⟨p(x)⟩ maximal iff irreducible. Unique Factorization in ℤ[x]

And he added, “We need to perform an endoscopic examination of the throat.” “It sounds fancy, formal, and as complicated as quantum physics or Galois algebra, but it is just a fucking tube put in my neck without real anaesthesia. Rough, cheap, and effective. Of course, anaesthesia is for sissies. It would be as painful as a mule kick to the growing,” I thought, feeling beyond apprehensive, Apocalypse, Anawim, #justtothepoint.

Recall that a field is a commutative ring with unity such that each nonzero element has a multiplicative inverse, e.g., every finite integral domain, ℤ_p (p prime), ℚ, ℝ, and ℂ. A field has characteristic zero or characteristic p with p prime and F[x] is a field, too.

Let A ⊂ R, A ≠ R, R be a ring, A is an ideal of R if the following conditions are satisfied:

1. a - b ∈ A, ∀a, b ∈ A.
2. ∀r ∈ R, ∀a ∈ A, ra ∈ A and ar ∈ A.

A maximal ideal A of a commutative ring R is a proper ideal of R that is maximal with respect to set inclusion amongst all proper ideals, i.e., whenever B is an ideal of R that lives between R and A, A ⊆ B ⊆ R, then B = A or B = R. In other words, we cannot fit any ideals between the ring and a maximal ideal.

Theorem. Let F be a field and let p(x) ∈ F[x]. Then, the ideal the polynomial p(x) generates, that is, ⟨p(x)⟩, is a maximal ideal in F[x] iff p(x) is irreducible over F.

Proof.

⇒) Suppose ⟨p(x)⟩ is a maximal ideal. A maximal ideal is an ideal that is maximal (with respect to inclusion) amongst all proper ideals ⇒ p(x) is neither the zero polynomial (trivial ideal) or a unit (the whole field F).

Suppose for the sake of contradiction, p(x) is reducible over F and say p(x) = g(x)·h(x) is a factorization of p(x) over the field F, where deg(g(x)) and deg(h(x)) < deg(p(x)) and g and h are non units ⇒ ⟨p(x)⟩ ⊆ ⟨g(x)⟩ ⊆ F[x] ⇒ [By assumption, ⟨p(x)⟩ is a maximal ideal] there are only two options, namely ⟨g(x)⟩ = F[x] or ⟨g(x)⟩ = ⟨p(x)⟩

1. ⟨g(x)⟩ = ⟨p(x)⟩ ⇒[Since 1∈ F because F is a field, p(x)=p(x)·1∈ ⟨p(x)⟩ = ⟨g(x)⟩ ⇒ ∃u(x) ∈ F[x] such that p(x) = g(x)·u(x) ⇒ deg(p) ≤ deg(g). Similar reasoning, deg(p) ≤ deg(g)] deg(g(x)) = deg(p(x)) ⊥
2. ⟨g(x)⟩ = F[x] ⇒ 1 ∈ ⟨g(x)⟩ ⇒[∃h(x) ∈ F[x]: 1 = g(x)·h(x), deg(1) = 0 = deg(g) + deg(h) = 0 + 0] deg(g(x)) = 0 ⊥ -g and h are non units-

⇐) Conversely, suppose that p(x) is irreducible over F, is ⟨p(x)⟩ maximal? Let I be any ideal of F[x] such that ⟨p(x)⟩ ⊆ I ⊆ F[x].

We have already shown that if F is a field, then F[x] is a principal ideal domain ⇒ Every ideal is principal, i.e., can be generated by a single element ⇒ ∃g(x) ∈ F[x]: I = ⟨g(x)⟩ ⇒ p(x) ∈ ⟨g(x)⟩ ⇒ p(x) = g(x)h(x), where h(x) ∈ F[x] ⇒ [p is irreducible over F] either g(x) is a constant (g(x) = a ≠ 0 ⇒[F field, a is invertible] 1 = a·a^-1 ∈ ⟨g(x)⟩ ⇒ I = ⟨g(x)⟩ = F[x]) or h(x) is a constant (⟨g(x)⟩ = ⟨p(x)⟩ = I) ⇒ ⟨p(x)⟩ is maximal in F[x] ∎

More generally, if I is an ideal containing a unit u, then 1 = u^-1u ∈ I and hence I = R (By the absorbing property for all r ∈ R, r = r·1 ∈ I ⇒ I = R). In our particular example, ∀f(x)∈ F[x], f(x) = f(x)·1∈ ⟨g(x)⟩.

