# Consequences Galois: Fundamental Theorem of Algebra

Lemma. Every nonnegative real number has a real square root.

Suppose f(x) = x2 -a with a > 0. f(0) < 0, f(u) > 0 for u sufficient large. ⇒ [Intermediate value theorem. If f is a continuous function whose domain contains the interval [a, b], then it takes on any given value between f(a) and f(b) at some point within the interval.] ∃c ∈ [0, u]: f(c) = 0, c = $\sqrt{a} ∈ \mathbb{R}$, i.e., f has a root in ℝ, that is, all complex numbers have real square roots.

Lemma. Every complex number has a complex square root.

Proof.

Let a ∈ ℂ, a = re, r ≥ 0. By the previous lemma, $\sqrt{r}∈ \mathbb{R}$, so b = $\sqrt{r}e^{\frac{iθ}{2}}∈\mathbb{C}$. Futhermore, b2 = $r(e^{\frac{iθ}{2}})^2=re^{iθ}=a$∎

Lemma. There are no quadratic extensions of ℂ.

Proof.

Let K be a degree 2 extension of ℂ ⇒ ∃α ∈ K \ ℂ, K = ℂ(α), degα = 2, let f ∈ ℂ[x] be the irreducible polynomial of α over ℂ, deg(f) = 2 so f(x) = x2 + bx + c, where b, c ∈ ℂ.

Of course, the quadratic formula gives the roots of f, $\frac{-b±\sqrt{b^2-4c}}{2}$, but the lemma (All complex numbers have complex square roots) ⇒ f has root in ℂ (deg(f) = 2, f irreducible) ⊥

Lemma. There are non-trivial finite extensions of ℝ of odd degree.

Proof.

Suppose for the sake of contradiction a field F of odd degree. Let α ∈ F.

[F : ℝ] = [F : ℝ(α)][ℝ(α) : ℝ]. Since [F : ℝ] is odd, then so is its divisor [ℝ(α) : ℝ] ⇒ the minimal irreducible polynomial of α over ℝ has odd degree.

Let p(x) = x2n+1 + ··· + a1x + a0 be the minimal polynomial for α ⇒ $\lim_{x \to ∞} p(x) = ∞, \lim_{x \to -∞} p(x) = -∞$ or viceversa (depending on the sign of the leading coefficient they can swap, -∞ and ∞)⇒ [By the intermediate value theorem] p(x) has a root c ∈ ℝ ⇒ [By the factor theorem] x -c divides p(x). Since p(x) is irreducible and monic (the leading coefficient is equal to one) ⇒ p(x) is linear, p(x) = c -x, so α = c ⇒ α ∈ ℝ.

∀α ∈ F ⇒ α ∈ ℝ ⇒ F ⊆ ℝ ⇒ F = ℝ.

Fundamental Theorem of Algebra. The field ℂ of the complex numbers is algebraically closed or, in other words, every non-constant polynomial in ℂ[x] has a root in ℂ.

Lemma. ℂ is algebraically closed ↭[By definition] Every non-constant polynomial in ℂ[x] has a root in ℂ ↭ If L/ℂ is a finite extension of fields, then L = ℂ.

Proof.

Suppose that every non-constant polynomial in ℂ[x] has a root in ℂ.

Let L/ℂ be a finite extension of fields, α ∈ L, f(x) ∈ ℂ[x] be the irreducible polynomial of α over ℂ, deg(f(x)) ≥ 1 ⇒ [By assumption] f has a root in ℂ ⇒ [f is irreducible] deg(f(x)) = 1 ⇒ α ∈ ℂ ⇒ L = ℂ

Suppose that if L/ℂ is a finite extension of fields, then L = ℂ, and let f ∈ ℂ[x] be a polynomial, deg(f) > 0. Consider L be the splitting field of f over ℂ ⇒ [Being L the splitting field of f over ℂ, then it is generated by the roots of f] Consequently, L over ℂ is a finite extension ⇒ [By assumption] L = ℂ ⇒ f has roots in (its splitting field L=)ℂ

Proof. (Fundamental Theorem of Algebra)

Let L/ℂ be a finite extension of ℂ

Since ℝ has characteristic zero, every finite extension is separable, [ℂ : ℝ] = 2 finite ⇒ ℂ/ℝ is separable. L/ℂ is a finite extension ⇒ L/ℝ is separable, too ⇒ [Galois extension II, Every finite separable field extension K/F can be extended to a Galois extension.] We can take an extension of L/ℝ, say K/ℝ which is Galois.

Let G be the Galois group of K/ℝ. Notice that 2 divides |G| (|G| = [K : ℝ] = [K : ℂ][ℂ : ℝ] = [K : ℂ]2), say |G| = 2em, m is odd, e ≥ 1

[For every prime factor p (= 2) with multiplicity n of the order of a finite group G, ∃ a Sylow p-subgroup of G, of order pn] Let H be a Sylow 2-subgroup of G of order 2e, so |G : H | = m = [KH : ℝ], |H| = 2e = [K : KH].

Since [Lemma. There are non-trivial finite extensions of ℝ of odd degree.], m = [KH : ℝ], m odd ⇒ m = 1, [KH : ℝ] = 1 ⇒ KH = ℝ and |G| = 2e

Futhermore, [K : ℝ] = 2e = [K : ℂ][ℂ : ℝ] = [K : ℂ]·2 ⇒ [K : ℂ] = 2e-1.

