# Monic Polynomials

1. Jun 17, 2007

### rookandpawn

1. The problem statement, all variables and given/known data

Show x^2 + 2 in Z_5[x] is irreducible. This is before the section on the factor theorem (j is a root -> (x-j) is a factor). So I'm not so sure I want to start checking for zero's since "its not available" per se.

2. Relevant equations

3. The attempt at a solution

Suppose it was reducible. Then x^2 + 2 = (ax + b) (a^-1x + c). First, I have a feeling that the products are monic, so i dont have to worry about the a, and a^-1.

But why would the products would be monic (if it were reducible, even though its not)?

I know if a poly can be factored into monics, the poly itself must be monic, (which is clear since the leading coefficient will turn out to be 1 upon collection of like powers.) When I say monic factors, i mean "just" monic factors, with no non-zero coefficients out in front.

But I can't say that x^2 + 2 *can be factored into monics* to begin with, since I can't assume this. Please help me with showing why its okay to assume if x^2 + 2 over Z_5 were reducible, it would have monic factors.

2. Jun 17, 2007

### rookandpawn

Nevermind, they are all units in Z_5. I can't believe i overlooked this.

x^2 + 2 = (ax + b) (a^-1x + c) = a(x + b/a) a^-1(x + ca) = (x + b/a)(x + ca).

3. Jun 17, 2007

### morphism

It doesn't have to have monic factors.

Take for example x^2 + 1. Over Z_5, it factors as (2x+1)(3x+1).

4. Jun 17, 2007

### rookandpawn

(2x + 1) = 2(x + 3)
(3x + 1) = 3(x + 2)
= 6(x + 3)(x + 2) = (x+3)(x+2) = x^2 + 5x + 6 = x^2 + 1

so x^2 + 1 = (x + 2)(x + 3).

5. Jun 17, 2007

### morphism

Right. You can assume without loss of generality that the factors are monic. I was just saying that you can write down a non-monic factorization.

6. Jun 17, 2007

### rookandpawn

Oh okay. Heh.