Group Characters. Fixed Fields

Pure mathematics is, in its way, the poetry of logical ideas, Albert Einstein

Group Characters

Let G be a group, and F be a field. F^x = F \ {0} is the set of non-zero elements of F. It is indeed a group under multiplication.

Definition. A multiplicative character, linear character, or simply character of a group G in a field F is a function or group homomorphism from G to the multiplicative group of a field, σ: G → F^x. By homomorphism is meant a mapping σ such that ∀a, b ∈ G: σ(a)·σ(β) = σ(α·β) where the group operation on the left side is that of F^x and on the right side is that of G.

An alternative definition is as follows. A character of a group G in a field F is a group homomorphism from G into G, such that ∀a, b ∈ G: σ(a)·σ(β) = σ(α·β) and σ(a) ≠ 0. There is no an element a ∈ G such that σ(a) ≠ 0. Otherwise σ(a·b) = σ(a)·σ(b) = 0 ∀b∈ G.

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Examples.

• Let G be a group of order two, G = {e, a}, F = ℚ. There are only two characters, namely the identity, and σ: G → ℚ^x, σ(e) = 1, σ(a) = -1. There are only two characters of G in ℚ and two characters of G in ℂ.

  Notice that σ(e) = σ(e·e) = σ(e)σ(e), you can cancel one of the σ(e) in the multiplicative group of the field, so σ(e) = 1. a² = e, so σ(a²) = (σ(a))² = σ(e) = 1 ⇒ σ(a) = ± 1.

• Let G be a cyclic group of order three, G = {e, a, a²}, F = ℚ. There is only one character of G in F, because a³ = e ⇒ σ(a³) = (σ(a))³ = σ(e) = 1 ⇒ σ(a)³ = 1, but the only rational that has this property is σ(a) = 1, and therefore σ = identity, the trivial character.
• Let G be a cyclic group of order three, G = {e, a, a²}, F = ℂ, there are three characters: id; σ: G → ℂ^x, σ(e) = 1, σ(a) = w, σ(a²) = w²; and σ²: G → ℂ^x, σ(e) = 1, σ(a) = w², σ(a²) = w.

Definition. The characters σ₁,σ₂, ···, σ_n are called dependent if there exist elements a₁, a₂, ···, a_n, not all zero in F such that a₁σ₁ + a₂σ₂ + ··· + a_nσ_n ≡ 0. Such a dependence relation is called non-trivial. The characters σ₁,σ₂, ···, σ_n are called independent if a₁σ₁ + a₂σ₂ + ··· + a_nσ_n ≡ 0 (≡ denotes equality as a function from G → F^x) for some a₁, a₂,..., a_n ∈ F ⇒ a₁ = a₂ = ... = a_n = 0. In other words, ∀g ∈ G: (a₁σ₁ + a₂σ₂ + ··· + a_nσ_n)(g) = a₁σ₁(g) + a₂σ₂(g) + ··· + a_nσ_n(g) = 0 ⇒ a₁ = a₂ = … = a_n = 0

Theorem. If G is a group and σ₁, σ₂, ···, σ_n are n mutually distinct characters of G in a field F, then σ₁, σ₂, ···, σ_n are independent.

Proof.

Let’s induct on n. Case base, n = 1, ∀g ∈ G, a₁σ₁(g) = 0 ⇒ [G is a group, let denote the identity 1 in both the group and field, 1 ∈ G, σ₁(1)=1] a₁σ₁(1) = 0 ⇒ a₁ = 0, so one character cannot be dependent.

Suppose the independence statement holds for n-1, that is, no set of less than n distinct characters are independent. Let σ₁, σ₂, ···, σ_n such that a₁σ₁ + a₂σ₂ + ··· + a_nσ_n ≡ 0

If a_n = 0 ⇒ a₁σ₁ + a₂σ₂ + ··· + a_n-1σ_n-1 ≡ 0 ⇒ [By our induction hypothesis] a₁ = a₂ = … = a_n-1 = 0.

