Splitting Field

A splitting field of a polynomial f is a ‘smallest’ field containing all the roots of f. This is only defined over a ‘base field’ F containing the coefficients of f. For example, if we regard x² + 1 as a real polynomial, its splitting field is ℝ(i) = ℂ; but if we regard it as a rational polynomial, its splitting field is the much smaller field ℚ(i) = {a + bi : a, b ∈ ℚ}. Our goal in this section is to show that splitting fields exist and are unique (over a specified base field), up to isomorphism.

If p(x) ∈ F[x] is a polynomial of positive degree n over the field F, there exists an extension field E of F that contains n roots of p(x).

Definition 8.2.1 (Splitting Field) Let E be an extension field of F and let f(x) ∈ F[x] with degree at least 1. We say that f(x) splits in E if there are elements a ∈ F and a₁, a₂, ... ,a_n ∈ E such that

We call E a splitting field for f(x) over F if

E is called the splitting field (or root field) of p(x) because it is the “smallest” field over which p(x) “splits” into first-degree factors, that is there are no proper subfields of E, over which the polynomial splits.  □

Note that a splitting field of a polynomial over a field depends not only on the polynomial but on the field as well. Indeed, a splitting field of p(x) over F is just a smallest extension field of F in which f(x) splits. Just as it makes no sense to say "p(x) is irreducible,” it makes no sense to say "E is a splitting field for p(x).” In each case, the underlying field must be specified; that is, one must say "p(x) is irreducible over F" and "E is a splitting field for p(x) over F."

Example 8.2.2 x² + 1 = (x + √−1) (x − √−1), splits in ℂ. When considering x² + 1 as a polynomial over the field ℝ of real numbers, then ℝ(i), or the field ℂ of complex numbers, is the splitting field for x² + 1 where

However, if x² + 1 is considered a polynomial over the field Q of rational numbers, then the splitting field of x² + 1 is ℚ(i) = {r + si | r, s ∈ ℚ}.  ■

Example 8.2.3. Consider the polynomial p(x) = x² − 5 ∈ ℚ[x]. The field ℚ(√5) is an extension of ℚ, contaning the roots ±√3 of f(x), thus p(x) splits over ℚ(√5) in linear factors

Moreover ℚ(√5) is the smallest field containing a complete set of roots of p(x) and thus ℚ(√5) is a splitting field for the the polynomial x² − 5. ■

Example 8.2.4. If we consider x² + 1 as a polynomial with rational coefficients, then to obtain a splitting field we only need to adjoin i to ℚ. Thus ℚ(i) is a splitting field for x² + 1 over ℚ. If we instead think to it as a polynomial with real coefficients, its splitting field is ℝ(i) = ℂ.  ■

As seen in the last examples it is important to specify over which field the polynomials splits. Further if of a polynomial p(x) ∈ F[x] we know the roots, the splitting field is easily found by adjoining the roots of p(x) to F. In other casese the task can be more difficult for example finding the splitting field of p(x) = x² + x + 1 over ℤ₂[x]. Such a polynomial has no roots in the field ℤ₂. Another instance is a polynomial of large degree n over ℚ, with no rational roots; finding a field containing all roots of such a polynomial can be a daunting task. Our task in this section is to determine the splitting field of a polynomial p(x) ∈ F we do not know explicitly the roots.

We start off by considering a simpler problem: given a polynomial find a field containing a root of the polynomial. Take again the polynomial x² − 5 ∈ ℚ[x]. Consider the ring

The following properties hold

1. K is a field since x² − 5 is irreducibile over ℚ.

2. K is an extension of ℚ; Consider the canonical projection

   Ψ: ℚ[x] ⟶ ℚ[x]/(x² − 5)
   f(x) ⟼ f(x) + (x² − 5)

   The set ℚ̅ = {a + (x² − 5) | a ∈ ℚ} is a field isomorphic to ℚ: indeed let a,b ∈ ℚ, a + (x² − 5) = b + (x² − 5) if and only if a = b. Thus

   K = ℚ[x]/(x² − 5) ⊇ ℚ̅ ≃ ℚ.

3. K contains a root α of x² − 5; let α = to the class x + (x² − 5), then

   α² − 5 = (x + (x² − 5))² − 5 = x² + (x² − 5) − 5
   = x² − 5 + (x² − 5) = (x² − 5)

   which is the null class of ℚ[x]/(x² − 5).

