‘Obvious’ is the most dangerous word in mathematics, Eric Temple Bell

**Degree of the fixed field extension is at least the number of homomorphisms**. Let K, L be two fields. If σ_{1},···, σ_{n}: K → L are mutually distinct field homomorphisms of K into L and let F be the fixed field of K, F = {a ∈ K| σ_{1}(a) = ··· = σ_{n}(a)}. Then, the degree of the fixed field extension [K : F] is at least the number of homomorphisms: [K : F] ≥ n.

The previous inequality can be strict, e.g., K = L = $\mathbb{Q}(\sqrt{2}, i)$, σ_{1}: K → K, σ_{1}: i → -i, $\sqrt{2}→\sqrt{2}$, σ_{2}: K → K, σ_{2}: i → i, $\sqrt{2}→-\sqrt{2}$. ℚ ⊆ K^{{id, σ1, σ3}} ⊆ K, and we know that [K : ℚ] = 4 and [K : K^{{id, σ1, σ2}}] ≥ 3 (there are three distinct field homomorphisms), but [K : K^{{id, σ1, σ2}}] divides 4, so it needs to be 4 and K^{{id, σ1, σ3}} = ℚ, therefore [K : F] = 4 > 3.

Recall. If G is a group and σ_{1}, σ_{2}, ···, σ_{n} are n mutually distinct characters of G in a field F, then σ_{1}, σ_{2}, ···, σ_{n} are independent. σ_{1}, σ_{2}, ···, σ_{n} are independent as characters of K^{x} in L^{x}.

Proof.

Let r = [K : F]. Let’s suppose for the sake of contradiction r < n, and take a basis α_{1}, α_{2}, ···, α_{r} of K as an F-vector space.

Consider the following homogeneous system of linear equations with coefficients in L:

$ \begin{cases} σ_1(α_1)x_1 + σ_2(α_1)x_2 + ··· + σ_n(α_1)x_n = 0 \\ σ_1(α_2)x_1 + σ_2(α_2)x_2 + ··· + σ_n(α_2)x_n = 0 \\ …………………………………… \\ σ_1(α_r)x_1 + σ_2(α_r)x_2 + ··· + σ_n(α_r)x_n = 0 \end{cases}$

Or, equivalently, Ax = 0,

$\left( \begin{smallmatrix}σ_1(α_1) & σ_2(α_1)& ··· & σ_n(α_1) \\ σ_1(α_2) & σ_2(α_2)&··· & σ_n(α_2)\\ ··· & ···&··· & ···\\ σ_1(α_r) & σ_2(α_r)&··· & σ_n(α_r)\end{smallmatrix} \right)\left( \begin{smallmatrix}x_1 \\ x_2\\···\\ x_n\end{smallmatrix} \right)=\left( \begin{smallmatrix}0\\ 0\\···\\ 0\end{smallmatrix} \right)$

where A = $\left( \begin{smallmatrix}σ_1(α_1) & σ_2(α_1)&··· & σ_n(α_1) \\ σ_1(α_2) & σ_2(α_2)&··· & σ_n(α_2)\\ ··· & ···&··· & ···\\ σ_1(α_r) & σ_2(α_r)&··· & σ_n(α_r)\end{smallmatrix} \right)$ is a r x n matrix with entries in L, so there are r equations and n variables ⇒ Since **r < n, this homogeneous system has a nontrivial solution**, say β_{1}, ··· β_{n} ∈ L where at least one β_{i} ≠ 0.

$\begin{cases} σ_1(α_1)β_1 + σ_2(α_1)β_2 + ··· σ_n(α_1)β_n = 0 \\ σ_1(α_2)β_1 + σ_2(α_2)β_2 + ··· σ_n(α_2)β_n = 0 \\ ··· \\ σ_1(α_r)β_1 + σ_2(α_2)β_2 + ··· σ_r(α_r)β_n = 0 \end{cases}$, so there are n equations

We claim that $β_1σ_1(α) + β_2σ_2(α) + ··· β_nσ_n(α) = 0, ∀α∈\mathbb{K}$. If this statement holds true, then $β_1σ_1 + β_2σ_2 + ··· + β_nσ_n≡0$. At least one β_{i} ≠ 0 (**r < n, this homogeneous system has a nontrivial solution**), so this claim violates the independence of σ_{1},···, σ_{n} ⊥

∀α ∈ K ⇒ [α_{1}, α_{2}, ···, α_{r} is a basis of K as an F-vector space] α can be written uniquely as α = a_{1}α_{1} + a_{2}α_{2} + ··· + a_{r}α_{r}, where a_{i} ∈ F.

