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Definition. Let D be an integral domain. An irreducible polynomial is, roughly speaking, a polynomial that cannot be factored into the product of two non-constant polynomials. A non-zero and non-unit polynomial f(x) ∈ D[x] is said to be irreducible if it cannot be written or expressed as the product of two non-units. In other words, whenever f(x) is expressed as a product of two polynomials g(x)h(x) with g(x) and h(x) ∈ D[x], then either g(x) or h(x) is a unit in D[x].

A non-constant polynomial f(x) ∈ F[x], F field, is said to be irreducible if f(x) cannot be expressed a as a product of two polynomials g(x), h(x) ∈ F[x] of lower degree, that is, f(x) = g(x)h(x), deg(g(x)) < deg(f(x)) and deg(h(x)) < deg(f(x)).

**Reducibility test for polynomials with degrees two or three**. Let F be a field. If f(x) ∈ F[x] and deg(f(x)) = 2 or 3, then f(x) is reducible over F iff f has a zero in F or **f(x) is irreducible over F iff f(α) ≠ 0 ∀α ∈ F**.

Theorem. Let f(x) ∈ ℤ[x]. If f(x) is reducible over ℚ, then it is reducible over ℤ.

Proof.

Suppose f is reducible over ℚ ⇒ ∃g(x), h(x) ∈ ℚ[x] such that f(x) = g(x)h(x) with both polynomials g(x) and h(x) having degree less than of f(x). We could assume that f is primitive because we can divide both f and g by the content of f.

Next, let a and b be the least common multiple of the denominators of the coefficients of g and h respectively, so abf(x) = ag(x)·bh(x), where ag(x) ∈ ℤ[x] and bh ∈ ℤ[x].

Now, let c and d be the content of ag(x) and bh(x) respectively ⇒ ag(x) = cg’(x) and bh(x) = dh’(x) where both g’(x) and h’(x) are primitives ⇒ abf(x) = (ag(x))·(bh(x)) = (cg’(x))(dh’(x)) = cdg’(x)h’(x), where both g’(x) and h’(x) are primitives polynomials in ℤ[x], so their product is primitive, and f is primitive, too ⇒ content(abf(x)) =[f is primitive] ab = content(cdg’(x)h’(x)) =[g’(h)h’(x) is primitive] cd ⇒[ab = cd & abf(x) = cdg’(x)h’(x) & Cancellation law applies] **f(x) = g’(x)h’(x) where both g’(x) and h’(x) are primitives polynomials in ℤ[x] and deg(g’(x)) = deg(g(x)) and deg(h’(x)) = deg(h(x))** and both polynomials g’ and h’ having degrees less than of f, therefore f is reducible over ℤ ∎

**Mod p Irreducibility Test**. Let p be a prime, f(x) ∈ ℤ[x], and deg(f(x)) ≥ 1. Let f'(x) be the polynomial in ℤ_{p}[x] obtained from f(x) by reducing all the coefficients of f(x) modulo p. If f'(x) is irreducible over ℤ_{p} and deg(f'(x)) = deg(f(x)), then f(x) is irreducible over ℚ.

Proof.

Suppose f’(x) is irreducible over ℤ_{p} and deg(f’(x)) = deg(f(x)). For the sake of contradiction, let us assume f is reducible over ℚ ⇒[Theorem. Let f(x) ∈ ℤ[x], then f(x) is irreducible over ℚ, then it is reducible over ℤ.] f is reducible over ℤ, that is, ∃g(x), h(x) ∈ ℤ[x], f(x) = g(x)h(x) with both polynomials g(x) and h(x) having degree less than of f(x) ⇒[f is reducible] 1 ≤ deg(g)< deg(f) and 1 ≤ deg(h) < deg(f). Then, **the idea is that the same relationship holds taking all coefficients modulo p.**

Let f’, g’, and h’ be the polynomials obtained from f, g, and h by reducing all the coefficients modulo p.

