Finite Fields 2

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Classification of finite fields. For each prime p and each positive integer n, there exist a unique finite field E of order pⁿ, up to isomorphism.It is isomorphic to the splitting field of x^pn-x over ℤ_p.

Proof.

Consider the polynomial f(x) = x^pn-x, and let F be the splitting field of f(x) over ℤ_p (and therefore char(F) = p).

f’(x) = pⁿx^pn-1-1 =[In a field of characteristic p] -1, and -1 is obviously relatively prime to f(x) ⇒ f(x) is separable, i.e., has pⁿ distinct zeros in F.

We claim that the roots of f(x) form a subfield of f.

1. 0, 1 are zeros of f(x).
2. If α, β are zeros of f(x), then α + β, α·β are also zeros of f(x) since f(α+β) = (α + β)^pn - (α + β) = [Freshman’s Dream or Exponentiation] α^pn+β^pn -(α + β)= α^pn -α + β^pn - β =[α, β are zeros of f(x)] 0 + 0 = 0. f(αβ) = (αβ)^pn - αβ = α^pnβ^pn -αβ = [α^pn - α = 0 ⇒ α^pn = α] αβ - αβ = 0.
3. If α is a root of f(x), -α is also a zero of f since f(-α)=(-α)^pn -(-α) = -α^pn +α = -(α^pn -α) = 0 (p is odd). Otherwise, p = 2, f(-α) = (-α)²ⁿ -(-α) =[In a field of characteristic two, everything is its own additive inverse, (-α)²ⁿ=α²ⁿ = -α = α] α + α = 0 (ℤ₂).
4. If α is a root of f, α ≠ 0, α^-1 is also a zero of f since f(α^-1)=(α^-1)^pn -α^-1 = (α^pn)^-1 -α^-1 = α^-1 -α^-1 = 0.

Therefore, the roots of f(x) form a subfield of F and F is the splitting field of f(x) ⇒[A splitting field of a polynomial is the smallest field extension of that field over which the polynomial splits] the subfield must be all of F, i.e., |F| = pⁿ.

To show that there is a unique field for each prime-power up to isomorphism, let E be any other field of order pⁿ, let’s show that E is isomorphic to F, we must prove that every element in E is a root of f(x).

0 is a root of f(x). Let α ≠ 0, α ∈ E, the order of the multiplicative group of non-zero elements of E is pⁿ-1, p prime ⇒ [The multiplicative group of non-zero elements of a finite field is cyclic -see the theorem just below-, hence the order of α is a divisor of pⁿ-1] α^pn-1 = 1 or α^pn - α = 0.

Since E contains pⁿ elements ⇒ E is the splitting field of f(x), and we know that the splitting field of any polynomial is unique up to isomorphism. Therefore, there is only one field for each prime-power pⁿ, we denote this field by $\mathbb{F}_{p^n}$ GF(pⁿ), and call it the Galois field of order pⁿ  

Theorem. Let p be a prime. Then, a field of order pⁿ contains a field of order p^m ($\mathbb{F}$_p^m ⊆ $\mathbb{F}$_pⁿ) iff m|n.

Proof. Assume $\mathbb{F}$_p^m ⊆ $\mathbb{F}$_pⁿ. Then, n = [$\mathbb{F}$_pⁿ:$\mathbb{F}_p$] = [$\mathbb{F}$_pⁿ:$\mathbb{F}$_p^m][$\mathbb{F}$_p^m:$\mathbb{F}$_p] = [$\mathbb{F}$_pⁿ:$\mathbb{F}$_p^m]m, thus m | n.

Conversely, assume m | n, that is, km = n. Let α ∈ $\mathbb{F}$_p^m ⇒ α^pm = α and consider f(x) = x^pn -x.

Reclaim that $\mathbb{F}$_pⁿ is the splitting field of x^pn -x. In other words, an element belongs to the field $\mathbb{F}$_pⁿ ↭ if this element is a root of the polynomial x^pn -x.

∀α ∈ $\mathbb{F}$_p^m, f(α) = α^pn -α = [n = km] α^pkm -α = α^(pm)k = α^{(pm)··_{k times}··(pm)} -α = $(α^{p^m})^{p^m}…^{p^m}$ -α = [α ∈ $\mathbb{F}$_p^m ⇒ α^pm = α] α - α = 0 ⇒ α ∈ $\mathbb{F}$_pⁿ ⇒ $\mathbb{F}$_p^m ⊆ $\mathbb{F}$_pⁿ ∎

Th. The Multiplicative Group of Finite Subgroups of a Field. Let F be a field and G a finite subgroup of F*, the multiplicative group of non-zero elements of F, then G is cyclic. In particular, the multiplicative group of all nonzero elements of a finite field is cyclic.

Proof.

Let |G| = n, a finite subgroup of F* of order n. F is a field ⇒ F* is an Abelian group ⇒ G is a finite Abelian group ⇒ [Fundamental Theorem of Finite Abelian Groups] $G ≋ ℤ_{p_1^{e_1}}x···xℤ_{p_k^{e_k}}$ where $n = p_1^{e_1}···{p_k^{e_k}}$ and the p_i are (not necessarily distinct) primes.

Let m be the least common multiple of the orders of the cyclic factors of G, $p_1^{e_1},···,{p_k^{e_k}}$ ⇒ m ≤ $p_1^{e_1}···{p_k^{e_k}}=n$. If a_i ∈ ℤ_(pi)^ei, then (a_i)^(p_ie_i) = 1 ⇒[m = lcm($p_1^{e_1},···,{p_k^{e_k}})$ ⇒ m is a multiple of p_i^e_i] a_i^m = 1 ⇒ ∀α ∈ G, α^m=1, that is, every element of G is a root of x^m = 1.

