1. ## irreducible polynomial

Prove that for every prime p, the polynomial x^p-x-1 is irreducible over Z_p

2. Originally Posted by mandy123
Prove that for every prime p, the polynomial x^p-x-1 is irreducible over Z_p

We will prove something stronger that if $a\in \mathbb{Z}_p^{\times}$ then $f(x) = x^p - x - a$ is irreducible over $\mathbb{Z}_p$. Notice that if $\alpha$ (in some larger field) is a root of $f(x)$ then $\alpha+1$ is a root. This is because $(\alpha + 1)^p - (\alpha + 1) - a = \alpha^p + 1 - \alpha - 1 - a = 0$. Consequently, this means $\alpha,\alpha+1,...,\alpha + (p-1)$ are the roots of $f(x)$ (since degree is $p$). Therefore, $\mathbb{Z}_p(\alpha)$ is a splitting field over $f(x)$ of $\mathbb{Z}_p$. Now say that $f(x) = g_1(x)...g_m(x)$ where $g_i(x)$ are irreducible over $\mathbb{Z}_p$. Let $\beta_i$ be any root of $g_i(x)$ and $\beta_j$ be any root of $g_j(x)$. Remember by above $\mathbb{Z}_p(\beta_i)$ and $\mathbb{Z}_p(\beta_j)$ are splitting fields over $f(x)$ of $\mathbb{Z}_p$. Thus, $\mathbb{Z}_p(\beta_i) = \mathbb{Z}_p(\beta_j)$. This means, obviously, that $[\mathbb{Z}_p(\beta_i) : \mathbb{Z}_p] = [\mathbb{Z}_p(\beta_j):\mathbb{Z}_p] \implies \deg (\beta_i, \mathbb{Z}_p) = \deg (\beta_j, \mathbb{Z}_p)$ and so $\deg g_i(x) = \deg g_j(x)$. In other words, all polynomials in the factorization of $f(x)$ have the same degree. But $\deg f(x) = p$ a prime! Thus, either all $g_i(x)$ are linear or else $f(x)$ is its own irreducible polynomial factorization. It cannot be that all $g_i(x)$ are linear for that would imply that the roots of $f(x)$ lie in $\mathbb{Z}_p$ which is impossible because $x_0^p - x_0 - a = -a\not = 0$ for all $x_0 \in \mathbb{Z}_p$ by Fermat's theorem. Thus, $f(x)$ is irreducible.

3. Hello,

I'd like to know if we could use this property :
$\text{In } \mathbb{Z}/p \mathbb{Z},~ \forall x,~ x^p-x= \prod_{k=0}^{n-1} (x-k)$
I cannot get the relationship yet, so if someone can enlighten me... Thanks :P

4. Originally Posted by Moo
I cannot get the relationship yet, so if someone can enlighten me... Thanks :P
Sure

A non-constant polynomial $f(x) \in F[x]$ (where $F$ is a field, like $\mathbb{Z}_p,\mathbb{Q},\mathbb{R}$) is irreducible iff we cannot write $f(x) = g(x)h(x)$ where $\deg g(x),\deg h(x) < \deg f(x)$ i.e. factorize $f(x)$ into lower degree polynomials.

If $f(x)$ is irreducible then $f(x)$ has no zeros in $F$. But the converse is not true. Consider $f(x) = x^4 + 2x^2 + 1$. Certainly $f(x)$ has no zeros in $\mathbb{Q}$. However, $f(x) = (x^2+1)(x^2+1)$. Thus, $f(x)$ is reducible.

Does that make sense?