Since deg(f) = deg(f’) ⇒ deg(g’) ≤ deg(g) < deg(f’) and deg(h’) ≤ deg(h) < deg(f’) and f’ = g’h’ ⇒ f’ = g’h’, deg(g’) < deg(f’) and deg(h’) < deg(f’) ⊥ Therefore, this contradicts our assumption that f’(x) is irreducible over ℤ_{p}, so f is irreducible over ℚ.

In fact, deg(g’) = deg(g) and deg(h’) = deg(h). Otherwise, deg(f’) = deg(g’h’) = deg(g’) + deg(h’) < deg(g) + deg(h) = deg(gh) = deg(f) ⊥ By assumption, deg(f) = deg(f’).

Examples.

- f(x) = x
^{3}+ x +1 ∈ ℤ[x], x^{3}+ x +1 ∈ ℤ/2ℤ[x], 1^{3}+1 +1 = 3 = 1, 0^{3}+0 +1 =1, deg(f) = 3 ⇒ It is irreducible over ℤ/2ℤ ⇒ It is irreducible over ℚ. - f(x) = x
^{5}+9x^{4}+12x^{2}+6. f’(x) = x^{5}+ x^{4}= x^{4}(x + 1) over ℤ_{2}, f’ is reducible over ℤ_{2}, so we cannot apply Mod p = 2 Irreducible Test. f’(x) = x^{5}= x^{3}·x^{2}over ℤ_{3}, f’ is reducible over ℤ_{3}, so we cannot apply Mod p = 3 Irreducible Test. f’(x) = x^{5}+4x^{4}+ 2x^{2}+ 1 over ℤ_{5}. Let’s evaluate f’(0) = 1 ≠ 0, f’(1) = 3 ≠ 0, f’(2) = 10 = 0 ⇒ f’(x) = (x-2) g(x) so f’ is reducible over ℤ_{5}, so we cannot apply Mod p = 5 Irreducible Test. - f(x) = x
^{3}+ 7x + 16. f’(x) = x^{3}+ 2x +1 over ℤ_{5}. Since it has degree 3 and has no root in ℤ_{5}[x], it is irreducible in ℤ_{5}[x] ⇒ f(x) is irreducible in ℚ[x]. - f(x) = 21x
^{3}-3x^{2}+2x +9. f’(x) = x^{3}+x^{2}+1 over ℤ_{2}. Since it has degree 3 and has no root in ℤ_{2}[x] (f’(0) = f’(1) = 1), it is irreducible in ℤ_{2}[x] ⇒ f(x) is irreducible in ℚ[x]. However, f’(x) = 2x over ℤ_{3}, f' is irreducible but 💀 we cannot apply p = 3 Irreducible Test because deg(f(x)) ≠ deg(f'(x)). - f(x)=(3/7)x
^{4}-(2/7)x^{2}+ (9/35)x + 3/5. Let h(x) = 35f(x) = 15x^{4}-10x^{2}+ 9x +21. h’(x) = x^{4}+x +1 in ℤ_{2}[x]. It is irreducible over ℤ_{2}[x] because it has no zeros in ℤ_{2}[x] and has no quadratic factor either (There are two options: a) x^{2}+ x + 1 is not a factor; b) and x^{2}+ 1 has a zero in x=1, but h’ does not, so it is not possible h’ = (x^{2}+ 1)·h’’) ⇒ h(x) is irreducible over ℚ ⇒ f(x) is irreducible over ℚ. (Figure 1.a.)

**Eisenstein Criterion**. Let p ∈ ℤ be a prime. If f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + ··· + a_{0} in ℤ[x] have degree ≥ 1 with

- p divides all the coefficient, except the leading one, that is, p | a
_{0}, p | a_{1}, ···, p | a_{n-1} - p does not divide the leading term, p ɫ a
_{n} - p
^{2}ɫ a_{0}

Then, f(x) is irreducible in ℚ[x].

