Pure mathematics is, in its way, the poetry of logical ideas, Albert Einstein
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.
Examples.
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. a^{2} = e, so σ(a^{2}) = (σ(a))^{2} = σ(e) = 1 ⇒ σ(a) = ± 1.
Definition. The characters σ_{1},σ_{2}, ···, σ_{n} are called dependent if there exist elements a_{1}, a_{2}, ···, a_{n}, not all zero in F such that a_{1}σ_{1} + a_{2}σ_{2} + ··· + a_{n}σ_{n} ≡ 0. Such a dependence relation is called non-trivial. The characters σ_{1},σ_{2}, ···, σ_{n} are called independent if a_{1}σ_{1} + a_{2}σ_{2} + ··· + a_{n}σ_{n} ≡ 0 (≡ denotes equality as a function from G → F^{x}) for some a_{1}, a_{2},..., a_{n} ∈ F ⇒ a_{1} = a_{2} = ... = a_{n} = 0. In other words, ∀g ∈ G: (a_{1}σ_{1} + a_{2}σ_{2} + ··· + a_{n}σ_{n})(g) = a_{1}σ_{1}(g) + a_{2}σ_{2}(g) + ··· + a_{n}σ_{n}(g) = 0 ⇒ a_{1} = a_{2} = … = a_{n} = 0
Theorem. 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.
Proof.
Let’s induct on n. Case base, n = 1, ∀g ∈ G, a_{1}σ_{1}(g) = 0 ⇒ [G is a group, let denote the identity 1 in both the group and field, 1 ∈ G, σ_{1}(1)=1] a_{1}σ_{1}(1) = 0 ⇒ a_{1} = 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 σ_{1}, σ_{2}, ···, σ_{n} such that a_{1}σ_{1} + a_{2}σ_{2} + ··· + a_{n}σ_{n} ≡ 0
If a_{n} = 0 ⇒ a_{1}σ_{1} + a_{2}σ_{2} + ··· + a_{n-1}σ_{n-1} ≡ 0 ⇒ [By our induction hypothesis] a_{1} = a_{2} = … = 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.
a_{1}σ_{1} + a_{2}σ_{2} + ··· + a_{n-1}σ_{n-1} + σ_{n} ≡ 0, that is, ∀g ∈ G: a_{1}σ_{1}(g) + a_{2}σ_{2}(g) + ··· + a_{n-1}σ_{n-1}(g) + σ_{n}(g) = 0 (I).
By assumption, σ_{i} are mutually distinct characters. In particular, σ_{1} ≠ σ_{n} ⇒ ∃ α ∈ G: σ_{1}(α) ≠ σ_{n}(α).
[Applying (I) to gα] a_{1}σ_{1}(gα) + a_{2}σ_{2}(gα) + ··· + a_{n-1}σ_{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] a_{1}σ_{n}(α)^{-1}σ_{1}(α)σ_{1}(g)+…+a_{n-1}σ_{n}(α)^{-1}σ_{n-1}(α)σ_{n-1}(g) + σ_{n}(g) = 0 (II)
Next, we compute I - II: [a_{1}-a_{1}σ_{n}(α)^{-1}σ_{1}(α)]σ_{1}(g) + ··· [a_{n-1}-a_{n-1}σ_{n}(α)^{-1}σ_{n-1}(α)]σ_{n-1}(g) = 0
Notice that the first coefficient, a_{1}-a_{1}σ_{n}(α)^{-1}σ_{1}(α) ≠ 0. Otherwise, a_{1} = a_{1}σ_{n}(α)^{-1}σ_{1}(α) ⇒ [a_{1} ≠ 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 a_{1}] 1 = σ_{n}(α)^{-1}σ_{1}(α) ⇒ [*σ_{n}(α)] σ_{n}(α) = σ_{1}(α) ⊥.
By relabelling or renaming our coefficients, we can write this equation as follows,
b_{1}σ_{1} + b_{2}σ_{2} + ··· + b_{n-1}σ_{n-1} ≡ 0, b_{1} ≠ 0 ⊥ (induction hypothesis, we have already assumed that no set of less than n distinct characters are independent). Therefore, a_{n} = 0 ⇒ a_{1} = a_{2} = … = a_{n-1} = 0.
