# Proving irreducibility

1. Feb 24, 2013

### Zondrina

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

Question 30 from contemporary abstract algebra :

http://gyazo.com/a2d039539754658b4485cf89fbddec8c

2. Relevant equations

I'm guessing these will be of assistance :

Mod p irreducibility test : http://gyazo.com/ac7deb2940c7a3c61192d793157ad2af
Einsteins Criterion : http://gyazo.com/c2328896c2cb3a58bfc5c319cb840641

3. The attempt at a solution

Let p be a prime and suppose that f(x) is in Z[x] with deg(f(x)) ≥ 1.

$f(x) = x^{p-1} - x^{p-2} + x^{p-3} - ... - x + 1$

Let $g(x) = f(x+1) = x^{p-1} - {p\choose 1}x^{p-2} + {p\choose 2}x^{p-3} - ... - {p\choose 1}$

Now by Einsteins criterion, notice that every term except the coefficient of $x^{p-1}$ is divisible by p and the constant term ${p\choose 1}$ is not divisible by p2. Hence g(x) is irreducible over Q and we are done.

My problem with this is it seems too... straightforward. I don't know if I'm over thinking this too much, or if I've missed something crucial, but if anyone could confirm this for me it would be much appreciated.

2. Feb 24, 2013

### Staff: Mentor

If I try to use the irreducibility test, I get $\overline{g}(x)=x^{p-1}$, as all other terms are divisible by p. This is reducible.

The Eisenstein (!=Einstein) criterion can be used here.

3. Feb 24, 2013

### Zondrina

Ohh I was treating it like it had no negative terms for a moment there. Also, sorry for the misspell.

Yes yes i see what you're going for now in the end. So using the irreducibility test on f(x) in Zp, you get g(x)=xp-1? That confused me a bit. I don't see any coefficients on the terms of f(x) so I don't see how you got g(x) = f(x)modp.

Last edited: Feb 24, 2013
4. Feb 24, 2013

### Zondrina

Sorry for the double post, but I think I got this now.

If : $f(x) = x^{p-1} - x^{p-2} + x^{p-3} - ... - x + 1$

I can re-write it as : $g(x) = x^{p-1} + (-1)x^{p-2} + x^{p-3} + ... + (-1)x + 1$ for ease.

In Zp[x] then, let h(x) denote the polynomial obtained by reducing all of the coefficients of g(x) modulo p.

Therefore $h(x) = x^{p-1} + (p-1)x^{p-2} + x^{p-3} - ... + (p-1)x + 1$ in Zp[x].

Waaaait a minute here... so notice that h(x) is non-zero for all choices of x in Zp. This means that h(x) has no roots and therefore must be irreducible over Zp[x]

So since h(x) is irreducible over in Zp[x], and deg(h(x)) = deg(f(x)) we know that f(x) is also irreducible over Q.

Is this a better attempt?

Last edited: Feb 24, 2013