Therefore, we can assume a_n ≠ 0. Then, by dividing by a_n, we can take for granted that a_n = 1.

a₁σ₁ + a₂σ₂ + ··· + a_n-1σ_n-1 + σ_n ≡ 0, that is, ∀g ∈ G: a₁σ₁(g) + a₂σ₂(g) + ··· + a_n-1σ_n-1(g) + σ_n(g) = 0 (I).

By assumption, σ_i are mutually distinct characters. In particular, σ₁ ≠ σ_n ⇒ ∃ α ∈ G: σ₁(α) ≠ σ_n(α).

[Applying (I) to gα] a₁σ₁(gα) + a₂σ₂(gα) + ··· + a_n-1σ_n-1(gα) + σ_n(gα) = 0 ⇒ [*σ_n(α)^-1, ∀i, σ_i: G → F^x, σ_n(α)≠0 by definition of a character, so it has an inverse] a₁σ_n(α)^-1σ₁(α)σ₁(g)+…+a_n-1σ_n(α)^-1σ_n-1(α)σ_n-1(g) + σ_n(g) = 0 (II)

Next, we compute I - II: [a₁-a₁σ_n(α)^-1σ₁(α)]σ₁(g) + ··· [a_n-1-a_n-1σ_n(α)^-1σ_n-1(α)]σ_n-1(g) = 0

Notice that the first coefficient, a₁-a₁σ_n(α)^-1σ₁(α) ≠ 0. Otherwise, a₁ = a₁σ_n(α)^-1σ₁(α) ⇒ [a₁ ≠ 0, otherwise we have n-1 mutually distinct characters so by our induction hypothesis, they are independent and we are done, by Groups’ law cancellation on a₁] 1 = σ_n(α)^-1σ₁(α) ⇒ [*σ_n(α)] σ_n(α) = σ₁(α) ⊥.

By relabelling or renaming our coefficients, we can write this equation as follows,

b₁σ₁ + b₂σ₂ + ··· + b_n-1σ_n-1 ≡ 0, b₁ ≠ 0 ⊥ (induction hypothesis, we have already assumed that no set of less than n distinct characters are independent). Therefore, a_n = 0 ⇒ a₁ = a₂ = … = a_n-1 = 0.

Independence of field homomorphisms. Let K, L be two fields, let σ₁, σ₂, ···, σ_n: K → L be distinct field homomorphisms (σ_i≠σ_j, ∀i, j: i≠j). If a₁σ₁(α) + a₁σ₁(α)+ ··· + a_nσ_n(α) = 0 ∀α ∈ K, then σ₁, σ₂, ···, σ_n are independent. Where independent means there is no non-trivial dependence a₁σ₁(α) + a₁σ₁(α)+ ··· + a_nσ_n(α) = 0 which holds for every α ∈ K, that is, a₁ = a₂ = ··· = a_n = 0.

Proof.

Each σ_i is a group homomorphism so σ_i(0) = 0, so we need to remove the identity element and by doing so, our hypothesis holds, and then we have, σ_i: K^x → L^x ⇒ σ₁, σ₂, ···, σ_n are mutually distinct characters of K^x in L^x and the previous theorem applies to conclude that σ₁, σ₂, ···, σ_n are independent.∎

Fixed fields

Let K, L be fields and let σ₁,···, σ_n: K → L be a collection of fields homomorphisms. An element a ∈ K is called a fixed point of K under σ₁,···, σ_n or just fixed by σ₁,···, σ_n if σ₁(a) = σ₂(a) = ··· = σ_n(a).

Lemma. The subset consisting of elements of K that are fixed by σ₁, ···, σ_n is a subfield of K. We shall called this subfield the fixed field or the fixed field of σ₁, σ₂, ···, σ_n. F = {a ∈ K | σ₁(a) = ··· = σ_n(a)} = K^{{σ₁, σ₂, ···, σ_n}} = K^S ⊆ K where S = {σ₁, σ₂, ···, σ_n}

Proof.