8.2.5. Example. From last example it is clear how to proceed to find a field containing all roots of the polynomial x² + x + 1 ∈ ℤ₂[x]. Being f(x) irreducible over ℤ₂, ℤ₂/(x² + x + 1) is a field. The quotient field ℤ₂[x]/(p(x)) consists of residue classes. By the division algorithm

Since f(x) = q(x)p(x) + r(x) and q(x)p(x) is already in the ideal (p(x)), then

It results, ℤ₂[x]/(x² + x + 1) has four elements, and represents an extension of ℤ₂:

The last equality ℤ₂(ξ) ≃ ℤ₂[x]/(x² + x + 1) by theorem 8.1.16 Theorem (c). Note how 0 + (p(x)) is the zero of the field and same element as x² + x + 1 + (p(x)). The following are also the same elements x² and x + 1 since x² − (x + 1) = x² + x + 1 as 1 = −1 in ℤ₂, so we write x² = x + 1 mod(x² + x + 1).
1 + (p(x)) is the multiplicative identity of the field i.e. the multiplicative inverse of (x + 1) + (p(x)) is x + (p(x)) since (x+1)x = x² + x = 1 mod (p(x)) or (x + 1 + (p(x))) (x + (p(x))) = 1 + (p(x))) ■

We can now consider the general case proving that for every polynomial f(x) ∈ F[x], with F an arbitrary field we are able to find a field containing a root of f(x). Then with an inductive method we shall prove that there exists a field containing all roots of f(x).

Lemma 8.2.6. Let p(x) ∈ F[x] an irreducible polynomial of degree n ≥ 1. Then there exists an extension E of F with [E:F] = n in which p(x) has a root.

Proof. We proceed analogously to what we have done in the example above. The field we are looking for is

Indeed it is a field because (p(x)) is maximal being p(x) irreducible (see theorem 1.6.10); It is also an extension of F since the homomorphism Ψ

it results that F[x]/(p(x)) contains a F̅ = {a + (p(x)) | a ∈ F} which is isomorphic to F. It is further a finite extension of degree n of F, since one of its basis is

Ultimately, the class α = x + (p(x)) is a root of p(x), since p(x) = xⁿ + a_{n − 1} + ... + a₁x + a₀, then

Lemma 8.2.7. Let f(x) ∈ F[x] an arbitrary polynomial of degree n ≥ 1. Then there exists a finite extension E of F in which f(x) has a root. We have further

Proof. Let p(x) an irriducible factor of f(x). By the previous lemma there exists an extension E of F in which p(x) has a root (hence f(x) as well has a root), and [E : F] = ∂p(x) ≤ n.  □

Example 8.2.8. Let f(x) = 2x⁴ − 2x² −4, it results f(x) = (2x² − 4)(x² + 1). Both factors of f are irreducible over ℚ. Let's consider 2x² − 4. We just showed that the field E = ℚ[x]/(2x² − 4) has a root of 2x² − 4, hence it possesses a root of f(x) as well. It results [E : ℚ] = 2 < 4 = ∂f(x). □

Recalling the definition of algebrically closed field, the properties expressed by the following theorem hold

Theorem 8.2.9. Let K a field. The following propositions are equivalent

1. K is algebrically closed.

2. every non-constant polynomial f(x) ∈ K[x] can be factorized in terms of linear factors in K[x].

3. every non-constant polynomial f(x) ∈ K[x] has a root in K.

Proof. (a) ⇒ (b). By induction on the degree n of f. If n = 1, f is linear and there's nothing to prove. Let now n > 1 and suppose the theorem true for all polynomial of degree n < 1. Adjoining to K a root α of f, we obtain the extension K(α) = K, that is α ∈ K. Then there exists a g(x) ∈ K[x] such that f(x) = (x − α)g(x). The result follows by induction.

(b) ⇒ (c) is obvious.

(c) ⇒ (a). Suppose by contradiction that there exists a proper algebraic extension E of F: let α ∈ E \ F an element (obviously algebraic over K). Then α satisfies an irreducible polynomial of degree > 1 in F[x]: but this contradicts that by hyphothesis such a polynomial must have a root in F (thus it cannot be irreducible). □

Theorem 8.2.10. Let f(x) ∈ F[x] a polynomial of degree n ≥ 1. Then there exists an extension E of F of degree ≤ n! where f(x) has all the roots.

Proof. We proceed by induction on the degree n of the polynomial. For n = 1 the polynomial has all roots (i.e. only one) in F and thus E = F. Suppose to have proved the theorem for all polynomials of degree n − 1, and we prove it for polynomials of degree n. By the previous lemma, there exists an extension E₀ of F, with [E₀ : F] ≤ n, where f(x) has a root α. The in E₀[x] we have the following factorization:

By induction there exists a finite extension E of E₀ with [E : E₀] ≤ (n − 1)!, in which g(x) has n − 1 roots. Since the roots of f(x) are either the root α or the roots of g(x) it follows that the E is an extension of F which contains all roots of f(x).  □

Once that is proved that there exist a finite extension of F containg all roots of f(x), there eixits for sure a minimal field extension with the same property. Thus the theorem just proved guarantees the existence of a splitting field for every polynoial f(x) ∈ F[x]. The same theorem guarantees that the degree [E: F] has the upper bound n!, with n the degree of the polynomial.