Corollary. Let F be a field, p(x) is an irreducible polynomial over F if and only if F[x]/⟨p(x)⟩ is a field.

Proof.

p(x) ∈ F[x] is irreducible over F ↭ ⟨p(x)⟩ is maximal [R/A is a field if and only if A is maximal] ↭ F[x]/⟨p(x)⟩ is a field.

Examples

• Consider F = ℤ₂, ℤ₂[x]/⟨x² +x +1⟩ is a field because x² +x +1 is irreducible over ℤ₂ (a polynomial of degree 2 or 3 is irreducible if it does not have any roots -this result will be demonstrated in another article-, 0 + 0 + 1 = 1 ≠ 0 mod 2, 1 + 1 + 1 = 1 ≠ 0 mod 2)

  R/I = {a + I | a ∈ R}. Notice that f(x) ∈ ℤ₂[x]/⟨x² +x +1⟩, f(x) = q(x)(x² +x +1) + r(x), and deg(r(x)) < 2. f(x) + ⟨x² +x +1⟩ = q(x)(x² +x +1) + r(x) + ⟨x² +x +1⟩ =[By the absorption property] = r(x) + ⟨x² +x +1⟩ and deg(r(x)) < 2.

How many element are there in ℤ₂[x]/⟨x² +x +1⟩? ℤ₂[x]/⟨x² +x +1⟩ = {ax + b + ⟨x² +x +1⟩ | a, b ∈ ℤ₂} is a field with four elements. We can represent 0 + ⟨x² +x +1⟩ by 0, 1 + ⟨x² +x +1⟩ by 1, x + ⟨x² +x +1⟩ by α, (x + 1) + ⟨x² +x +1⟩ by α + 1.

Notice that α² = x² + ⟨x² +x +1⟩ =[By the absorption property] x² + (x² +x + 1) + ⟨x² +x +1⟩ =[2x² = 0 in ℤ₂] x +1 + ⟨x² +x +1⟩ = α + 1. ℤ₂[x]/⟨x² +x +1⟩ = {a + bα | a, b ∈ ℤ₂, α² = α + 1}  

• Consider F = ℤ₂, ℤ₂[x]/⟨x³ +x +1⟩ is a field because x³ +x +1 is irreducible over ℤ₂ (a polynomial of degree 2 or 3 is irreducible if it does not have any roots -this result will be demonstrated in another article-, 0 + 0 + 1 = 1 ≠ 0 mod 2, 1 + 1 + 1 = 1 ≠ 0 mod 2)

  R/I = {a + I | a ∈ R}. Notice that f(x) ∈ ℤ₂[x]/⟨x³ +x +1⟩, f(x) = q(x)(x³ +x +1) + r(x), and deg(r(x)) < 3. f(x) + ⟨x³ +x +1⟩ = q(x)(x³ +x +1) + r(x) + ⟨x³ +x +1⟩ =[By the absorption property] = r(x) + ⟨x³ +x +1⟩ and deg(r(x)) < 3.

How many elements are in ℤ₂[x]/⟨x³ +x +1⟩? ℤ₂[x]/⟨x³ +x +1⟩ = {ax² + bx + c + ⟨x³ +x +1⟩ | a, b, c ∈ ℤ₂} is a field with eight elements, 0 + ⟨x³ +x +1⟩, 1 + ⟨x³ +x +1⟩, x + ⟨x³ +x +1⟩, x +1 + ⟨x³ +x +1⟩, ··· x² +x +1 + ⟨x³ +x +1⟩.

