Exercise 10.2.2

Proof. (Ex 10.2.2)

(a)

As in Example 9.1.6, using Proposition 9.1.5, we obtain $$x^p-1={\Phi }_1(x){\Phi }_p(x)\mathrm{and}{x}^{p^2}-1={\Phi }_1(x){\Phi }_p(x){\Phi }_{p^2}(x).$$

Thus

$$x^{p^2}-1=(x^p-1){\Phi }_{p^2}(x).$$

The substitution $x\to x+1$ gives

$${(x+1)}^{p^2}-1=({(x+1)}^p-1){\Phi }_{p^2}(x+1).$$

(b)

We know that $p\mid \left(\genfrac{}{}{0.0pt}{}{p}{k}\right),1\le k\le p-1$, so $${(x+1)}^p\equiv x^p+1(\mathit{mod}p)$$

Therefore

$${(x+1)}^{p^2}={[{(x+1)}^p]}^p\equiv {(x^p+1)}^p\equiv x^{p^2}+1(\mathit{mod}p).$$

The reduction modulo $p$ of the equality proved in part (a) gives

$$x^{p^2}=x^p{\bar{\Phi }}_{p^2}(x+1).$$

(c)

As ${\bar{\Phi }}_{p^2}(x+1)=x^{p^2-p}$, all the the coefficients of ${\Phi }_{p^2}(x+1)$ are divisible by $p$, except the leading coefficient. Moreover ${\Phi }_{p^2}(1)=p$, so the constant coefficient of ${\Phi }_{p^2}(x+1)$ is not divisible by $p^2$, so the Schönemann-Eisenstein criterion shows that ${\Phi }_{p^2}(x+1)$ is irreducible over $\mathbb{Q}$, and this is equivalent to the irreducibility of ${\Phi }_{p^2}(x)$ over $\mathbb{Q}$.