Exercise 6.4

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}}

A polynomial $f(x) \in \Z[x]$ is said to be primitive if the greatest common divisor of its coefficients is 1. Prove that the product of primitive polynomials is also primitive.

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 Solution 1}
\begin{proof}
Let $p(x) =\sum_{i=0}^n a_i x^i, q(x) = \sum_{j=0}^m b_j x^j$ two primitive polynomials, and $p$ a prime number. There exist a coefficient of $p(x)$ (and of $q(x)$) not divisible by $p$.
Let 
\begin{align*}
i_0 &= \min\{i \in [0,n]\ \  \vert \ a_i 
ot \equiv 0 \ [p]\}\
j_0 &= \min\{j \in [0,m]\ \vert\  b_j 
ot \equiv 0 \ [p]\}\
\end{align*}
Let $p(x) q(x) = \sum_{k=0}^{n+m} c_k x^k$. Then $c_k = \sum\limits_{i+j = k} a_i b_{j}, \ k=0,\ldots n+m$.
Then $$c_{i_0+j_0} = \sum_{i+j = i_0+j_0} a_i b_j.$$

$\bullet$ If $i<i_0$, then $a_i \equiv 0 \pmod p$.

$\bullet$ If $i>i_0$, then $j<j_0$ and $b_j \equiv 0 \pmod p$.

In the two cases $a_i b_j \equiv 0 \pmod p$, so $c_{i_0+j_0} \equiv a_{i_0} b_{j_0} \pmod p$, so $c_{j_0} 
ot \equiv 0 \pmod p$. As it's true for all primes $p$, the polynomial $p(x)q(x)$ is primitive.
\end{proof}

\bigskip

{\bf Solution 2}
\begin{proof}
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.

As $\F_p[x]$ is an integrity domain, if $p(x), q(x)$ are both primitive,
$$\overline{p(x)} 
e 0, \overline{q(x)} 
e 0 \Rightarrow \overline{p(x)q(x)} =\overline{p(x)}\,   \overline{q(x)}  
e 0.$$
As $\overline{p(x)q(x)} 
e 0$ in all fields $\F_p$, $p(x)q(x)$ is a primitive polynomial.
\end{proof}

Exercise 6.4

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Exercise 6.4

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