• ℤ₅[x]/⟨x⁴+3x³+x+4⟩ is not a field because x⁴+3x³+x+4 is not irreducible. It can be proved that p(x) has no solutions or roots in ℤ₅, but it does not mean it is irreducible (deg(x⁴+3x³+x+4) = 4). Obviously, since it has no roots, it cannot be factored into a cubic and a linear term, so we can aim to factor it into two quadratics. x⁴+3x³+x+4 = (x² +4x +2)² ⇒ x⁴+3x³+x+4 is reducible ⇒ ℤ₅[x]/⟨x⁴+3x³+x+4⟩ is not a field.

• Construct fields with…

1. 9 elements: ℤ₃[x]/⟨x² +1⟩ because x² + 1 is irreducible over ℤ₃. ℤ₃[x]/⟨x² +1⟩ = {ax + b | a, b ∈ ℤ₃}
2. 16 elements: ℤ₂[x]/⟨x⁴ + x + 1⟩, 2⁴ elements, namely {0, 1, x, x², x³, 1 + x, 1 + x²,1 + x³, x +x², x + x³, x² + x³, x + x² +x³, 1 + x + x², 1 +x +x³, 1 +x² +x³, 1 +x +x² + x³}. Notice that f(x) = x⁴ +x² +1 has no roots in ℤ₂[x] (f(0) = f(1) = 1). Since the only quadratic irreducible polynomial over ℤ₂[x] is x² + x +1 and x² + x +1 is not a divisor of x⁴ +x +1 -Just notice that (x²+x+1) · (x²+x+1) = x⁴+ x³ +x³ +x² +x² +x² +x +x +1=x⁴+x²+1)-, x⁴ +x +1 is irreducible over ℤ₂[x].
3. 25 elements: ℤ₅[x]/⟨x² + x + 1⟩ is a field of order 5².

• x² - 2 ∈ ℚ[x] is irreducible ⇒ ℚ[x]/⟨x² -2⟩ is a field. Futhermore, ℚ[x]/⟨x² -2⟩ ≋ ℚ($\sqrt{2}$). Take Φ: ℚ[x] → ℝ, p(x) → p($\sqrt{2}$), p ∈ Ker(Φ) ↭ p($\sqrt{2}$) = 0 ↭ p ∈ ⟨x² -2⟩ ⇒ Ker(Φ) = ⟨x² -2⟩. By the first isomorphism theorem, ℚ[x]/⟨x² -2⟩ ≋ Φ(ℚ[x]) = ℚ($\sqrt{2}$).

Corollary. Let F be a field and let p(x), a(x), b(x) ∈ F[x]. If p(x) is irreducible over F and p(x)|a(x)b(x), then p(x)|a(x) or p(x)|b(x).

Proof.

Since p(x) is irreducible over F ⇒[Corollary. Let F be a field, p(x) is an irreducible polynomial over F if and only if F[x]/⟨p(x)⟩ is a field.] F[x]/⟨p(x)⟩ is a field ⇒[Every field is an integral domain] F[x]/⟨p(x)⟩ is an integral domain ⇒ [Let R be a commutative ring with unity and A an ideal of R. R/A is an integral domain if and only if A is prime] ⟨p(x)⟩ is a prime ideal.

Recall that an ideal P of R is prime if ∀ab ∈ P ⇒ a ∈ P or b ∈ P.

By assumption, p(x)|a(x)b(x) ⇒[∃u(x): a(x)b(x) = p(x)u(x)] a(x)b(x) ∈ ⟨p(x)⟩ ⇒[⟨p(x)⟩ is a prime ideal] a(x) ∈ ⟨p(x)⟩ or b(x) ∈ ⟨p(x)⟩ ⇒ p(x)|a(x) or p(x)|b(x)∎

Unique Factorization in ℤ[x]. Every non-zero, non-unit polynomial in ℤ[x] can be expressed or written in the form b₁b₂···b_sp₁(x)p₂(x)···p_m(x), where the b_i's are irreducible polynomials of degree 0, and the p_i(x)'s are irreducible polynomials of positive degree. Futhermore, this factorization is unique, that is, if b₁b₂···b_sp₁(x)p₂(x)···p_m(x) = c₁c₂···c_tq₁(x)q₂(x)···q_n(x), then s = t, m = n, and b_i = ± c_i ∀i: 1 ≤ i ≤ s, p_i(x) = ± q_i(x) ∀i: 1 ≤ i ≤ m.