Since x^m-1 has at most m roots in F ⇒ n ≤ m ⇒[Before we have established that m ≤ n] n = m ⇒ m = lcm($p_1^{e_1},···,{p_k^{e_k}})$ = n =[$G ≋ ℤ_{p_1^{e_1}}x···xℤ_{p_k^{e_k}}$] $p_1^{e_1}···{p_k^{e_k}}$ ⇒ lcm($p_1^{e_1},···,{p_k^{e_k}})$ = $p_1^{e_1}···{p_k^{e_k}}$. Therefore, pi’s are distinct primes, and the group G is isomorphic to the cyclic group ℤ_m ∎

In particular, I insist, the multiplicative group of all nonzero elements of a finite field is cyclic.

Besides, if F is a finite field, then there exist some element α ∈ F (it is called a primitive root of F) which is a generator of F*, that is, all nonzero elements of F are expressible as a power of α. In particular, if characteristic(F) = p, then F = $\mathbb{F}_p(α)$ which is a simple extension.

Examples.

• f(x) = x² + x + 1, $\mathbb{F}_2$[x]/⟨(x² + x + 1)⟩ = [Notation] $\mathbb{F}_2$[w]/⟨(w² + w + 1)⟩ is a field of order 4. Its elements are 0, 1, w -Abusing notation, x + ⟨(x² + x + 1)⟩-, w +1, where w² = w + 1.

• f(x) = x³ + x + 1, $\mathbb{F}_2$[x]/⟨(x³ + x + 1)⟩ is a field of order 8. The eight polynomials of degree less than 3 in ℤ₂[x] form a field with 8 elements, namely, 0, 1, x, x +1, x², x² + 1, x² + x, x² + x + 1.

• f(x) = 1 + x + x⁴ ∈ ℤ₂[x]. We claim f(x) is irreducible because f(0) = f(1) = 1 and f(x) ≠ (x² +x +1)² and this is the only irreducible quadratic. Therefore, $\mathbb{F}_2$[x]/⟨f(x)⟩ ≋ F(α) is a field of order 2⁴ = 16 -its base is {1, α, α², α³} in ℤ₂, α root of f-, ℤ₂(α)= $\mathbb{F}$₁₆ ≋ ℤ₂⁴ as vector spaces. Besides, $\mathbb{F}$^*₁₆ ≋ ℤ₁₅ as groups.

α⁴ + α + 1 = 0 ⇒[ℤ₂(α)] α⁴ = α + 1.

1. α¹ = α.
2. α² = α²
3. α³ = α³
4. α⁴ = [α⁴ = α + 1.] 1 + α
5. α⁵ = αα⁴ = α + α²
6. α⁶ = α² + α³
7. α⁷ = α³ + α⁴ = [α⁴ = α + 1.] 1 + α + α³
8. α⁸ = α + α² + α⁴ = α + α² + 1 + α = 1 + α²
9. α⁹ = α + α³
10. α¹⁰ = α² + α⁴ = 1 + α + α²
11. α¹¹ = α + α² + α³
12. α¹² = α² + α³ + α⁴ = 1 + α + α² + α³
13. α¹³ = α + α² + α³ + α⁴ = α + α² + α³ + 1 + α = 1 + α² + α³
14. α¹⁴ = α + α³ + α⁴ = α + α³ + 1 + α = 1 + α³
15. α¹⁵ = α + α⁴ = α + 1 + α = 1

Theorem. Let F be a finite field of order q = pⁿ. Then, F is the splitting field of the polynomial x^q-x ∈ $\mathbb{F}_p[x]$.

Proof.

First notice that 0^q = 0. Let u ∈ F^* ⇒ [By Lagrange’s Theorem] u^|F*| = 1 ⇒ u^q-1 = 1 ⇒ [*u] u^q = u. Then, every element of F is a root of f(x) = x^q -x.

As f(x) has at most q roots, we see that F contains all the roots of f(x) ⇒ f(x) splits over F. As obviously F = $\mathbb{F}_p$(F), F is necessarily the splitting field of f(x).

Structure theorem for finite fields. Let p be a prime, n be a positive integer, q = pⁿ. Then,

1. There exists a unique finite field of order q, up to isomorphism (Classification of finite fields)
2. Let F be a field of order q. Then, F^x = F \ {0} is a cyclic group under the multiplication operation (The Multiplicative Group of Finite Subgroups of a Field)
3. Let F be a finite field of order q = pⁿ. F is the splitting field of the polynomial x^q-x ∈ $\mathbb{F}_p[x]$.
4. A field of order pⁿ contains a field of order p^m ($\mathbb{F}$_p^m ⊆ $\mathbb{F}$_pⁿ) iff m|n.

$\mathbb{F_4}$ = {0, 1, α, 1 + α}, α is a root of x² +x + 1 (the only irreducible polynomial of degree 2 in $\mathbb{F_2}$).

A polynomial p(x) of degree 2 or 3 is irreducible ↭ it does not have linear factors. Therefore, it suffices to show that p(0)=p(1)=1. This quickly tells us that x²+x+1 is the only irreducible polynomial of degree 2 (0+0+1=1≠0, 1+1+1=1≠0), and x³+x²+1 and x³+x+1 are the only irreducible polynomials of degree 3.

By the Structure Theorem for finite fields (3), $\mathbb{F_4}$ is the splitting field of the polynomial x⁴ -x = [This equality holds in $\mathbb{F_4}$[x] not in $\mathbb{F_2}$[x]] x(x -1)(x -α)(x+α) = x(x³-1) = x(x -1)(x²+x+1).

x(x -1)(x²+x+1) are the three irreducible polynomials in $\mathbb{F_2}$[x] whose order divides 2.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
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3. Mathematics LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.