Exercise 2.7.5 (Complete factorization)

Question

With the assumptions and notation of the preceding problem, prove that if the degree of $f(x)$ is $j$, then $q(x)$ is a constant and can be taken as the leading coefficient of $f(x)$.

Richard Ganaye · Accepted Answer

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

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

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

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

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

Note: A polynomial $q(x)$ such that $f(x)\equiv (x-a_1)\cdots (x-a_j) q(x) \pmod p$, where $j = \deg(p)$ may be a non constant polynomial. For instance
\begin{align*}
x^2 -3x+ 2 = (x-1)(x-2) \equiv (x-1)(x-2) (1 + 5 x^7) \pmod 5.
\end{align*}
Here $q(x) = 1 + 5 x^7$ and $\deg(q)  = 7 > 0$. So we must interpret the question: it is possible to find such a polynomial $q$ verifying $\deg(q) =0$.
\begin{proof} 
It is hard to reason with congruences in $\Z$. I prefer here to use the field $\F_p = \Z/p\Z$ with $p$ elements, and write $\overline{f} \in \F_p[x]$ the reduced polynomial of $f$: if $f(x) =\sum_{i=0}^d a_i x^i$ then $\overline{f}(x) = \sum_{i=0}^d [a_i]_p x^i \in \F_p[x]$.

The result of problem 4 gives:

If $\overline{f} \in \F_p[x]$ satisfies $\overline{f}(\overline{a_1}) = \overline{f}(\overline{a_2}) = \cdots = \overline{f}(\overline{a_j}) = 0$ (where the $\overline{a_i} =[a_i]_p \in \F_p$ are distinct), then there exists $\overline{q} \in \F_p[x]$ such that
$$\overline{f}(x) = (x -\overline{a_1})\cdots  (x -\overline{a_j})\overline{q}(x).$$
(This result is true if we replace $\F_p$ by any field.)

We suppose that $c$ is the leading coefficient of $f$, and that $p 
mid c$, so that $j = \deg(f) = \deg(\overline{f})$.

Then $j = \deg(f(x)) =  \deg((x -\overline{a_1})\cdots  (x -\overline{a_j})\overline{q}(x)) = j + \deg(\overline{q})$, thus $\deg(\overline{q}) = 0$, so $\overline{q} = \overline{c} \in \F_p^*$, and
$$\overline{f}(x) = \overline{c}(x -\overline{a_1})\cdots  (x -\overline{a_j}).$$
Therefore $\overline{f}(x)  =\overline{c} x^j + \cdots$, so $\overline{c}$ is the leading coefficient of $\overline{f}(x)$. We can choose $c \in \Z$ such that $[c]_p = \overline{c}$ and $c$ is the leading coefficient of $f$.

To conclude, if $\deg(f) = j$, and $c$ is the leading coefficient of $f$, where $p 
mid c$, then $f(a_1) \equiv \cdots \equiv f(a_j) \equiv 0 \pmod p$, $a_i 
ot \equiv a_j \pmod p$ if $i 
e j$, imply
$$f(x) \equiv c (x-a_1)\cdots (x-a_j) \pmod p.$$

\end{proof}

Exercise 2.7.5 (Complete factorization)

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Exercise 2.7.5 (Complete factorization)

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