• ∀a, b ∈ F, σ_i(a + b) = [σ_i is a field homomorphism] σ_i(a) + σ_i(b) = [a, b ∈ F] σ_j(a) + σ_j(b) = [σ_j is a field homomorphism] σ_j(a + b)

• σ_i(a · b) = σ_i(a) · σ_i(b) = σ_j(a) · σ_j(b) = σ_j(a · b). Therefore, both the sum and product of two fixed points, a + b, a·b are fixed points.

• σ_i(a^-1) = [Lemma. A group homomorphism preserves inverses] (σ_i(a))^-1 = [σ_i(a) = σ_j(a)] (σ_j(a))^-1 = σ_j(a^-1) ⇒ a^-1 is a fixed point, -a are both inverses under different operations.∎

Examples

Recall. Let F be a field, let K and L be two extension fields of F. Suppose α ∈ K is algebraic over F. Let σ: K → L be an F-homomorphism of fields. Then, σ(α) is algebraic over F, too, and its irreducible polynomial over F is the same as the irreducible polynomial of α over F.

• K = $\mathbb{Q}(\sqrt[3]{2})$, L = ℂ, $\sqrt[3]{2}$ is the real cube root of 2, and w a primitive third root of unity.

If σ: K → L is a field homomorphism, then σ($\sqrt[3]{2}$) must be $\sqrt[3]{2}, \sqrt[3]{2}w,~ or~ \sqrt[3]{2}w^2$. $σ(\sqrt[3]{2})$ has the same irreducible polynomial over ℚ as $\sqrt[3]{2}$, which is x³-2. Futhermore, to define a field homomorphism, σ: K = $\mathbb{Q}(\sqrt[3]{2})$ → ℂ we only need to specify the image of $\sqrt[3]{2}.$

That’s because {1, $\sqrt[3]{2}$, and $\sqrt[3]{2}^2$} is a basis of $\mathbb{Q}(\sqrt[3]{2})$ as a ℚ-vector space, so every element of K can be written as $a + b\sqrt[3]{2}+ c\sqrt[3]{2}^2$ → $a + bσ(\sqrt[3]{2}) + cσ(\sqrt[3]{2})^2$

Therefore, there are only three possible homomorphisms K → L, namely the identity, σ₁ defined by σ₁$(\sqrt[3]{2}) = \sqrt[3]{2}w$, and σ₂$(\sqrt[3]{2}) = \sqrt[3]{2}w^2$

Let S = {id, σ₁, σ₂}, K^S = {α ∈ K: id(α) = σ₁(α) = σ₂(α) = α} = ℚ.

∀α ∈ K, $α = a + b\sqrt[3]{2}+ c\sqrt[3]{2}^2$ → σ₁(α) = $a + bσ_1(\sqrt[3]{2}) + cσ_1(\sqrt[3]{2})^2 = a + b(\sqrt[3]{2})w + c(\sqrt[3]{2})^2w^2, σ_2(α) = a + bσ_2(\sqrt[3]{2}) + cσ_2(\sqrt[3]{2})^2 = a + b(\sqrt[3]{2})w^2 + c(\sqrt[3]{2})^2w.⇒~ σ_1(α)=σ_2(α)⇒b(\sqrt[3]{2})=c(\sqrt[3]{2})^2 ⇒ ~ α = σ_1(α)⇒b(\sqrt[3]{2})= - c(\sqrt[3]{2})^2⇒ c(\sqrt[3]{2})^2 = - c(\sqrt[3]{2})^2$ ⇒ c = 0 ⇒ b = 0.

• Let K = L = $\mathbb{Q}(\sqrt[3]{2})⊆\mathbb{R}$, there is only one homomorphism σ: K → L (=K), namely the identity because $\sqrt[3]{2}$ is the only (real) root of x³ -2 in K ⊆ ℝ.

