They came up through never ending discussions, painstaking research, in-depth analysis, some tragic assassinations and mysterious disappearances, and trial and error experimentation with abducted human beings involving (for some unknown reason) their genitalia with the Ten Immutable, Absolute, and Transcendent Laws of Physics, Psychology, Philosophy, and Reality… Apocalypse, Anawim, #justtothepoint.

Definition. A ring R is a non-empty set with two binary operations. It is an Abelian group under addition. A ring is associative, but not necessarily commutative under multiplication. When it is, we say that the ring is **commutative**. A ring does not necessarily have an identity under multiplication. When it does have an identity, it is called as **unity** or **identity**. Examples: ℤ, ℤ/ℤ_{n}.

Definition. The characteristic of a ring R is the smallest number of times one must use the ring's multiplicative identity in a sum to get the additive identity, that is the smallest n such that: 1 + ···_{n times} + 1 = 0 if such a number exist, ∞ otherwise.

A finite field is a field that contains a finite number of elements, e.g., $\mathbb{F_p}=ℤ/pℤ$. It is also called as Galöis field. The order is either a prime number or a prime power, i.e., it has the form p^{n}, where p is a prime number and n is a positive integer.

The simplest example of a finite fields are given by the integers mod p when p is a prime number.

Let F be a field. A subset K that is itself a field under the operations of F is called a subfield of F. The field F is called an extension field of K. If K ≠ F, K is called a proper subfield of F.

Definitions. A field is called a prime field if it has no proper subfields. An obvious prime field is the field which is obtained by the intersection of the family of all subfields of a given field. ℚ and ℤ/pℤ where p is a prime number are examples of prime fields.

**$\mathbb{F_p}=ℤ/pℤ$ is a prime field**. Let P denotes any nontrivial subfield of ℤ_{p}. We know that P ⊆ ℤ_{p}and claim that P = ℤ_{p}.

Let’s show that ℤ_{p} ⊆ P. [1] ∈ P [If P is a subfield, then [1] is the multiplicative identity] and ∀[a] ∈ ℤ_{p}, [a] = [1] + ··_{a times}·· + [1] ∈ P ⇒ P = ℤ_{p} ∎

**ℚ is a prime field.**Let P denotes any nontrivial subfield of ℚ = {m/n: m, n ∈ ℤ, n ≠ 0, gcd(m, n) = 1}

1 ∈ P [1 is the multiplicative identity] and P is closed with respect to addition and inverses so 1 + 1 + ··_{m times}·· + 1 = m ∈ P, 1 + 1 + ··_{n times}·· + 1 = n ∈ P ⇒[P is closed with respect to multiplication and **inverses**] m/n ∈ P ⇒ Q ⊆ P∎

Lemma. The characteristic of a finite field have to be a prime number.

Proof.

Let n, m be positive integers and F a finite field. Define the operations: n·1_{F} = 1 + 1 + ···_{n times} + 1, (-n)·1_{F} = -(n·1_{F}), 0·1_{F} = 0.

Suppose char(F) = n = p·q, 1< p, q< n, then

(1 + 1 + ···_{p times} + 1)(1 + 1 + ···_{q times} + 1) =[char(F) = n = p·q] (1 + 1 + ···_{n times} + 1) = 0 ⇒ p · q = 0, i.e., p and q are zero divisors ⇒ p^{-1} and q^{-1} do not exist (Suppose for the sake of contradiction, p^{-1} do exist, p≠0, q≠0, pq=0 ⇒ q = (p^{-1}p)q =[Associative] p^{-1}(pq) = p^{-1}0 = 0 ⇒ q = 0⊥) ⊥ Every nonzero element in a finite field has an inverse.

Theorem. Let F be a finite field of characteristic p. Then |F|= p^{n} for some n ∈ ℕ

Proof.

By our previous lemma, **finite fields have characteristic p > 0 where p is a prime**. Let Φ: ℤ → F be a ring homomorphism defined by Φ(n) = n·1. Since char(F)=p ⇒ Ker(Φ) = pℤ ⇒[By the first isomorphism theorem] ℤ/pℤ ≋ Φ(ℤ) ⊆ F, i.e.,the image of Φ is a subfield of F (say K) isomorphic to ℤ_{p}.

So every field F of characteristic p has an isomorphic copy of ℤ_{p} ⊆ F, so F can be viewed as a finite extension of $\mathbb{F_p} = ℤ/pℤ$, say [F : $\mathbb{F_p}$] = n, or, in other words, as a ℤ_{p} or $\mathbb{F_p}$ vector-space, [$\mathbb{F}:\mathbb{F_p}$]=dim_{$\mathbb{F_p}$}($\mathbb{F}$) = n.

Let {e_{1}, e_{2}, ···, e_{n}} be a basis of F over $\mathbb{F_p}$. Any element x ∈ F can be written uniquely as x = a_{1}e_{1} + a_{2}e_{2}+ ··· + a_{n}e_{n}, a_{i} ∈ $\mathbb{F_p}$. The question is, how many ways can we represent the element “x”? As a_{i} ∈ $\mathbb{F_p}$, a_{i} can be selected in “only” p-ways, thus the total of ways in which x is selected is p^{n}.

In other words, there are p^{n} possible linear combinations of the a_{i}, F ≋ $ \mathbb{F_p^n}$ as vector spaces ⇒ |F|=|$\mathbb{F_p^n}$|=|$\mathbb{F_p}$|^{n} = p^{n}.

