j=0 n p1s p,jt jp(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,jRs_{p, j} \in R. We claim that the set Σ\Sigma of all s p,js_{p, j} with pp ranging over SS generates a proper ideal in RR. For given an RR-linear combination g=g 1s 1++g ks kg = g_1 s_1 + \ldots + g_k s_k with s iΣs_i \in \Sigma, where s is_i is associated with a polynomial p ip_i, let EE be the splitting field of p=p 1p kp = p_1 \ldots p_k, and define a homomorphism k[X]Ek[X] \to E that sends distinct elements of the form x=x p i,jx = x_{p_i, j} to distinct roots r jr_j of p ip_i, and otherwise sends x=x f,jx = x_{f, j} with f{p 1,,p k}f \notin \{p_1, \ldots, p_k\} to 00. Then gg is sent to 00 in EE, hence gg cannot be 11.

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

In particular, taking SS to be the set of all monic polynomials in k[x]k[x], this gives an efficient construction of the algebraic closure of kk 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 SS 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,FE, F be splitting fields for SS; without loss of generality we may suppose SS is closed under multiplication of polynomials. For pSp \in S let E p,F pE_p, F_p denote the corresponding subfields obtained by adjoining all roots of pp that occur in E,FE, F to the ground field kk. Any isomorphism EFE \to F restricts to an isomorphism E pF pE_p \to F_p. Notice that the E pE_p form a directed system of fields, with an inclusion morphism E pE qE_p \to E_q if pp divides qq, and EE is the colimit of the E pE_p over this directed system. We are required to show that the set of isomorphisms Iso(E,F)Iso(E, F) is nonempty; we have restriction maps Iso(E,F)Iso(E p,F p)Iso(E, F) \to Iso(E_p, F_p) which exhibit Iso(E,F)Iso(E, F) as the inverse limit of Iso(E p,F p)Iso(E_p, F_p) over the corresponding system, with transition maps Res pq:Iso(E q,F q)Iso(E p,F p)Res_{p q}: Iso(E_q, F_q) \to Iso(E_p, F_p) for p|qp|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)Iso(E_p, F_p) is inhabited, and it is also finite (being a torsor of the group Aut(E p)Aut(E_p) which has cardinality at most n!n! where nn is the number of roots). It can then be proven from the ultrafilter theorem that Φ= pSIso(E p,F p)\Phi = \prod_{p \in S} Iso(E_p, F_p) is nonempty; see here for a proof. Similarly, the subset

Φ pq{ϕ=(ϕ p) pSΦ:Res pq(ϕ q)=ϕ p}\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 Φ pq\Phi_{p q} (which actually is another Φ pq\Phi_{p' q'}, by considering the product of pp‘s and qq’s).

Now Φ\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 p|qΦ pq\bigcap_{p|q} \Phi_{p q} is inhabited. But this intersection is just the inverse limit which is Iso(E,F)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):

  • 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):

Last revised on October 25, 2023 at 12:15:31. See the history of this page for a list of all contributions to it.