Proving Irreducibility of Polynomial Using Mod p Test and Einstein's Criterion

[STEMucator]

Homework Statement

Question 30 from contemporary abstract algebra :

http://gyazo.com/a2d039539754658b4485cf89fbddec8c

Homework Equations

I'm guessing these will be of assistance :

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

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.

 

[mfb]

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.

 

[STEMucator]

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.

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 Z_p, you get g(x)=x^p-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.

 

[STEMucator]

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 Z_p[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 Z_p[x].

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

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

Is this a better attempt?