Proof:

This proof is by contradiction. Let’s assume that f is not irreducible. If f(x) is reducible over ℚ ⇒[Theorem. Let f(x) ∈ ℤ[x], then f(x) is irreducible over ℚ, then it is reducible over ℤ.] f is reducible over ℤ, that is, ∃g(x), h(x) ∈ ℤ[x]: f(x) = g(x)h(x) and the degrees of g and h are less than n.

Say g(x) = b_{r}x^{r} + ··· + b_{0}, h(x) = c_{s}x^{s} + ··· + c_{0}.

Since p | a_{0}, p^{2} ɫ a_{0}, and a_{0} = b_{0}c_{0} ⇒ p divides b_{0} or c_{0}, but not both. Let’s suppose **p | b _{0} and p ɫ c_{0} without any loss of generality**.

Since p ɫ a_{n} = b_{r}c_{s} ⇒ p ɫ b_{r}, but p | b_{0}, hence there is a least integer t such that p ɫ b_{t} (p | b_{t-1}, ··· p | b_{1}, p| b_{0}). Next, let us consider a_{t} = b_{t}c_{0} + b_{t-1}c_{1} + ··· + b_{0}c_{t}. Then, p | a_{t} (Notice that r = deg(g(x)) = deg(b_{r}x^{r} + ··· + b_{0}) < n ⇒ t ≤ r < n ), and p divides every summand on the right after the first term b_{t}c_{0}, that is, (b_{t-1}c_{1} + ··· + b_{0}c_{t}) ⇒ p | b_{t}c_{0}, p prime ⊥ p ɫ b_{t}, p ɫ c_{0}

- f(x) = x
^{5}+9x^{4}+ 12x^{2}+6, 3 ɫ 1, 3 | 9, 3 | 12, 3 | 6, 3^{2}= 9 ɫ 6, then f(x) is irreducible over ℚ. - f(x) = x
^{4}+ 3x^{2}+ 3, 3 ɫ 1, 3 | 3, 3 | 3, 3^{2}ɫ 3, then f(x) is irreducible over ℚ. - f(x) = 5x
^{10}-3x^{7}-6x^{3}+15x^{2}-9x +6 ∈ ℤ[x], then f(x) is irreducible over ℚ (p = 3). - x
^{n}-2 ∈ ℤ[x], n ≥ 2 is irreducible over ℚ (Eisenstein criteria, p=2) ∀n ≥ 2, so there are**irreducible polynomials over ℚ of arbitrary high degree.**This is not true over ℝ. We know that any irreducible polynomial over ℝ has degree 1 or 2. - f(x) = x
^{5}+3x^{3}-3x +6, 3 ɫ 1, 3 | 3, 3 | -3, 3 | 6, 3^{2}= 9 ɫ 6, then f is irreducible over Q. - f(x) = x
^{2}+x +2 ∈ ℤ[x]. By changing x → x + 3, f’(x) = x^{2}+ 6x + 9 + x + 3 + 2 = x^{2}+ 7x + 14 is irreducible over ℚ (Eisenstein p = 7)Eisenstein theorem applies after substitution x → x + a. If the polynomial after substitution (is is just an automorphism of the ring ℚ([x]), and automorphisms preserve algebraic properties) is irreducible then allows concluding that the original polynomial is irreducible as well.

Corollary. For any prime p, the pth cyclotomic polynomial Φ_{p}(x) = $\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+···+x+1$ is irreducible over ℚ.

Proof. Let f(x) = Φ_{p}(x+1) = $\frac{(x+1)^p-1}{(x+1)-1}=x^{p-1}+(\begin{smallmatrix}p\\ 1\end{smallmatrix})x^{p-2}+(\begin{smallmatrix}p\\ 2\end{smallmatrix})x^{p-3}+···+(\begin{smallmatrix}p\\ p-2\end{smallmatrix})x+(\begin{smallmatrix}p\\ p-1\end{smallmatrix})1$

f(x) does satisfy the conditions of Eisenstein’s criterion because $(\begin{smallmatrix}p\\ k\end{smallmatrix})$ is divisible by p for 1 ≤ k ≤ p−1, p ɫ 1, and $(\begin{smallmatrix}p\\ p-1\end{smallmatrix})$ is not divisible by p^{2} ⇒ Φ_{p}(x+1) is irreducible.

