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:
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)⟩
⇐) 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^{-1}u ∈ 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.
R/I = {a + I | a ∈ R}. Notice that f(x) ∈ ℤ_{2}[x]/⟨x^{2} +x +1⟩, f(x) = q(x)(x^{2} +x +1) + r(x), and deg(r(x)) < 2. f(x) + ⟨x^{2} +x +1⟩ = q(x)(x^{2} +x +1) + r(x) + ⟨x^{2} +x +1⟩ =[By the absorption property] = r(x) + ⟨x^{2} +x +1⟩ and deg(r(x)) < 2.
How many element are there in ℤ_{2}[x]/⟨x^{2} +x +1⟩? ℤ_{2}[x]/⟨x^{2} +x +1⟩ = {ax + b + ⟨x^{2} +x +1⟩ | a, b ∈ ℤ_{2}} is a field with four elements. We can represent 0 + ⟨x^{2} +x +1⟩ by 0, 1 + ⟨x^{2} +x +1⟩ by 1, x + ⟨x^{2} +x +1⟩ by α, (x + 1) + ⟨x^{2} +x +1⟩ by α + 1.
Notice that α^{2} = x^{2} + ⟨x^{2} +x +1⟩ =[By the absorption property] x^{2} + (x^{2} +x + 1) + ⟨x^{2} +x +1⟩ =[2x^{2} = 0 in ℤ_{2}] x +1 + ⟨x^{2} +x +1⟩ = α + 1. ℤ_{2}[x]/⟨x^{2} +x +1⟩ = {a + bα | a, b ∈ ℤ_{2}, α^{2} = α + 1}
R/I = {a + I | a ∈ R}. Notice that f(x) ∈ ℤ_{2}[x]/⟨x^{3} +x +1⟩, f(x) = q(x)(x^{3} +x +1) + r(x), and deg(r(x)) < 3. f(x) + ⟨x^{3} +x +1⟩ = q(x)(x^{3} +x +1) + r(x) + ⟨x^{3} +x +1⟩ =[By the absorption property] = r(x) + ⟨x^{3} +x +1⟩ and deg(r(x)) < 3.
How many elements are in ℤ_{2}[x]/⟨x^{3} +x +1⟩? ℤ_{2}[x]/⟨x^{3} +x +1⟩ = {ax^{2} + bx + c + ⟨x^{3} +x +1⟩ | a, b, c ∈ ℤ_{2}} is a field with eight elements, 0 + ⟨x^{3} +x +1⟩, 1 + ⟨x^{3} +x +1⟩, x + ⟨x^{3} +x +1⟩, x +1 + ⟨x^{3} +x +1⟩, ··· x^{2} +x +1 + ⟨x^{3} +x +1⟩.
ℤ_{5}[x]/⟨x^{4}+3x^{3}+x+4⟩ is not a field because x^{4}+3x^{3}+x+4 is not irreducible. It can be proved that p(x) has no solutions or roots in ℤ_{5}, but it does not mean it is irreducible (deg(x^{4}+3x^{3}+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. x^{4}+3x^{3}+x+4 = (x^{2} +4x +2)^{2} ⇒ x^{4}+3x^{3}+x+4 is reducible ⇒ ℤ_{5}[x]/⟨x^{4}+3x^{3}+x+4⟩ is not a field.
Construct fields with…
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 b_{1}b_{2}···b_{s}p_{1}(x)p_{2}(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 b_{1}b_{2}···b_{s}p_{1}(x)p_{2}(x)···p_{m}(x) = c_{1}c_{2}···c_{t}q_{1}(x)q_{2}(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 = b_{1}b_{2}···b_{s} ⇒ f(x) = b_{1}b_{2}···b_{s}f’(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)).
To prove the uniqueness of the factorization, suppose that f(x) = b_{1}b_{2}···b_{s}p_{1}(x)p_{2}(x)···p_{m}(x) = c_{1}c_{2}···c_{t}q_{1}(x)q_{2}(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 = b_{1}b_{2}···b_{s} and c = c_{1}c_{2}···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 p_{1}(x)p_{2}(x)···p_{m}(x) and q_{1}(x)q_{2}(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), p_{1}(x)p_{2}(x)···p_{m}(x) = ± q_{1}(x)q_{2}(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 p_{1}(x) divides q_{1}(x)q_{2}(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)] p_{1}(x) divides q_{i}(n) for some i. Without any loss of generality, we can assume p_{1}(x) | q_{1}(x), and since q_{1} is irreducible (it cannot be factored into nontrivial polynomials over the same field) ⇒ ∃r, s ∈ ℤ: q_{1}(x) = (r/s)p_{1}(x) ⇒ [Since q_{1} and p_{1} are primitive, r/s = ±1] q_{1}(x) = ±p_{1}(x). Besides, after cancelling we have p_{2}(x)p_{3}(x)···p_{m}(x) = ± q_{2}(x)q_{3}(x)···q_{n}(x).
Next, we should repeat the same argument above with p_{2}(x), p_{3}(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 ∎