Exercise 6.23

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

If $f(x) = x^n + a_1 x^{n-1} + \ldots + a_n$, $a_i \in \Z$, and $p$ is prime such that $p \mid a_i$ for $i = 1, \ldots, n$, and $p^2 
mid a_n$, show that $f(x)$ is irreducible over $\Q$ (Eisenstein's irreducibility criterion).

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\bigskip

{\bf Lemma.} If $f \in \Z[x], \deg(f) \geq 1,$ is not irreducible in $\Q[x]$, then there exist $g,h \in \Z[x] , \deg(g) \geq 1, \deg(h) \geq 1$ such that $f = gh$.
\begin{proof}(lemma)
Suppose that  $f(x)  = \sum\limits_{k=0}^n a_k x^k, a_k \in \mathbb{Z}$, is not irreducible in $\mathbb{Q}[x]$.

Then $f(x)= f_1(x)f_2(x) $, with $f_1,f_2 \in \mathbb{Q}[X]$, and $\deg(f_1)\geq 1, \deg(f_2) \geq 1$. As in Ex. 6.5, we can write $f_1(x) = \lambda p(x), f_2(x) = \mu q(x)$ where $\lambda, \mu \in \mathbb{Q}$, and $p,q \in \mathbb{Z}[X]$ are primitive. Let $
u = \lambda \mu \in \Q$ : write $
u = u/v, u\wedge v = 1, v \geq 1$. Then  $r(x) =  p(x)q(x) = \sum\limits_{k=0}^n c_k x^k$ is primitive (Ex. 6.4), and $f(x) = \frac{u}{v} r(x) =  \frac{u}{v} p(x)q(x)$.

As $vf(x) = ur(x)$, $v\mid uc_i,\ i=0,1,\ldots,n$, with $u \wedge v = 1$, so   $u \mid c_i$ for all $i$. The polynomial $r$ being primitive, $v \mid 1$, so $v = \varepsilon = \pm 1$.

Let $g(x) = \varepsilon u p(x), h(x)  = q(x)$.Then $g,h \in \Z[x]$,  $\deg(g)  \geq 1, \deg(h)  \geq 1$, and $f = gh$ is the product of  two non constant polynomials in  $\Z[x]$.

\end{proof}
\begin{proof}(Ex. 6.23)

Let
$$
\varphi : 
\left\{
\begin{array}{ccc}
  \Z[x]&   	o & \F_p[x]  \
  p(x)= a_0 +\cdots+a_n x^n& \mapsto  & \overline{p}(x) =  \overline{a_0}+\cdots+\overline{a_n}x^n,
\end{array}
ight.
$$
where $\overline{a_i}$ is the class of $a_i$ in $\F_p$.  $\varphi$ is a ring homomorphism.

We show that $f(x) = g(x) h(x),\  g,h \in \Z[x], \deg(g) \geq 1, \deg(h) \geq 1$ is impossible. Indeed in such a situation,
$$\overline{f}(x) = x^n = \overline{g}(x)\overline{h}(x).$$ 
As the only irreducible factor of $x^n$ is $x$, the unicity of the decomposition of a polynomial in irreducible factors in $\F_p[x]$ gives
$$\overline{g}(x) = \lambda x^i,\,  \overline{h}(x) = \mu x^j,\ \lambda, \mu \in \F_p, i,j \in \N.$$

As $\deg(\overline{g}) \leq \deg(g), \deg(\overline{h}) \leq \deg(h)$ and $\deg(\overline{g}) + \deg(\overline{h}) = n = \deg(f) + \deg(g)$, this implies that $i = \deg(\overline{f}) = \deg(f)$, $j=\deg(\overline{g}) = \deg(g)$, so $i\geq 1, j \geq 1$. Therefore $p \mid g(0), p \mid h(0)$, so $p^2 \mid a_n = g(0)h(0)$, which is in contradiction with the hypothesis.

From the lemma we deduce that $f(x)$ is irreducible in $\Q[x]$.
\end{proof}

Exercise 6.23

Answers

Comments

Exercise 6.23

Answers

Comments

Add answer