$σ_1(α_1)β_1 + ··· σ_n(α_1)β_n = 0$ ⇒ [Let’s multiply the first equation by σ_{1}(a_{1})] $σ_1(a_1)σ_1(α_1)β_1 + σ_1(a_1)σ_2(α_1)β_2 + ··· + σ_1(a_1)σ_n(α_1)β_n = 0$ ⇒ [a_{1} ∈ F = {a ∈ K| σ_{1}(a) = ··· = σ_{n}(a)} so σ_{1}(a_{1}) = σ_{2}(a_{1})··· = σ_{n}(a_{1})] $σ_1(a_1α_1)β_1 + σ_2(a_1α_1)β_2 + ··· + σ_n(a_1α_1)β_n = 0$

Notice that $σ_1(a_1)σ_2(α_1)β_2$ = [a_{1} ∈ F] $σ_2(a_1)σ_2(α_1)β_2$ = [σ_{2} is an homomorphism] $σ_2(a_1α_1)β_2$

In a similar fashion, take the second equation… $σ_1(α_2)β_1 + ··· σ_n(α_2)β_n = 0$ ⇒ [… and multiply it by σ_{1}(a_{2})] $σ_1(a_2)σ_1(α_2)β_1 + σ_1(a_2)σ_2(α_2)β_2 + ··· + σ_1(a_2)σ_n(α_2)β_n = 0$ ⇒ [a_{2} ∈ F = {a ∈ K| σ_{1}(a) = ··· = σ_{n}(a)} so σ_{1}(a_{2}) = ··· = σ_{n}(a_{2})] $σ_1(a_2α_2)β_1 + σ_2(a_2α_2)β_2 + ··· + σ_n(a_2α_2)β_n = 0$

And so on, and so forth, we multiply the last rth equation by σ_{1}(a_{r}) and use the fact that a_{r} ∈ F, σ_{1}(a_{r}) = ··· = σ_{n}(a_{r}) ⇒ $σ_1(a_rα_r)β_1 + σ_2(a_rα_r)β_2 + ··· + σ_n(a_rα_r)β_n = 0$

To sum up,

$ \begin{cases} σ_1(a_1α_1)β_1 + σ_2(a_1α_1)β_1 + ··· + σ_n(a_1α_1)β_n = 0 \\ σ_1(a_2α_2)β_1 + σ_2(a_2α_2)β_2 + ··· + σ_n(a_2α_2)β_n = 0 \\ … \\ σ_1(a_rα_r)β_1 + σ_2(a_rα_r)β_2 + ··· + σ_n(a_rα_r)β_n = 0 \end{cases}$

Let’s add all these equations and take into consideration that σ_{i} are homomorphisms, we get, $σ_1(a_1α_1 + a_2α_2+···+a_rα_r)β_1 + σ_2(a_1α_1 + a_2α_2+···+a_rα_r)β_2 + ··· + σ_n(a_1α_1 + a_2α_2+···+a_rα_r)β_n = 0 ⇒ σ_1(α)β_1 + σ_2(α)β_2 + ··· + σ_n(α)β_n = 0$ ∎

Lemma. Let K/F be a field extension. The set of all F-isomorphism K → K or automorphisms (field isomorphism from F to itself) which leave F fixed (or fix F point-wise) forms a group under composition.

Proof.

Gal(K/F) = {σ: K → K | σ is an automorphism (field homomorphism, bijection) of K such that σ fixes K point-wise}

Gal(K/F) ⊆ Aut(K) = $\mathbb{K}$, i.e., the Galois group is a set of the group of automorphisms.

By the **Two-Step Subgroup Test** we only need to prove that:

- Gal(K/F) ≠ ∅, and that is the case because id ∈ Gal(K/F). It is obviously an automorphism K→ K that fixes F point-wise.
- It is closed under composition of mappings. ∀σ, τ ∈ Gal(K/F), σ∘τ: K → K, a composition of isomorphisms is an isomorphism, too, such that ∀a ∈ F, σ∘τ(a) = σ(τ(a)) = [τ ∈ Gal(K/F), τ fixes F] σ(a) = [σ ∈ Gal(K/F), σ fixes F] a.
- ∀σ ∈ Gal(K/F), σ
^{-1}: K → K is an automorphism, too. Futhermore, ∀a ∈ F, σ(a) = a ⇒ σ^{-1}(a) = a ⇒ σ^{-1}∈ Gal(K/F).

