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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.


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.


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.


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.


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

G a l o i s K L | | A 2 f i n i t e e x t e n s i o n

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].

| H | = o 2 d ^ K d e H m / \ K | 2 = K H

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.)

2 ^ e - 1 K | M | 2 2 ^ ( e - 2 ) = | M |

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.


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

2 K F F ( δ ) F [ ( K δ : ) F ] = = K 2 o r = 2 F · ( 1 & ) o r = 1 F · 2

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.


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.

6 K K F ^ H 2 3 = F ( α )

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 2 K F F | F ( ( α α 3 1 1 , ) α 2 , α 3 )

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 ∎


This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory.
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Fields and Galois Theory. Howie, John. M. Springer Undergraduate Mathematics Series.
  5. Fields and Galois Theory. Morandi. P., Springer.
  6. Fields and Galois Theory. By Evan Dummit, 2020.
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