I need to show that if I have a reducible polynomial $f\in \mathbb{F}_{3}[X] $ of degree $5$ with no roots. Then there exists a monic irreducible polynomial of degree 2 dividing f.
My attempt:
$\mathbb{F}_{3}[X]$ is a unique factorization domain, so f can be factored uniquely by irreducible elements.
If f is factored by two polynomials $f=q_1 q_2$, then it needs to satisfy $\deg(f)=\deg(q_1)+\deg(q_2)=5$.
So if $p_n$ is an irreducible polynomial of degree n, then f can be factored in two ways:
$f=p_1p_4$ or $f=p_2p_3$
(But do we don't know anything about the multiplicity of the roots? So couldn't it in theory be factored as a multiple of the same degree polynomial? For example 1 one degree and 2 second degree polynomials?)
I'm somehow supposed to conclude that the only way it can be factored is by $f=p_2p_3$ and thus there must be a second-degree polynomial dividing f (since it is a factor of a second degree). But how do I draw this conclusion?
Best Answer
If the polynomial is reducible, it can be factored as $f=pq$ with $0\leq \deg(p),\deg(q)\lt \deg(f)$. In this case, $\deg(f)=5$. Moreover, you know that $f$ does not have any degree $1$ factors, since it has no roots and roots correspond to degree $1$ factors.
Now, the factorization of $f$ into irreducibles will correspond to a partition of $5$, given by the degrees of the irreducibles. The partitions of $5$ are: $$\begin{align*} 5 &= 5\\ 5 &= 4+1\\ 5 &= 3+2\\ 5 &= 3+1+1\\ 5 &= 2+2+1\\ 5 &= 2+1+1+1+1\\ 5 &= 1+1+1+1+1. \end{align*}$$
Of these, only two of them correspond to factorizations with no factor of degree $1$, and one of those two corresponds to the case where $f$ is irreducible. That leaves only one possibility.