Exercise 4.2.2

Proof.

(a)

Let $f(x)={(x-a)}^n+\mathit{pF}(x)$, where $a\in \mathbb{Z}$, and $p$ is prime. We show that $f$ is irreducible. If we suppose on the contrary that $f$ is reducible over $\mathbb{Q}$, then by Corollary 4.2.1 $$f=\mathit{gh},g,h\in \mathbb{Z}[x],k=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)\ge 1,l=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)\ge 1.$$

As $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(F)\le n$, $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)\le n$, and as the coefficient of $x^n$ in $f$ is congruent to 1 modulo $p$, it is nonzero, so $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=n$, and $k+l=n$.

Write $\bar{f}\in {𝔽}_p[x]$ the reduction modulo $p$ of $f$, and write $\bar{a}={[a]}_p$ the class of $a\in \mathbb{Z}$ modulo $p$.

The application

$$\begin{array}{llll}\hfill \phi :\mathbb{Z}[x]& \to {𝔽}_p[x]& \hfill & \\ \hfill q=\underset{i=0}{\overset{d}{\sum }}{a}_i{x}^i& \mapsto \bar{q}=\underset{i=0}{\overset{d}{\sum }}\overline{a_i}{x}^i& \hfill & \end{array}$$

is a ring homomorphism, so $\bar{f}=\overline{\mathit{gh}}=\bar{g}\bar{h}$.

Thus

$$\bar{f}={(x-\bar{a})}^n=\bar{g}\bar{h}$$

As $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{g})\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g),\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{h})\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)$ and as $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{g})+\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{h})=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({(x-\bar{a})}^n)=n=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)+\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)$, we conclude that $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{g})=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=k,\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{h})=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)=l$.

$x-\bar{a}$ is irreducible in ${𝔽}_p[x]$, as every polynomial of degree 1. ${𝔽}_p$ being a field, the unicity of the decomposition in irreducible factors in the principal ideal domain ${𝔽}_p[x]$ shows that the only irreducible factors of $\bar{g},\bar{h}$ are associate to powers of $x-\bar{a}$:

$$\bar{g}=\bar{u}{(x-\bar{a})}^k,\bar{h}=\bar{v}{(x-\bar{a})}^l,\bar{u},\bar{v}\in {𝔽}_p^{\text{∗}}.$$

Hence there exist polynomials $G,H\in \mathbb{Z}[x]$ such that

$$g=u{(x-a)}^k+\mathit{pG}(x),h=v{(x-a)}^l+\mathit{pH}(x).$$

Consequently

$$f(x)={(x-a)}^n+\mathit{pF}(x)=[u{(x-a)}^k+\mathit{pG}(x)][v{(x-a)}^l+\mathit{pH}(x)].$$

As $k\ge 1,l\ge 1$, ${(x-a)}^k$ and ${(x-a)}^l$ have $a$ as a root, thus

$$f(a)=\mathit{pF}(a)=p^{2}G(a)H(a).$$

Then $F(a)=\mathit{pG}(a)H(a)$ is divisible by $p$, in contradiction with the hypothesis $p\nmid F(a)$.

Conclusion: $f\in \mathbb{Z}[x]$ is not a product of nonconstant polynomials in $\mathbb{Z}[x]$. By Corollary 4.2.1, $f$ is irreducible over $\mathbb{Q}$.

(b)

More generally, suppose that $u\in \mathbb{Z}[x]$ is such that $\bar{u}$ is irreducible over ${𝔽}_p$, and that $f(x)=u{(x)}^n+\mathit{pF}(x)$, $F(x)\in \mathbb{Z}[x],\bar{u}\wedge \bar{F}=1$ and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(F)\le n\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)$.

We must suppose also that the leading coefficient of $u$ is not divisible by $p$, so $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{u})=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)$.

Then $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)\le n\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)$, and the coefficient of the monomial of degree $n\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)$ being nonzero modulo $p$, $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=n\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)=n\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{u})=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{f})$.

If we suppose $f$ reducible, then $f=\mathit{gh},k=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)\ge 1,l=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)\ge 1$, which implies as in (a)

$$\bar{f}={\bar{u}}^n=\bar{g}\bar{h}.$$

Since $\bar{u}$ is irreducible,

$$\bar{g}=\bar{c}{\bar{u}}^i,\bar{h}=\bar{d}{\bar{u}}^j,i,j\in \mathbb{N},\bar{c},\bar{d}\in {𝔽}_p$$

As $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{g})\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g),\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{h})\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)$, and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{g})+\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{h})=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{f})=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)+\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)$, we conclude $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{g})=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)\ge 1,\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\bar{h})=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)\ge 1$. Consequently $i\ge 1,j\ge 1$.

There exist polynomials $G,H\in \mathbb{Z}[x]$ such that

$$g=c{u}^i+\mathit{pG},h=d{u}^j+\mathit{pH}.$$

Thus

$$f=u^n+\mathit{pF}=(c{u}^i+\mathit{pG})(d{u}^j+\mathit{pH}).$$

As $i\ge 1,j\ge 1$, $u$ divides $\mathit{pF}-p^2\mathit{GH}$ in $\mathbb{Z}[x]$, so there exists $v\in \mathbb{Z}[x]$ such that

$$\mathit{uv}=p(F-\mathit{pGH}).$$

As $\bar{u}\bar{v}=0$, and $\bar{u}\ne 0$ in the integral domain ${𝔽}_p[x]$, then $\bar{v}=0$: all the coefficients of $v$ are divisible by $p$, thus $w=v∕p\in \mathbb{Z}[x]$, and

$$\mathit{uw}=F-\mathit{pGH},\bar{u}\bar{w}=\bar{F}.$$

Hence $\bar{u}\mid \bar{F}$, in contradiction with the hypothesis $\bar{u}\wedge \bar{F}=1$.

$f=u^n+\mathit{pF}$ is so irreducible.