• Let K = ℚ(i, $\sqrt{2}$). A basis of K over ℚ is {1, i, $\sqrt{2}, -\sqrt{2}$} ⇒ There are only 4 homomorphisms, the image of i needs to be ±i (a root of x²+1=0) and the image of $\sqrt{2}→±\sqrt{2}$ (a root of x²-2 = 0).

1. identity, $i → i, \sqrt{2}→\sqrt{2}$
2. σ₁, $i → -i, \sqrt{2}→\sqrt{2}$
3. σ₂, $i → i, \sqrt{2}→-\sqrt{2}$
4. σ₃, $i → -i, \sqrt{2}→-\sqrt{2}$

Let G = {1, σ₁, σ₂, σ₃}, and G is indeed a group under composition, e.g., (closure under composition) σ₁σ₂=σ₃, $σ_1∘σ_2(i)=σ_1(i)=-i=σ_3(i), σ_1∘σ_2(\sqrt{2})=σ_1(-\sqrt{2})=-\sqrt{2}=σ_3(\sqrt{2})$.

Claim: K^G = ℚ. ∀α ∈ K, α = $a + bi + c\sqrt{2} + di\sqrt{2}, σ_1(α) = a -bi + c\sqrt{2} - di\sqrt{2}, σ_2(α) = a +bi - c\sqrt{2} - di\sqrt{2}, σ_3(α) = a -bi - c\sqrt{2} + di\sqrt{2}$.

K^G = {α ∈ K| σ₁(α) = σ₂(α) = σ₃(α) = α}

σ₁(α) = σ₂(α) ⇒ b = -b, c = -c ⇒ b= 0, c = 0, σ₃(α) = σ₁(α) ⇒ d = -d ⇒ d = 0 ⇒ K^G = ℚ ∎

Futhermore, K^{{1, σ₁}} = $ℚ(\sqrt{2})$, K^{{1, σ₂}} = ℚ(i), K^{{1, σ₃}} = $ℚ(i\sqrt{2})$.

• Let p be a prime, consider the case where the ground field is the prime field $\mathbb{F}_p$. Let K = $\mathbb{F}_q$ be the finite field of q elements, where q = p^r. This is the only finite field of order p^r up to isomorphism.

Consider the mapping σ: K → K called the Frobenius homomorphism defined by σ(α) = α^p ∀α ∈ K, σ is indeed a field homomorphism.

σ(α + β) = (α + β)^p = $α^p+{p \choose 1}α^{p-1}β + {p \choose 2}α^{p-2}β^2 + ··· + pαβ^{p-1} + β^p = α^p + β^p$ and that’s because all the intermediate terms are zero.

σ(α·β) = (αβ)^p = α^pβ^p = σ(α)σ(β)

σ, σ², σ³, ···, σ^r-1 are all distinct homomorphism: K → K, and σ^r = id. G = {1, σ, σ², ···, σ^r-1}, and $K^G = \mathbb{F_p}$

Since $\mathbb{F_p}^*$ has p - 1 elements ⇒ ∀a ∈ $\mathbb{F_p}$ satisfies a^p-1 = 1 (the identity trivially satisfies it, too) ⇒ σ(a) = a^p = a ⇒ $\mathbb{F_p}⊆ \mathbb{K}^G$

Futhermore, σ is injective since a^p = 0 ↭ a = 0. Since we have a finite field $\mathbb{F_q}$, an injective map is also surjective, so σ is bijective.

Therefore, $\mathbb{F_q}^*$ is a cyclic group with p^r -1 elements ⇒ [Let G be a cyclic group of order n, G = ⟨a⟩. Then a^k = e iff (if and only if) n divides k. Let G be a group, a ∈ G. If a has infinite order, then aⁱ=a^j ↭ i=j. If a has finite order n, then ⟨a⟩ = {e, a, a²,..., a^n-1} and aⁱ=a^j ↭ n|(i-j),] a ∈ $\mathbb{F_q}$, $a^p, a^{p^2},···,a^{p^r}$ are all distinct elements, which is really just to say that {σ, σ², ···, σ^r-1, id} are distinct elements.

Bibliography

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).