10 Splitting Fields. 2. The splitting field for x 3 2 over Q is Q( 3 2,ω), where ω is a primitive third root of 1 in C. Thus, since ω = 1+ 3

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10 Splitting Fields

We have seen how to construct a ﬁeld K ⊇ F such that K contains a root α of a given (irreducible) polynomial p(x)∈ F [x], namely K ∼= F [x]/(p(x)).

We can extendthe procedure to build a ﬁeld that contains all roots of the given polynomial.

Definition 10.1. The extension K/F is a splitting field for the polynomial f (x) ∈ F [x] if f(x) splits completely (i.e. factors into linear factors) over K, but does not factor completely over any proper subfield of K containing F . We will show later that any two splitting fields for a fixed polynomial are isomorphic, so in fact we may talk about the splitting field for f (x) over F . Theorem 10.2. Given any f (x) ∈ F [x], there exists an extension K ⊇ F which is a splitting field for f (x) over F .

Proof. By iterating the procedure for ﬁnding an extension containing one root of the polynomial, we can see there exists an extension E of F containing all roots of f (x). Let K be the intersection of all subﬁelds of E containing F and all roots of f (x).

Deﬁnition 10.3. If K/F is an algebraic extension and K is the splitting ﬁeld over F for a collection of polynomials over F , then K is a normal extension of F .

Note: Some people use the term “normal” for what we will call “Galois.”

Notice that if F ⊆ E ⊆ K and K is a splitting ﬁeld over F for some collection of polynomials, then K is also a splitting ﬁeld over E for the same collection.

Examples:

1. The splitting ﬁeld for x² − 2 over Q is Q(√ 2).

2. The splitting ﬁeld for x³−2 over Q is Q(√³

2, ω), where ω is a primitive third root of 1 in C. Thus, since ω = ⁻¹⁺₂^√⁻³, this ﬁeld is Q(√³

2,√

−3).

We have [Q(√³ 2,√

−3) : Q] = 6. The diagram of intermediate ﬁeld extensions is given below. (We don’t yet have the results to prove this is correct.)

3. The ﬁeld Q(√ 2,√

3) is the splitting ﬁeld for (x² − 2)(x² − 3) over Q.

Again the diagram of intermediate extensions is given below.

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Q(√³ 2,√

−3)

Q(√

−3)

Q(√³

2) Q(ω√³

2)

Q(ω²√³ 2)

Q

Q(√ 2,√

3)

Q(√ 3)

Q(√

6) Q(√

2)

Q

How big is the splitting ﬁeld of a polynomial of degree n, relative to the original ﬁeld?

Proposition 10.4. A splitting ﬁeld of a polynomial of degree n over F is of degree ≤ n! over F .

Proof. Let F1 be obtained from F by adjoining one root of f (x) to F . Then if f (x) has degree n, we have [F1 : F ]≤ n. The polynomial f(x) has at least one linear factor over F1, so any other root satisﬁes a polynomial of degree at most n− 1 over F1. The result follows by induction.

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Example: Cyclotomic Fields

The roots of the polynomial xⁿ−1 over Q are the n^throots of 1, e^2πkiⁿ , 0≤ k ≤ n−1. The subset of n^th roots of 1 form a ﬁnite subgroup of the multiplicative group of any ﬁeld that contains them, thus it is a cyclic group.

Definition 10.5. A generator of the cyclic group of all n^th roots of 1 is a primitive n^th root of unity. If ζn is a primitive n^th root of 1, then the others are given by ζ_nâ where 1 ≤ a < n is an integer relatively prime to n. All n^th roots of unity are powers of ζ_n, so the splitting field of xⁿ− 1 over Q is Q(ζn). This is called the n^th cyclotomic field. (The word “cyclotomic”means

“circle-cutting”.)

Note that if n = p is prime, then x^p− 1 = (x − 1)(x^p⁻¹+· · · + x + 1) is a factorization into irreducible polynomials over Q, so [Q(ζp) : Q] = p− 1.