Definition. The Galois group of the extension K/F, denoted by Gal(K/F), is the group of all F-automorphisms of K, the group of all automorphisms that fix or leave alone the base field, Gal(K/F) = {σ ∈ Aut(K)| σ(a) = a ∀a ∈ F}.

- The complex conjugation, σ: ℂ → ℂ, defined by a + bi → a - bi is an
**automorphism**of the complex numbers. - Let consider the fields $\mathbb{Q} ⊆ \mathbb{Q}(\sqrt{5})⊆\mathbb{Q}(\sqrt{3},\sqrt{5})$, $σ(a + b\sqrt{3}) = a -b\sqrt{3}$ is an automorphism of $\mathbb{Q}(\sqrt{3},\sqrt{5})$ leaving $\mathbb{Q}(\sqrt{5})$ fixed. Analogously, $τ(a + b\sqrt{5}) = a -b\sqrt{5}$ is an automorphism of $\mathbb{Q}(\sqrt{3},\sqrt{5})$ leaving $\mathbb{Q}(\sqrt{3})$ fixed. Their composition μ = στ moves both $\sqrt{3},~ and~ \sqrt{5}$. Gal($\mathbb{Q}(\sqrt{3},\sqrt{5})/\mathbb{Q}$) = {id, σ, τ, μ} ≋ ℤ/ℤ
_{2}xℤ/ℤ_{2}∘ id σ τ μ id id σ τ μ σ σ id μ τ τ τ μ id σ μ μ τ σ id - $Gal(\mathbb{Q}(i)/\mathbb{Q})≋\mathbb{Z}/2\mathbb{Z}$.
- $Gal(\mathbb{Q}(i, \sqrt{2})/\mathbb{Q})≋\mathbb{Z}/2\mathbb{Z} x \mathbb{Z}/2\mathbb{Z}$
- $Gal(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})≋$ {e}
- $Gal(\mathbb{F_{p^r}}/\mathbb{F_p})≋\mathbb{Z}/r\mathbb{Z}$

Some authors always refer to Aut(K/F) as the Galois group of K over F (or of the extension), whether or not the extension is a Galois extension. Other books use the generic term automorphism group to refer to Aut(K/F) in the general case, that is, the group of field automorphisms of K that leave each element of F fixed, and reserve the term Galois group **exclusively** for the situation in which K is a Galois extension of F, i.e., finite, separable, and normal.

**Degree of Fixed Fields**. Let K be any field, and let σ_{1}, ···, σ_{n}: K → K be distinct field automorphisms. Suppose that σ_{1}, ···, σ_{n} forms a group under composition. If F is a fixed field of σ_{1}, ···, σ_{n}. Then, [K: F] = n

$\mathbb{K} = \mathbb{Q}(\sqrt{2},i)$, S = {id, σ_{1}, σ_{2}}, we did have [K : K^{S}] = 4 > 3, K^{S} = ℚ. However, **S is not a group.**

Proof.

By the previous theorem, we have already shown that [K : F] ≥ n. Generally, it can happen that [K : F] > n.

Suppose for the sake of contradiction, [K:F] > n. Let’s take α_{1}, α_{2}, ···, α_{n+1} ∈ K which are linear independent over F (there could be more).

Consider the following homogeneous system of linear equations with coefficients in K,

$\left( \begin{smallmatrix}σ_1(α_1) & σ_1(α_2)&··· & σ_1(α_{n+1}) \\ σ_2(α_1) & σ_2(α_2)&··· & σ_2(α_{n+1})\\ ··· & ···&··· & ···\\ σ_n(α_1) & σ_n(α_2)&··· & σ_n(α_{n+1})\end{smallmatrix} \right)\left( \begin{smallmatrix}x_1 \\ x_2\\···\\ x_{n+1}\end{smallmatrix} \right)=\left( \begin{smallmatrix}0\\ 0\\···\\ 0\end{smallmatrix} \right)$

There are n equations and (n+1) variables in this homogeneous system of linear equations ⇒ [#equations < #variables] **∃ a non-trivial solution**. Let us choose a non-trivial solution with **the least number of non-zero coordinates**. Let’s denote as (a_{1}, a_{2}, ···, a_{r}, 0, ···, 0) ∈ K^{n+1}, where a_{1} ≠ 0, a_{2} ≠ 0, ···, a_{r} ≠ 0, and there is non-trivial solution with fewer than r non-zero entries, r ≥ 1.