Proof.

Let f be a non-zero, non-unit polynomial from ℤ[x].

If deg(f(x)) = 0, then f(x) is constant and the theorem’s statement is obtained by the Fundamental Theorem of Arithmetic (For every integer n such that n > 1, n can be expressed or written as the product of one or more primes, uniquely up to the order in which they appear. 📖It was known to Euclid, one of the most famous mathematician of the ancient world, and first proved formally by Gauss.)

If deg(f(x)) > 0, let b be the content of f(x) -the greatest common divisor of its coefficients-, and let the factorization of b as a product of prime numbers be as follows b = b₁b₂···b_s ⇒ f(x) = b₁b₂···b_sf’(x) where f’(x) ∈ ℤ[x] is primitive -its content is equal to 1- and deg(f’(x)) = deg(f(x)).

To prove the existence portion of the theorem, it suffices to show that a primitive polynomial f(x) can be written as a product of irreducible polynomials of positive degree.

Let’s procede by induction on deg(f(x)).

• If deg(f(x))=1, then f(x) is already irreducible ∎
• Suppose that every primitive polynomial of degree less than deg(f(x)) can be written as as product of irreducible polynomials of positive degrees (we are going to use strong induction).

1. If f(x) is irreducible, there is nothing else to do or prove and we are done ∎
2. Otherwise, f(x) can be written or expressed as a product of two polynomials of lesser degrees, say f(x) = g(x)h(x), where both g(x) and h(x) are primitive (their contents are equal to one), deg(g(x)) and deg(h(x)) < deg(f(x)) ⇒ [By the induction hypothesis] Both g(x) and h(x) can be written as a product of irreducible polynomials of positive degrees, and therefore f(x) can be written as a product of irreducible polynomials of positive degrees.

To prove the uniqueness of the factorization, suppose that f(x) = b₁b₂···b_sp₁(x)p₂(x)···p_m(x) = c₁c₂···c_tq₁(x)q₂(x)···q_n(x) where the b’s and c’s are irreducible polynomials of degree 0 and the p’s and q’s are irreducible polynomials of positive degree.

Set b = b₁b₂···b_s and c = c₁c₂···c_t

Gauss's Lemma. The product of two primitive polynomials is primitive ⇒ Since the polynomials p’s and q’s are all primitive polynomials, their product p₁(x)p₂(x)···p_m(x) and q₁(x)q₂(x)···q_n(x) are both primitive ⇒ b and c must equal plus or minus the content of f(x), that is, b = ± content(f), c = ± content(f) ⇒ |b| = |c| ⇒ [By the uniqueness of the Fundamental Theorem of Arithmetic, after renaming and/or renumbering] s = t, b_i = ± c_i ∀i: 1 ≤ i ≤ s ⇒ By cancelling the constant term in the two factorization for f(x), p₁(x)p₂(x)···p_m(x) = ± q₁(x)q₂(x)···q_n(x).

Now, viewing the p(x)’s and q(x)’s as elements of ℚ[x] (the field of fraction of ℤ) and noticing that p₁(x) divides q₁(x)q₂(x)···q_n(x) ⇒ [By the previous corollary and applying induction. p(x) irreducible over a field F and p(x)|a(x)b(x) ⇒ p(x)|a(x) or p(x)|b(x)] p₁(x) divides q_i(n) for some i. Without any loss of generality, we can assume p₁(x) | q₁(x), and since q₁ is irreducible (it cannot be factored into nontrivial polynomials over the same field) ⇒ ∃r, s ∈ ℤ: q₁(x) = (r/s)p₁(x) ⇒ [Since q₁ and p₁ are primitive, r/s = ±1] q₁(x) = ±p₁(x). Besides, after cancelling we have p₂(x)p₃(x)···p_m(x) = ± q₂(x)q₃(x)···q_n(x).

Next, we should repeat the same argument above with p₂(x), p₃(x), and so on. If m < n, we would have 1 on the left and a non-constant polynomial on the right, that is impossible. Analogously, mutatis mutandis m > n is also impossible. Therefore, m = n and p_i(x) = ± q_i(x) ∀i: 1 ≤ i ≤ m ∎

Bibliography

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