In general, we will show that [Q(ζn) : Q] = φ(n) where φ denotes the Euler φ-function.

We want to show that any two splitting ﬁelds of the same polynomial over F are isomorphic. We need the following result.

Theorem 10.6. Let φ : F → F be an isomorphism of ﬁelds, f (x) ∈ F [x]

a polynomial,and f(x) the image of f (x) under the natural extension of φ to an isomorphism F [x] → F[x]. Let E be a splitting ﬁeld for f (x) over F , and E a splitting ﬁeld for f(x) over F. Then φ extends to an isomorphism σ : E → E.

Proof. Induct on the degree n of f (x). If f (x) splits completely over F , then f(x) splits completely over F, so take E = F, E = F, σ = φ. Thus it is true for n = 1, and more generally when all irreducible factors of f (x) are of degree 1. Assume the result is true whenever deg f (x) < n. Let deg f (x) = n and let p(x) be an irreducible factor of f (x) of degree ≥ 2. Let p(x) be the corresponding factor of f(x). Let α be a root of p(x) and β a root of p(x).

Then φ extends to an isomorphism σ : F (α) → F(β) as we have already seen. Write f (x) = (x− α)f1(x), f(x) = (x− β)f1. Then E is a splitting ﬁeld for f1(x) over F (α), and E is a splitting ﬁeld for f₁(x) over F (β).

The degrees of f1(x) and f₁(x) are n − 1, so by induction there exists an isomorphism σ : E → E extending σ : F (α) → F (β), and this σ is the desired extension of φ.

Corollary 10.7. Any two splitting ﬁelds for f (x) ∈ F [x] over F are iso- morphic.

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Proof. Take φ to be the identity, F = F, and E, E the two splitting ﬁelds for f (x) in the theorem above.

Next we look at the notion of an algebraic closure, when one adds all roots of all polynomials over a ﬁeld.

Definition 10.8. The field ¯F is an algebraic closure of F if ¯F is algebraic over F and if every polynomial f (x) ∈ F [x] splits completely over ¯F . A field K is algebraically closed if every polynomial with coefficients in K has a root in K. An algebraically closed field is thus its own algebraic closure. (Indeed, K = ¯K if and only if K is algebraicaly closed.)

Note that it is not a priori clear that a given ﬁeld F has an algebraic closure.

Proposition 10.9. Let ¯F be an algebraic closure of F . Then ¯F is alge- braically closed.

Proof. Let f (x)∈ ¯F [x], α a root of f (x). Then ¯F (α) is algebraic over ¯F , and F is algebraic over F , so ¯¯ F (α) is algebraic over F . Thus α is algebraic over F , and hence α∈ ¯F . Therefore ¯F is algebraically closed.

To show that every field F has an algebraic closure, we first construct an algebraically closed field K containing F , which may be much bigger than an algebraic closure of F , and then take the subfield of elements that are algebraic over F . This will be ¯F .

Lemma 10.10. Let K be a ﬁeld. Then there exists an algebraic extension K1 of K such that every polynomial in K[x] has a root in K1.

Proof. It suffices to construct a field K1 such that every monic irreducible polynomial of degree > 1 in K[x] has a root in K1. For each such monic irreducible f (x), define a variable xf, and let S be the set of all such xf. Substituting xf for x, we have the polynomial f (xf) ∈ K[S]. Let A denote the ideal of K[S] generated by {f(xf) | xf ∈ S}. We claim A is a proper ideal of K[S]. For if not, 1 ∈ A, and so there exist monic polynomials f1, . . . , fr of degree > 1 in K[x] and polynomials g1, . . . , gr ∈ K[S], such that 1 = g1f1(xf1) +· · · + grfr(xfr). Now there exists a (finite) extension L of K in which each of f1(x), . . . , fr(x) has a root; say αi ∈ L is a root of fi(x), 1≤ i≤ r. Then substituting αi for xi, we have 1 = g1(α1, . . . , αr)f1(α1) +· · · + gr(α1, . . . , αr)fr(α) = 0, a contradiction.

