# How do I show that a 5 degree reducible polynomial must be divisible by some second degree irreducible polynomial?

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?

## 1 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.

• Thanks, that makes sense. I think I forgot that f doesn't have any roots along the way. – sjm23 Nov 29 '20 at 0:16
• Just to be sure. What are the reason that a 1st degree factor corresponds to a root? If I think on regular polynomials, then it makes sense. Because from the fundamental theorem of algebra, a polynomial can be factored into it's roots. and the factors will be of the form $(x-\alpha)$ which is first degree polynomial. But is there another explanation? Especially since in algebra we are dealing with all kinds of polynomials (Or should I say all kind of rings). – sjm23 Nov 29 '20 at 0:30
• @sjm23: What is a “regular polynomial”, and how does it differ from a polynomial? You have polynomials over a field. The Factor Theorem holds, which tells you that $f(a)=0$ if and only if $x-a$ divides $f(x)$. And if $bx-c$ is a factor of $f$, $b\neq 0$, then $f(x) = (bx-c)q(x)$ for some $q(x)$, and then what happens when you evaluate at $\frac{c}{b}$? – Arturo Magidin Nov 29 '20 at 0:32