# Group Characters. Fixed Fields

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

# Group Characters

Recall A set F with addition and multiplication operations is a field if the following three conditions holds: (i) F is an Abelian group under addition. (ii) F \ {0} is an Abelian group under multiplication. (iii) The distributive law holds: a (b + c) = ab + ac. ℚ, ℝ, ℂ, and ℤp (p prime) are fields, but ℕ, ℤ, and ℤp (composite) are not fields.

Let G be a group, and F be a field. Fx = 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 → Fx. 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 Fx 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.

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 → ℚx, 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 → ℂx, σ(e) = 1, σ(a) = w, σ(a2) = w2; and σ2: G → ℂx, σ(e) = 1, σ(a) = w2, σ(a2) = w.

Definition. The characters σ12, ···, σn are called dependent if there exist elements a1, a2, ···, an, not all zero in F such that a1σ1 + a2σ2 + ··· + anσn ≡ 0. Such a dependence relation is called non-trivial. The characters σ12, ···, σn are called independent if a1σ1 + a2σ2 + ··· + anσn ≡ 0 (≡ denotes equality as a function from G → Fx) for some a1, a2,..., an ∈ F ⇒ a1 = a2 = ... = an = 0. In other words, ∀g ∈ G: (a1σ1 + a2σ2 + ··· + anσn)(g) = a1σ1(g) + a2σ2(g) + ··· + anσn(g) = 0 ⇒ a1 = a2 = … = an = 0

Theorem. If G is a group and σ1, σ2, ···, σn are n mutually distinct characters of G in a field F, then σ1, σ2, ···, σn 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, ···, σn such that a1σ1 + a2σ2 + ··· + anσn ≡ 0

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

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

To keep our notation sweet and simple, we keep the same symbols but they are not really the same.

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

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

[Applying (I) to gα] a1σ1(gα) + a2σ2(gα) + ··· + an-1σn-1(gα) + σn(gα) = 0 ⇒ [*σn(α)-1, ∀i, σi: G → Fx, σn(α)≠0 by definition of a character, so it has an inverse] a1σn(α)-1σ1(α)σ1(g)+…+an-1σn(α)-1σn-1(α)σn-1(g) + σn(g) = 0 (II)

Next, we compute I - II: [a1-a1σn(α)-1σ1(α)]σ1(g) + ··· [an-1-an-1σn(α)-1σn-1(α)]σn-1(g) = 0

Notice that the first coefficient, a1-a1σn(α)-1σ1(α) ≠ 0. Otherwise, a1 = a1σn(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 = σn(α)-1σ1(α) ⇒ [*σn(α)] σn(α) = σ1(α) ⊥.

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

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

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

Proof.

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

# Fixed fields

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

Lemma. A group homomorphism preserves inverses: σ(x)σ(x-1) =[σ is a group homomorphism] σ(x·x-1) = σ(e) = e. Similarly, σ(x-1)σ(x) = e. Every element x ∈ G, group, has exactly one inverse ⇒ σ(x-1) = σ(x)-1

Lemma. The subset consisting of elements of K that are fixed by σ1, ···, σn is a subfield of K. We shall called this subfield the fixed field or the fixed field of σ1, σ2, ···, σn. F = {a ∈ K | σ1(a) = ··· = σn(a)} = K1, σ2, ···, σn} = KS ⊆ K where S = {σ1, σ2, ···, σn}

Proof.

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

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

• σi(a-1) = [Lemma. A group homomorphism preserves inverses] (σi(a))-1 = [σi(a) = σj(a)] (σj(a))-1 = σj(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}, KS = {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) σ1σ23, $σ_1∘σ_2(i)=σ_1(i)=-i=σ_3(i), σ_1∘σ_2(\sqrt{2})=σ_1(-\sqrt{2})=-\sqrt{2}=σ_3(\sqrt{2})$.

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

KG = {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 ⇒ KG = 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 = pr. This is the only finite field of order pr up to isomorphism.

Consider the mapping σ: K → K called the Frobenius homomorphism defined by σ(α) = αp ∀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 = αpβp = σ(α)σ(β)

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

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

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

The multiplicative group of a finite field is cyclic.

Therefore, $\mathbb{F_q}^*$ is a cyclic group with pr -1 elements ⇒ [Let G be a cyclic group of order n, G = ⟨a⟩. Then ak = e iff (if and only if) n divides k. Let G be a group, a ∈ G. If a has infinite order, then ai=aj ↭ i=j. If a has finite order n, then ⟨a⟩ = {e, a, a2,..., an-1} and ai=aj ↭ 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, ···, σr-1, id} are distinct elements.

# Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
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3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
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9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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