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Thus A is a proper ideal of K[S] and there exists a maximal ideal M of K[S] with M ⊇ A. Let K1 = K[S]/M. Let βf denote the image in K1 of xf under the map π : K[s] → K1. Then f (βf) = π(f (xf)) = 0, so each monic irreducible polynomial f (x) ∈ K[x] of degree > 1 has a root in K1

as claimed. Each βf is algebraic over K1, and clearly K → K1. We have K1 = K[βf | xf ∈ S] is an algebraic extension of K.

Proposition 10.11. Let K be a ﬁeld. Then K has an algebraic closure.

Proof. Using the previous lemma, we can construct a sequence K = K0 ⊆ K1 ⊆ · · · ⊆ Kn⊆ . . . of ﬁeld extensions such that for each i ≥ 0, Ki+1/Kiis algebraic, and each polynomial in Ki[x] has a root in Ki+1. Then Ki/K is algebraic for all i. Usning set theory, one can construct a set in which the Ki all embed compatibly with their embeddings in each other. (For example, one can take the direct limit lim_→Ki. Let L = ∩iKi in this set.

Each element of L lies in some Ki, and for any finite set of elements in L, we can find a Ki containing all of them. Therefore L is a field. Each element of L is algebraic over K since it lies in some KI algebraic over K.

If f (x) ∈ L[x], then there exists Ki containing all the coeﬃcients of f . In particular, f (x) ∈ Ki[x]⊆ L[x], so f(x) has a root in Ki+1 and hence in L.

Thus L is algebraically closed, and hence is an algebraic closure of K. (We will prove uniqueness later – probably.)

Among algebraic extensions of F we distinguish two types, separable and inseparable.

Deﬁnition 10.12. A polynomial over F is called separable if it has no mul- tiple roots. Otherwise it is called inseparable.

We proved already that a polynomial f (x) has a multiple root α if and only if α is also a root of its formal derivative f(x), if and only if both f (x) and f(x) are divisible by mα,F(x). Thus f (x) is separable if and only if it is relatively prime to its derivative. From this we have the following:

Corollary 10.13. Every irreducible polynomial over a ﬁeld of characteristic 0 (e.g. Q) is separable.

Proof. If p(x)∈ F [x] is irreducible of degree n, then p(x) is of degree n− 1, and the only factors of p(x) are 1 and p(x), so p(x), p(x) are relatively prime.

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Then the question becomes, where does this proof fail in characteristic p? If the derivative of p(x)= 0, then it is relatively prime to p(x), and so we ask when does p(x) = 0.

Proposition 10.14. Let char(F ) = p. Then for all a, b ∈ F , we have (a + b)^p = a^p+ b^p, (ab)^p = a^pb^p, so the map φ : F → F, a → a^p is an injective field homomorphism. If F is finite, this map is an automorphism of F . (Thus every element of F is a p^th power: F = F^p. For F either finite or infinite, the elements fixed by φ are precisely the elements of Z/pZ = Fp, the prime subfield of F .

Proof. The statements are obvious except for the final one. Since φ(x) = x^p = x for any element fixed by φ, we see that φ can only fix the roots of x^p − x, of which there can be at most p. But every element of Fp satisfies this equation, so those are precisely the roots.

Definition 10.15. The map φ is called the Frobenius homomorphism. (It is defined for any commutative ring with 1, of characteristic p.) A field of characteristic p is called perfect if φ : F → F is onto, hence an automorphism of F . In particular, finite fields are perfect. We also say that any field of characteristic 0 is perfect.

Proposition 10.16. Every irreducible polynomial over a perfect ﬁeld is sep- arable.