If r = 1, then the first equation is a_{1}σ_{1}(α_{1}) = 0 ⇒ [α_{1}, α_{2}, ···, α_{n+1} are linear independent over F, and in particular α_{1}≠ 0, σ_{1}(α_{1}) ≠ 0 -homomorphisms map non-zero elements to non-zero elements] a_{1} = 0 ⊥

Therefore, r ≥ 2 and without losing any generosity, since a_{r} ≠ 0, we can multiply by a_{r}^{-1} and assume that a_{r} = 1. The system of equations can be rewritten like this: (we don’t have any terms after the rth term)

(I) = $\begin{cases} a_1σ_1(α_1) + a_2σ_1(α_2) + ··· + a_{r-1}σ_1(α_{r-1}) + σ_1(α_r) = 0 \\ a_1σ_2(α_1) + a_2σ_2(α_2) + ··· + a_{r-1}σ_2(α_{r-1}) + σ_2(α_r) = 0 \\ ··· \\ a_1σ_n(α_1) + a_2σ_n(α_2) + ··· + a_{r-1}σ_n(α_{r-1}) + σ_n(α_r) = 0 \end{cases}$

By assumption, {σ_{1}, σ_{2}, ···, σ_{n}} is a group of automorphisms of K ⇒ One of them is the identity, say σ_{2} = identity. So the second equation of (I) will look like a_{1}α_{1} + a_{2}α_{2} + ··· + a_{r-1}α_{r-1} + α_{r} = 0 ⇒ [α_{1}, α_{2}, ···, α_{r} are linear independent over F] If ∀i, a_{i} ∈ F ⇒ there exist a non-trivial linear relationship among α_{r} ⊥

Therefore, at least one a_{i} is not in F. Without losing any generosity, we could assume a_{1} ∈ K \ F ⇒ [a_{1} is not in the fixed field] ∃k ∈ {1, 2, ··· n}: σ_{k}(a_{1}) ≠ a_{1}.

(I) can be rewritten in a more compact way as:

$a_1σ_j(α_1)+a_2σ_j(α_2) + ··· +a_{r-1}σ_j(α_{r-1}) + σ_j(α_r) =0,~ ∀1 ≤ j ≤ n$. Next, let’s apply σ_{k} to these equations.

$σ_k(a_1)σ_kσ_j(α_1)+σ_k(a_2)σ_kσ_j(α_2) + ··· +σ_k(a_{r-1})σ_kσ_j(α_{r-1}) + σ_kσ_j(α_r) =0,~ ∀1 ≤ j ≤ n$ (II)

By assumption {σ_{1},··· σ_{n}} is a group, then {σ_{k}σ_{1},··· σ_{k}σ_{n}} is just a permutation of the group {σ_{1},··· σ_{n}}.

We can see this because every element σ_{k}σ_{i} is a member of the group by closure, and they are all distinct because if σ_{k}σ_{i} = σ_{k}σ_{j} ⇒ [By cancellation laws] σ_{i} = σ_{j}

In other words, (II) can be rewritten as $σ_k(a_1)σ_i(α_1)+σ_k(a_2)σ_i(α_2) + ··· +σ_k(a_{r-1})σ_i(α_{r-1}) + σ_i(α_r) =0,~ ∀1 ≤ i ≤ n$ (III)

(I) can be rewritten in a compact way: $a_1σ_i(α_1)+a_2σ_i(α_2) + ··· +a_{r-1}σ_i(α_{r-1}) + σ_i(α_r) =0,~ ∀1 ≤ i ≤ n$

(I) - (III): $(a_1-σ_k(a_1))σ_i(α_1)+(a_2-σ_k(a_2))σ_i(α_2) + ··· +(a_{r-1}-σ_k(a_{r-1}))σ_i(α_{r-1}) =0,~ ∀1 ≤ i ≤ n$

We have found a new solution to the original homogeneous system of linear equations, namely $(a_1-σ_k(a_1), a_2-σ_k(a_2),··· , a_{r-1}-σ_k(a_{r-1}), 0, ···, 0)$, so this is a non-trivial solution (a_{1} -σ_{k}(a_{1}) ≠ 0) and it has at most r -1 consecutive non-zero coordinates ⊥ (it contradicts the choice of r).

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.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).