Independence of field homomorphisms. Let K, L be two fields, let σ_{1}, σ_{2}, ···, σ_{n}: K → L be distinct field homomorphisms (σ_{i}≠σ_{j}, ∀i, j: i≠j). If a_{1}σ_{1}(α) + a_{1}σ_{1}(α)+ ··· + a_{n}σ_{n}(α) = 0 ∀α ∈ K, then σ_{1}, σ_{2}, ···, σ_{n} are independent. Where independent means there is no non-trivial dependence a_{1}σ_{1}(α) + a_{1}σ_{1}(α)+ ··· + a_{n}σ_{n}(α) = 0 which holds for every α ∈ K, that is, a_{1} = a_{2} = ··· = 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} ⇒ σ_{1}, σ_{2}, ···, σ_{n} are mutually distinct characters of K^{x} in L^{x} and the previous theorem applies to conclude that σ_{1}, σ_{2}, ···, σ_{n} are independent.∎
Let K, L be fields and let σ_{1},···, σ_{n}: K → L be a collection of fields homomorphisms. An element a ∈ K is called a fixed point of K under σ_{1},···, σ_{n} or just fixed by σ_{1},···, σ_{n} if σ_{1}(a) = σ_{2}(a) = ··· = σ_{n}(a).
Lemma. The subset consisting of elements of K that are fixed by σ_{1}, ···, σ_{n} is a subfield of K. We shall called this subfield the fixed field or the fixed field of σ_{1}, σ_{2}, ···, σ_{n}. F = {a ∈ K | σ_{1}(a) = ··· = σ_{n}(a)} = K^{{σ1, σ2, ···, σn}} = K^{S} ⊆ K where S = {σ_{1}, σ_{2}, ···, 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.∎
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.
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 x^{3}-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, σ_{1} defined by σ_{1}$(\sqrt[3]{2}) = \sqrt[3]{2}w$, and σ_{2}$(\sqrt[3]{2}) = \sqrt[3]{2}w^2$
Let S = {id, σ_{1}, σ_{2}}, K^{S} = {α ∈ K: id(α) = σ_{1}(α) = σ_{2}(α) = α} = ℚ.
∀α ∈ K, $α = a + b\sqrt[3]{2}+ c\sqrt[3]{2}^2$ → σ_{1}(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 x^{3} -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 x^{2}+1=0) and the image of $\sqrt{2}→±\sqrt{2}$ (a root of x^{2}-2 = 0).
Let G = {1, σ_{1}, σ_{2}, σ_{3}}, and G is indeed a group under composition, e.g., (closure under composition) σ_{1}σ_{2}=σ_{3}, $σ_1∘σ_2(i)=σ_1(i)=-i=σ_3(i), σ_1∘σ_2(\sqrt{2})=σ_1(-\sqrt{2})=-\sqrt{2}=σ_3(\sqrt{2})$.
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| σ_{1}(α) = σ_{2}(α) = σ_{3}(α) = α}
σ_{1}(α) = σ_{2}(α) ⇒ b = -b, c = -c ⇒ b= 0, c = 0, σ_{3}(α) = σ_{1}(α) ⇒ d = -d ⇒ d = 0 ⇒ K^{G} = ℚ ∎
Futhermore, K^{{1, σ1}} = $ℚ(\sqrt{2})$, K^{{1, σ2}} = ℚ(i), K^{{1, σ3}} = $ℚ(i\sqrt{2})$.
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^{p}β^{p} = σ(α)σ(β)
σ, σ^{2}, σ^{3}, ···, σ^{r-1} are all distinct homomorphism: K → K, and σ^{r} = id. G = {1, σ, σ^{2}, ···, 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 a^{i}=a^{j} ↭ i=j. If a has finite order n, then ⟨a⟩ = {e, a, a^{2},..., a^{n-1}} and a^{i}=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, σ^{2}, ···, σ^{r-1}, id} are distinct elements.