Proof. It suﬃces to show this for a ﬁeld F of characteristic p such that F^p = F . Let p(x) ∈ F [x] be irreducible. If p(x) were inseparable, then p(x) = q(x^p) for some polynomial q(x) = amx^m+· · · + a1x + a0. Then since F = F^p, we may write ai = b^p_i for some bi ∈ F . We have p(x) = q(x^p) = amx^mp+· · ·+a1x^p+a0 = b^p_mx^mp+· · ·+b^p1x^p+b^p₀ = (bmx^m)^p+· · ·+(b1x)^p+b^p₀ = (bmx^m+· · · + b1x + b0)^p. Thus p(x) is the p^th power of a polynomial in F [x], contradicting irreducibility.

We then ask, what do inseparable polynomials over a ﬁeld F of charac- teristic p look like?

Proposition 10.17. Let p(x) be an irreducible polynomial over a ﬁeld F of characteristic p. Then there exists a unique integer k ≥ 0 and a unique separable polynomial p_sep(x)∈ F [x] such that p(x) = psep(x^p^k).

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Proof. If p(x) is separable, we are done. If p(x) is inseparable, then p(x) = p1(x^p) for some polynomial p1(x). If p1(x) is separable, stop. If not, con- tinue. Eventually we get a uniquely determined power p^k of p such that p(x) = pk(x^p^k), where pk(x) is separable. Also pk(x) is irreducible, since any factorization of pk(x) gives a factorization of p(x).

Deﬁnition 10.18. The degree of psep(x) is the separable degree of p(x), denoted deg_s(p(x)). The integer p^k is the inseparable degree of p(x), denoted deg_i(p(x)). We have deg(p(x)) = deg_s(p(x))· degi(p(x)).

Definition 10.19. The field K is said to be separable over F if every ele- ment of K is the root of a separable polynomial over F . A field which is not separable is inseparable.

Corollary 10.20. Every finite extension of a perfect field is separable. In particular, every finite extension of Q or a finite field is separable.

We will return to the topic of separability later.

Finite Fields and Cyclotomic Fields

We begin with a brief consideration of ﬁnite ﬁelds. We will have much more to say about them once we have the Fundamental Theorem of Galois Theory at our disposal.

Let n > 0, and consider the splitting ﬁeld of x^pⁿ − x over Fp. This polynomial is separable (its derivative is −1 = 0), and so it has pⁿ distinct roots. Let α, β be two roots of this polynomial. Then α^pⁿ = α, β^pⁿ = β.

This then means (α· β)^pⁿ = α^pⁿβ^pⁿ = α· β, (α⁻¹)^pⁿ = α⁻¹, and (α + β)^pⁿ = α^pⁿ + β^pⁿ = α + β, so the set of roots to this polynomial forms a field, and hence must be the splitting field F . Then |F | = pⁿ and [F : Fp] = n. This shows that there exist finite fields of degree n over Fp for every positive integer n.

Now if F is any ﬁnite ﬁeld of characteristic p, say [F : Fp] = n, then

|F | = pⁿ. The multiplicative group ˙F is cyclic of order pⁿ − 1, so every nonzero element satisfies x^pⁿ⁻¹ − 1 = 0, and hence every element satisfies x^pⁿ− x = 0. By cardinality considerations, these must be all roots, so this is the splitting field for x^pⁿ − x over Fp, and hence this field is unique (up to isomorphism).

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Next we turn our attention to the so called cyclotomic ﬁelds. The word cyclotomic comes from roots meaning “circle cutting,” and they are the ﬁelds obtained by adjoining n^th roots of 1 to Q.

Definition 10.21. Let µn denote the group of n^th roots of 1 over Q. Then the multiplicative group µn is isomorphic to the additive group Z/nZ where the isomorphism is given by (ζn)â→ a ∈ Z/nZ, where ζn denotes a primitive n^th root of 1. There are precisely φ(n) primitive n^th roots of 1, corresponding to (ζn)â where (a, n) = 1. (Here φ denotes the Euler phi-function.) Observe that if d| n, then µd⊆ µn, since a d^th root of 1 is also an n^th root of 1.