As it was previously stated, a finite field is also called a Galöis field, and denote by GF(p^{n})

Theorem. Every field F is either an extension of ℚ or an extension of ℤ_{p}, for some prime p.

Proof.

Let Φ: ℤ → F be a ring homomorphism defined by Φ(n) = n·1.

If char(F) = p ⇒ p·1 = 0 ⇒ Ker(Φ) = pℤ ⇒[By the first isomorphism theorem] ℤ/pℤ ≋ Φ(ℤ) ⊆ F, i.e.,the image of Φ is a subfield of F (say K) isomorphic to ℤ_{p}, F is an extension of ℤ/pℤ, and we are done.

Suppose char(F) = 0. In this case, Ker(Φ) = {0} ⇒ Φ is one-to-one. It induces a **well-defined, one-to-one homomorphism** ψ: ℚ → F, defined by: ψ(^{a}⁄_{b}) = Φ(a)Φ(b)^{-1}, thus F has a subfield which is isomorphic to ℚ (ℚ/ker(ψ) = ℚ ≋ ψ(ℚ) ⊆_{subfield} F)∎

**Freshman’s Dream or Exponentiation.** Let R be a commutative ring with unity of characteristic p. Then, (a + b)^{pn}=a^{pn} + b^{pn} ∀a, b ∈ R, n ∈ ℕ.

Proof.

Let’s prove it by induction on n. n = 1 holds by the Binomial Theorem. It is valid in any commutative ring with unity.

n=1, we use the binomial formula, (a+b)^{p} = $\sum_{k=0}^p \binom{p}{k}a^kb^{p-k}$ where $\binom{p}{k} = \frac{p!}{k!(p-k)!}$

p | $\binom{p}{k}$ ∀0 < k < p, and R is a commutative ring of characteristic p, so all but the first and last terms are equal to zero ⇒ (a+b)^{p} = a^{p} + b^{p}.

Let’s suppose the result holds for all k, where 1 ≤ k ≤ n, ⇒ (a+b)^{pn+1} = ((a+b)^{p})^{pn} = [If n=1, the result holds] (a^{p}+b^{p})^{pn} = [By the induction hypothesis, the result holds for k = n] (a^{p})^{pn} + (b^{p})^{pn} = a^{pn+1} + b^{pn+1} ⇒ It holds for n+1 ∎

Definition. Let F be a field. A polynomial f(x) ∈ F[x] is called separable, if its roots are distinct in the splitting field of f(x), that is, each root has multiplicity 1, i.e., the number of distinct roots is equal to the degree of the polynomial and f(x) factors into distinct linear factors over the splitting field of f. If f(x) has a multiple root then f(x) is called inseparable.

An extension E of F is a separable extension of F is every element in E is the root of a separable polynomial in F[x].

Example: In ℝ[x], the polynomial x^{2}-x is separable since its roots are 0 and 1. The polynomial x^{2} -2 is separable over ℚ since it factors as $(x -\sqrt{2})(x + \sqrt{2})$. Futhermore, ℚ($\sqrt{2}$) is a separable extension of ℚ.

Proof. ∀α ∈ ℚ($\sqrt{2}$), α = a + b$\sqrt{2}$.

- b = 0, then α is the unique root of x -a
- b ≠ 0, then α is the root of x
^{2}-2ax +a^{2}-2b^{2}= $(x -(a + b\sqrt{2}))(x -(a - b\sqrt{2}))$ and its both roots are obviously distinct.

Theorem. Let F be a field, f(x)∈F[x]. Then f(x) is separable ↭ f(x) and f'(x) are coprime or relatively prime.

Proof.

⇒) Suppose f(x) is separable ⇒ Let L be a splitting field of f, and pick any root of f, say α ∈ L/K ⇒ f(x) = (x -α)g(x) but (x -α) ɫ g(x) since g is separable. Besides, applying derivation’s product rules, f’(x) = g(x) + (x -α)g’(x) ⇒[(x -α) ɫ g(x)] (x -α) ɫ f’(x). Since α was taken as an arbitrary root of f, we’ve shown than any linear factor of f cannot be a factor of f’. Thus, f and f’ share no nontrivial factors, hence they are coprime.

⇐) Conversely, suppose that f(x) and f’(x) are relatively prime, but f(x) has a multiple root α, f(x) = (x -α)^{k}g(x), where k > 1. f’(x) = k(x - α)^{k-1}g(x) + (x-α)^{k}g’(x) and x - α is a common factor of f(x) and f’(x) and α is both a zero of f(x) and f’(x)⊥

Example: is f(x) = x^{7} +9x + 6 ∈ ℚ[x] separable? f’(x) = 7x^{6} + 9. Let (x -α) be common factor of f and f’, then it is also a common factor of any linear combination of f and f’, in particular 7f -xf’ = 7x^{7} +63x + 42 -7x^{7} -9x = 54x + 42 = 6(9x +7) ⇒ there are two options: (x -α) must be either a constant times (9x +7) ↭ α is a root of 9x +7 (⊥ -7/9 is not a root of f) or simply a constant ⇒ f and f’ are coprime ⇒ f is separable.

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, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn (Abstract Algebra), and MathMajor.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- Andrew Misseldine: College Algebra and Abstract Algebra.