Theorem 8.2.11. All splitting fields of a polynomial f(x) ∈ F[x] are isomorphic.

Before proving the theorem we need some additional definitions.

8.2.12 Definition. Let φ an isomorphism between two fields F and F'. If R and R' are two rings (or two fields) such that R ⊇ F and R' ⊇ F', then a isomorphism ψ between R and R' is called an extension of the isomorphism between F and F' if ψ|_F coincides with φ.

Let now F and F' two isomorphic fields by an isomorphism τ. It is possible to extend such an isomorphism τ to an isomorphism τ̅ between F[x] and F'[x] by letting

If f(x) = ∑ⁿ_i=1 a_ixⁱ, then τ̅̅(f(x)) = ∑ⁿ_i=1τ(a_i)xⁱ.

Theorem 8.2.13. Let τ a isomorphism between two fields F and F' and let f(x) and f'(x) two polynomials such that f'(x) = τ̅̅(f(x)), with τ̅̅ the isomorphism defined above. Let E a splitting field for f(x) and E' a splitting field for f'(x). Then the isomorphism τ between F and F' can be extended to a isomorphism Ψ between E and E'.

Proof. We proceed by induction on k, the number of roots of f(x) not belonging to F. If k = 0, then all roots of f(x) are in F, hence f(x) is splitted in F into linear factors and E = F. By the isomorphism τ̅̅, f(x) = τ̅̅f)(x) is splitted as well in F', hence E' = F' and Ψ = τ.
Suppose now to have proved the theorem for every polynomial with at last k − 1 roots not in F, then we have to prove it for the case of f(x) having k roots not in F. By factorizing f(x) in irreducible factors on F:

and let

the corresponding (through τ̅̅) factorization of f'(x). At least on of the two factors p_i(x) (let it p₁(x)) must have degree strictly greater than 1, otherwise all roots would in F and k = 0, a case which has already been proved. Let α a root of p₁(x). Then f'(x) = τ̅̅f((x)) as well, has an irreducible factor, p'(x), of degree strictly greater than 1, and let α' a root of p'(x). Let n = deg(p₁(x)), the mapping

is an isomorphism (verify it). On the other hand the following isomprhisms hold

and

By the transitivity property of the isomorphism, it follows that

The last isomorphism that we call Ψ is an extension of τ. Let's sart now by the field F(α) and F'(α') = Ψ(F(α)). Then f(x) has at last k − 1 roots not in F'(a'). Thus Ψ, by the inductive hyphotesis can be externded to an isomorphism Ψ̅̅ between E and E'. This isomorphim Ψ̅̅ is clearly as well an extension of the isomorphism τ.  □

Corollary 8.2.14. All splitting fields of a polynomial f(x) ∈ F[x] are isomorphic to an isomorphism that fixes F.

Proof. In the previous theorem let F' = F and τ is the identity mapping. If E and E' are two splitting fields of f(x) then E ≃ E' by the previous theorem.  □

Corollary 8.2.15. Let p(x) ∈ F[x] an irreducible polynomial and let α and β two roots of p(x). Then

by an isomorphism that maps α to β leaving fixed the remaining elements of F.

Proof. The isomoprhisms of Theorem 8.2.13 become

For example the three roots of the irreducible polynomial x³ − 2 over ℚ are ∛2, ∛2ζ, ∛2ζ², with ζ primitive root of unity. It results

Every isomorpshim fixes the elements of ℚ and maps a root to another root. Note that the hyphotesis that p(x) is essential and also that all the three extensions do not represent a splitting field for x³ − 2, which instead is ℚ(∛2, ζ) and has degree 6: looking at the subfield tower (Theorem 8.1.4) we compute their degrees [ℚ(∛2, ζ) : ℚ] = [ℚ(∛2, ζ) : ℚ(∛2)] ⋅ [ℚ(∛2) : ℚ], the last term is 3 by Theorem 8.1.17 point d), whilst the first is 2.

Proposition 8.2.16. For every positive integer n ∈ ℕ, the splitting field of xⁿ − 1 over ℚ, is ℚ(ζ) with ζ an arbritary primitive n-th root of unity. It results [ℚ(ζ): ℚ] = φ(n), with φ(n) Euler function of n.

Proof. Let ζ be a primitive root of unity, so it satisfies xⁿ − 1 = 0 for some n, hence it is algebraic over ℚ. All the n-th roots of units are powers of a primitice root of unity, thus to obtain the splitting field it suffices to adjoin to ℚ an arbitrary primitive n-th root of unity. Further it results [ℚ(ζ): ℚ] = φ(n) since the minimal polynomial of ζ is the n-th polynomial, Φ_n(x), which is irreducible over ℚ and has degree φ(n).  □