Deﬁnition 10.22. The n^th cyclotomic polynomial Φn(x) is the polynomial whose roots are the primitive n^th roots of unity:

Φn(x) =

ζ∈µn,ζprimitive

(x− ζ) =

1≤a<n,(a,n)=1

(x− ζn^a),

which is a polynomial of degree φ(n).

Observe that xⁿ− 1 =

ζ∈µn(x− ζ), and if we group the factors (x − ζ) where ζ is a primitive d^th root of 1, d| n, we have

xⁿ− 1 =

d|n

ζ∈µd,ζprimitive

(x− ζ) =

d|n

Φd(x).

(Note that n =

d|nφ(d).) Thus we can compute Φn(x) recursively for any n. This also shows the following.

Lemma 10.23. Φn(x) is a monic polynomial of degree φ(n)∈ Z[x].

Proof. Clearly φ(n) is a monic polynomial of degree φ(n) in Q(x). One can show inductively that the coeﬃcients lie in Z[x]. It is clearly true for n = 1. Assume Φd(x) ∈ Z[x] for all 1 ≤ d < n. Then xⁿ − 1 = (

d|n,d<nΦd(x))(Φn(x)). The ﬁrst factor is a monic polynomial in Z[x], and so by Gauss’s Lemma, Φn(x) is also a monic polynomial in Z[x].

Theorem 10.24. Φn(x) is irreducible over Q.

Proof. Suppose Φn(x) = f (x)g(x), f, g ∈ Z[x], monic. We may assume f(x) is irreducible. Let ζ ∈ µnbe a root of f (x), so f (x) is the minimal polynomial for ζ over Q. Let p be a prime such that p n. Then ζ^p is a primitive n^throot

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of 1, so ζ^p is a root of f or g. Suppose g(ζ^p) = 0. Then ζ is a root of g(x^p), and since f (x) is the minimal polynomial for ζ, we have f (x) | g(x^p), say g(x^p) = f (x)h(x), h∈ Z[x]. Then consider this equation reduced modulo p:

[¯g(x)]^p = ¯g(x^p) = ¯f (x)¯h(x) ∈ Fp[x]. Now Fp[x] is a UFD, so ¯f (x) and ¯g(x) have a common factor in Fp[x]. Then ¯Φn(x) = ¯f (x)¯g(x), so ¯Φn(x) has a multiple root. Then xⁿ−1 has a multiple root over Fp. But we have already seen that xⁿ− 1 has n distinct roots over Fp if p n. This is a contradiction, so ζ^p cannot be a root of g(x), and so ζ^p would have to be a root of f 9x).

This applies to every root ζ of f (x), and so ζ^a is a root of f (x) for every integer a relatively prime to n. Thus every primitive n^th root of 1 is a root of f (x), and hence f (x) = Φn(x) is irreducible.

Corollary 10.25. [Q(ζn) : Q] = φ(n).

We conclude this section with the calculation of some examples.

We have [Q(ζ8) : Q] = 4 = φ(8). But 1 ∈ Q(ζ8), and also ζ8 + ζ₈⁷ =√ 2, so

√2 ∈ Q(ζ8). Since [Q(i,√

2) : Q] = 4, we have equality. Note Φ8(x) = x⁴+ 1 = (x²+i)(x²−i), but this also factors over R as (x²+√

2x+1)(x²−√

2x+1).

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Some other Φn(x):

1. If n = p a prime, Φp(x) = x^p⁻¹+ x^p⁻²+· · · + x + 1.

2. Φ6(x): We have φ(6) = 2, and x⁶ − 1 = (x³ + 1)(x³ − 1) = (x − 1)(x + 1)(x² + x + 1)(x² − x + 1), so Φ6(x) = x² − x + 1). We have ζ6 = ¹⁺^√₂⁻³ =−ζ3.

3. Φ9(x) = x⁶+ x³+ 1

4. Φ10(x) = x⁴− x³+ x²− x + 1, since again ζ10=−ζ5. 5. Φ12(x) = x⁴− x²+ 1. Again, note ζ12 =√

ζ6, since Φ6(x²) = Φ12(x).