If e = 1 ⇒ [K : ℂ] = 2e-1 = 1 ⇒ K = ℂ ⇒ K = L = ℂ ∎

Suppose [K : ℂ] = 2e-1 = |Gal(K/ℂ)|, e > 1, let’s apply Sylow theorem to Gal(K/ℂ) ⇒ Gal(K/ℂ) has a subgroup of order 2i for every 1 ≤ i ≤ e-1. Take a subgroup of order 2e-2 and let M be its fixed field ⇒ there is a quadratic extension of ℂ ⊥ (Lemma. There are no quadratic extensions of ℂ.)

Theorem. Suppose F is a field, char(F) ≠ 2. Let K/F be a degree 2 extension. Then, K/F is Galois and K = F(α) where α2 ∈ F. In other words, a degree 2 extension can be obtained by adding a square root of an element of F.

Counterexample: Consider the splitting field of p(x) = x2 -t ∈ $\mathbb{F_2(t)}$. It can be demonstrated that p(x) = $(x + \sqrt{t})^2$ so the extension is not separable, and not Galois.

Solution:

K/F is a degree 2 extension ⇒ K ≠ F ⇒ ∃α ∈ K \ F. Then, K = F(α). Since [K : F] = 2, the degree(f(x)) = 2 where f(x) ∈ F[x] is the irreducible polynomial of α, say f(x) = x2 + bx + c, where b, c ∈ F.

The roots are $\frac{-b±\sqrt{b^2-4c}}{2}$ (that’s true because char(F) ≠ 2). Let δ be the discriminant of the irreducible polynomial, δ = $\sqrt{b^2-4c}$

Claim: K = F(δ) and δ2 ∈ F.

δ2 = b2 -4 ∈ F, and [α = $\frac{-b±δ}{2} ∈ K ⇒δ ∈ K$] δ ∈ K

Two cases:

1. F(δ) = F ⇒ δ ∈ F ⇒ f has roots in F and f is not irreducible over F ⊥
2. F(δ) = K. Futhermore, K/F is normal because K is a splitting field of a polynomial, namely f, the roots are $\frac{-b+δ}{2},\frac{-b-δ}{2}∈\mathbb{K}$ and f is separable because f’ = 2x + b ≠ 0 (again we are using char(F)≠2) ⇒ [Theorem. If α separable over F then F(α)/F is a separable extension.] K/F separable ⇒ K/F Galois.

Reclaim: if f is an irreducible polynomial such that its derivative f′ is not zero then f has only simple roots, i.e., f is separable.

Theorem. Let K/F be a Galois extension such that G = Gal(K/F) ≋ S3. Then, K is the splitting field of an irreducible cubic polynomial over F.

Solution.

G = Gal(K/F) ≋ S3 ⇒ ∃H ≤ G of order 2 and H is not normal ⇒ [Fundamental theorem of Galois fields] KH/F is not Galois.

Since [KH:F] = 3 (prime), we know ∃α ∈ KH\F: F(α) = KH, and let f(x)∈ F[x] be the irreducible polynomial of α over F. [KH:F] = 3 ⇒ deg(f(x)) = 3.

Claim: K is the splitting field of f over F (f is an irreducible polynomial of degree 3).

K/F Galois extension ⇒ [Generally a Galois extension is defined to be a field extension which is both normal and separable] K/F normal and f ∈ F[x] irreducible has a root in K (α ∈ K) ⇒ [K/F is normal: Every irreducible polynomial p∈F[X] with some root α∈K splits completely in K.] Therefore, f splits completely in K[x].

Let α = α1 = α, α2, α3 ∈ K be the distinct (K/F Galois ⇒ α ∈ K is separable over F, too) roots of f in K.

Next, let’s prove that it is generated by this roots, i.e., K = F(α1, α2, α3).

6 = |S3| = |Gal(K/F)| = because K/F is Galois [K : F]

Therefore, there are two options:

1. K = F(α1, α2, α3)∎ K is the splitting field of a separable polynomial and therefore is Galois.
2. F(α1) = F(α1, α2, α3) ⇒ F(α1) is the splitting field of a separable polynomial and therefore is Galois, but this is not the case (KH/F = F(α)/F is not Galois ↭ H is not normal ⊥)

Proposition. Let L1, L2 be two Galois extensions of a field F, then their intersection L1∩L2 is also Galois over F.

Solution. First notice that L1∩L2 is a field (it is left as an exercise).

L1/F Galois ⇒ L1/F separable ⇒ [A separable extension of a field is one in which every element’s algebraic minimal polynomial does not have multiple roots.] L1∩L2/F is separable.

Let f ∈ F[x] be an irreducible polynomial with a root α in L1∩L2. In particular, f has a root α in L1 and since L1/F is Galois (normal), f splits completely in L1, and obviously f has the same root α in L2, f splits completely in L2, too. Therefore, all the roots of f lie in L1∩L2 or, in other words, f splits completely in L1∩L2, L1∩L2/F is normal ⇒ [L1∩L2/F is separable]L1∩L2/F is Galois ∎