Thread: 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 is irreducible in

After this they challenge the reader with the following exercise:

Show that is irreducible in . taking your lead from the discussion of above. (see attached)

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

Peter

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 is irreducible in

After this they challenge the reader with the following exercise:


Show that is irreducible in . taking your lead from the discussion of above. (see attached)

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

Peter

You have the polynomial



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

where are all of the factors of p and q. so the possible zeros are

If you check these in 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



If we expand the left hand side we get



This gives us 3 equations that must hold



The first equation and the third equation force but this contradicts the 2nd equation. So the equation does not factor

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.

Thanks Deveno

Will work through your post now

Peter