But if Φ_{p}(x) = g(x)h(x) were a nontrivial factorization of Φ_{p}(x) over ℚ, then so would Φ_{p}(x+1) be no trivially factored, Φ_{p}(x+1) = g(x+1)h(x+1) ⊥

Examples: p = 2, $\frac{x^2-1}{x-1}=x+1$ is irreducible over ℚ. p = 3, $\frac{x^3-1}{x-1}=$ x^{2} + x + 1 is irreducible over ℚ.

**Rational root test**. Let f = a_{n}x^{n} +a_{n-1}x^{n-1} + ··· + a_{0} ∈ ℤ[x], a_{0}a_{n} ≠ 0. Each rational solution r/s written in lowest terms so that r and s are relatively prime, satisfies that r is a factor of the constant term a_{0} (r | a_{0}) and s is a factor of the leading coefficient a_{n} (s | a_{n}).

Proof.

Let f = a_{n}x^{n} +a_{n-1}x^{n-1} + ··· + a_{0} ∈ ℤ[x], let r/s be a rational solution.

a_{n}^{rn}⁄_{sn} + ··· + a_{1}^{r}⁄_{s} + a_{0} = 0 ⇒ [Multiply by s^{n}] a_{n}r^{n} + a_{n-1}r^{n-1}s + ··· + a_{1}rs^{n-1} + a_{0}s^{n} = 0 ⇒ r(a_{n}r^{n-1} + ··· + a_{1}s^{n-1}) = -a_{0}s^{n} ⇒ r | a_{0}s^{n} ⇒ [r and s are coprime] r | a_{0}.

Similarly, a_{n}r^{n} + a_{n-1}r^{n-1}s + ··· + a_{1}rs^{n-1} + a_{0}s^{n} = 0 ⇒ s(a_{n-1}r^{n-1} + ··· + a_{1}rs^{n-2} +a_{0}s^{n-1}) = -a_{n}r^{n} ⇒ s | a_{n}r^{n} ⇒ [r and s are coprime] s | a_{n}∎

- x
^{3}-x-1, x^{3}-3x +1 are irreducibles because any root r/s must have r = ±1, s = ±1, but they are not roots. - 3x
^{3}-11x^{2}+5x-3. Possible roots can be ±1, ±3, ±^{1}⁄_{3}. The actual rational roots of the function are 1, 3, and^{-1}⁄_{3}. 3x^{3}-11x^{2}+5x-3 = (x-1)·(x-3)·(3x+1)

- f(x) = x
^{4}+3x^{2}-2x + 1 ∈ ℚ[x] is irreducible. There are two options, f(x) = linear·cubic or quadratic·quadratic.

- f(x) = linear·cubic ⇒ there exist a zero or root associated with the linear factor, but any root r/s must have r = ±1, s = ±1, but they are not roots of f(x). f(1) = 3 ≠ 0, f(-1) = 7 ≠ 0.
- x
^{4}+3x^{2}-2x + 1 = quadratic·quadratic =[f(x) is monic] (x^{2}+ax +b)(x^{2}+cx +d) = x^{4}+ (a + c)x^{3}+ (b + ac + d)x^{2}+ (ad +bc)x + bd ⇒ bd = 1, a + c = 0, b + ac + d = 3, ad + bc = -2.

bd = 1 ⇒ b = d = 1 or b = d = -1.

If b = d = 1 ⇒[ad + bc = -2] a + c = -2 ⊥ a + c = 0. Otherwise, b = d = -1 ⇒[ad + bc = -2] -a -c = -2 ⇒ a + c = 2 ⊥ a + c = 0.

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.