Algebra Theory - Reducibility/Irreducibility of a polynomial

• Irreducibility of Polynomial

x2−4=(x−2)(x+2)over R,Q,Zx^2 - 4 = (x - 2)(x + 2) \quad \text{over } \mathbb{R}, \mathbb{Q}, \mathbb{Z}
i2=−1i^2 = -1 x2+4=x2−(−4)=x2−(2i)2=(x−2i)(x+2i)over Complexx^2 + 4 = x^2 - (-4) = x^2 - (2i)^2 = (x - 2i)(x + 2i) \quad \text{over Complex}

  • x2−4x^2 - 4 is reducible over R\mathbb{R}

  • x2+4x^2 + 4 is not reducible over R\mathbb{R}

  • x2+4x^2 + 4 is reducible over C\mathbb{C}

(Example 1)

Check reducibility of x4+x2+1over Rx^4 + x^2 + 1 \quad \text{over } \mathbb{R}

(Example 2)

Check reducibility of x4+x2+1over Complexx^4 + x^2 + 1 \quad \text{over Complex}

Factorization:

x4+x2+1=x4+2x2+1−x2=(x2+1)2−x2=(x2+1−x)(x2+1+x)=(x2−x+1)(x2+x+1)x^4 + x^2 + 1 = x^4 + 2x^2 + 1 - x^2 = (x^2 + 1)^2 - x^2 = (x^2 + 1 - x)(x^2 + 1 + x) = (x^2 - x + 1)(x^2 + x + 1) ⇒x4+x2+1=(x2−x+1)(x2+x+1)→ Reducible over R\Rightarrow x^4 + x^2 + 1 = (x^2 - x + 1)(x^2 + x + 1) \quad \text{→ Reducible over } \mathbb{R}

Roots of:

x2−x+1=0⇒x=1±3i2x^2 - x + 1 = 0 \Rightarrow x = \frac{1 \pm \sqrt{3}i}{2} x2+x+1=0⇒x=−1±3i2x^2 + x + 1 = 0 \Rightarrow x = \frac{-1 \pm \sqrt{3}i}{2}

So,

x4+x2+1=(x−1+3i2)(x−1−3i2)(x−−1+3i2)(x−−1−3i2)x^4 + x^2 + 1 = \left(x - \frac{1 + \sqrt{3}i}{2}\right) \left(x - \frac{1 - \sqrt{3}i}{2}\right) \left(x - \frac{-1 + \sqrt{3}i}{2}\right) \left(x - \frac{-1 - \sqrt{3}i}{2}\right)

So x^4 + x^2 + 1 is reducible over real numbers and in fact over integers.
But the resulting factors are further reducible over only complex numbers.

What does reducibility over rational,real,integer,complex numbers mean?
It means that the resulting factors have only rational,real,integer,complex coefficients.

Example 3:

Is x4+x2−x−1x^4 + x^2 - x - 1 reducible over R\mathbb{R}?

Consider:

x4+x2−x−1x^4 + x^2 - x - 1

Group and factor:

=x3(x+1)−1(x+1)=(x+1)(x3−1)=(x+1)(x−1)(x2+x+1)= x^3(x + 1) - 1(x + 1) = (x + 1)(x^3 - 1) = (x + 1)(x - 1)(x^2 + x + 1)

So the factorization is:

x4+x2−x−1=(x+1)(x−1)(x2+x+1)x^4 + x^2 - x - 1 = (x + 1)(x - 1)(x^2 + x + 1)

Conclusion:
Yes, it is reducible over R\mathbb{R}.

Example 4:


Prove that

x4+x3−x+1x^4 + x^3 - x + 1

is not reducible over Z\mathbb{Z} (integers)

Assume it can be reduced:

x4+x3−x+1=(x2+ax+b)(x2+cx+d)with a,b,c,d∈Zx^4 + x^3 - x + 1 = (x^2 + ax + b)(x^2 + cx + d) \quad \text{with } a, b, c, d \in \mathbb{Z}

Multiply the factors:

=x4+x3(a+c)+x2(d+ac+b)+x(cd+bc)+bd= x^4 + x^3(a + c) + x^2(d + ac + b) + x(cd + bc) + bd

Equating coefficients with original:

{a+c=1d+b+ac=0cd+bc=−1bd=1\begin{cases} a + c = 1 \\ d + b + ac = 0 \\ cd + bc = -1 \\ bd = 1 \end{cases}

From bd=1bd = 1, the possible integer pairs are:

  • b=1,d=1b = 1, d = 1

  • b=−1,d=−1b = -1, d = -1

Case 1: b=1,d=1b = 1, d = 1
Then:

  • a+c=1a + c = 1

  • ac=−2ac = -2 from d+b+ac=0d + b + ac = 0

But this gives inconsistent values (not possible for integer a,ca, c)

Case 2: b=−1,d=−1b = -1, d = -1
Then:

  • a+c=1a + c = 1

  • ac=2ac = 2

Again not possible with integer a,ca, c

Conclusion:

⇒We can’t reduce x4+x3−x+1 over Z

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