Factorization of Polynomials - Irreducibles

Printable View

• Nov 3rd 2012, 04:49 PM
Bernhard
Factorization of Polynomials - Irreducibles
I am reading Anderson and Feil - A First Course in Abstract Algebra.

On page 56 (see attached) ANderson and Feil show that the polynomial $\displaystyle f = x^2 + 2$ is irreducible in $\displaystyle \mathbb{Q} [x]$

After this they challenge the reader with the following exercise:

Show that $\displaystyle x^4 + 2$ is irreducible in $\displaystyle \mathbb{Q} [x]$. taking your lead from the discussion of $\displaystyle x^2 + 2$ above. (see attached)

Can anyone help me to show this in the manner requested. Would appreciate the help.

Peter
• Nov 3rd 2012, 05:34 PM
TheEmptySet
Re: Factorization of Polynomials - Irreducibles
Quote:

Originally Posted by Bernhard
I am reading Anderson and Feil - A First Course in Abstract Algebra.

On page 56 (see attached) ANderson and Feil show that the polynomial $\displaystyle f = x^2 + 2$ is irreducible in $\displaystyle \mathbb{Q} [x]$

After this they challenge the reader with the following exercise:

Show that $\displaystyle x^4 + 2$ is irreducible in $\displaystyle \mathbb{Q} [x]$. taking your lead from the discussion of $\displaystyle x^2 + 2$ above. (see attached)

Can anyone help me to show this in the manner requested. Would appreciate the help.

Peter

You have the polynomial

$\displaystyle f(x)=\underbrace{x^4}_{p}+\underbrace{2}_{q}$

Then by the rational roots theorem if it has any linear factors they have to be of the form

$\displaystyle \frac{q_{i}}{p_{i}}$ where $\displaystyle p_i,q_i$ are all of the factors of p and q. so the possible zeros are $\displaystyle \pm 1, \pm 2$

If you check these in $\displaystyle f(x)$ they are not zero, it does not have any linear factors.

Now comes the harder work (it isn't too bad :) )

If it has a non trivial factorization it must have two quadratic factors. Since the leading coefficent is 1 and 2 only has two factors it would have to look like

$\displaystyle (x^2+bx+1)(x^2+cx+2)=x^4+2$

If we expand the left hand side we get

$\displaystyle x^4+(b+c)x^3+(2+b+c)x^2+(2b+c)x+2$

This gives us 3 equations that must hold

$\displaystyle b+c=0 \quad 2+b+c=0 \quad 2b+c=0$

The first equation and the third equation force $\displaystyle b=-c=0$ but this contradicts the 2nd equation. So the equation does not factor
• Nov 4th 2012, 12:00 AM
Deveno
Re: Factorization of Polynomials - Irreducibles
one need not appeal to the rational roots theorem, because Q is an ORDERED field.

therefore: q4 = (q2)2 ≥ 0 for all rational q.

if q is a root of x4 + 2, we have: q4 = -2 < 0, a contradiction.

this shows x4 + 2 has no rational roots (indeed, no REAL roots), thus no linear factors.

now suppose we had a factorization into quadratic factors:

x4 + 2 = (x2 + ax + b)(x2 + cx + d) = x4 + (a+c)x3 + (b+d+ac)x2 + (ad+bc)x + bd, with a,b,c,d rational.

this gives:

a + c = 0
b + d + ac = 0
ad + bc = 0
bd = 2

the first equation tells us c = -a. this gives by substitution into the other 3 equations:

b + d = a2
a(d - b) = 0
bd = 2

the equation a(d - b) = 0 tells us either a = 0, or b = d. we can look at each of these in turn.

a = 0 leads to: b = -d, in which case, bd = -b2 ≤ 0, and thus cannot equal 2.

on the other hand, if b = d, then we have bd = b2 = 2, in which case b = ±√2, which cannot happen because b is rational.

**********

with all due respect to TheEmptySet, his proof is incomplete (although what is missing isn't hard to show).

it may be that we have factors of the form:

(x2 + bx - 1)(x2 + cx - 2), as well.

also, the expansion seems to be incorrect, i get:

(x2 + bx + 1)(x2 + cx + 2) = x4 + (b+c)x3 + (3+bc)x2 + (2b+c)x + 2

leading to the equations:

b + c = 0
3 + bc = 0
2b + c = 0

the same contradiction does hold, though. and the first and third equations are still the same, so we have b = 2b + c - (b + c) = 0 - 0 = 0, and thus c = 0, so that:

3 + 0 = 0, which certainly is never true.
• Nov 4th 2012, 12:37 AM
Bernhard
Re: Factorization of Polynomials - Irreducibles
Thanks Deveno

Will work through your post now

Peter