Show that there are no $f_1, f_2$ such that $f_1f_2 = f$.

Question

Let $\mathbb{F}$ a finite field. Show that there's $f\in\mathbb{F}[x]$ such that $\deg(f)=2$ and there are no linear polynomials $f_1,f_2\in \mathbb{F}[x]$ such that $f_1f_2 = f$.

Hint: Define $\mathbb{F}\times \mathbb{F} \to \mathbb{F}\times \mathbb{F}$ such that:
if $a\ne b$ then $(a,b) \mapsto (c,d)$ where $a,b$ are the roots of $x^2 +cx +d$. Otherwise, $(a,a) \mapsto (-2a,a^2)$. Show this map isn't injection and use the fact that if $f\in\mathbb{F}[x]$ is a polynomial with the degree of $2$ with a single root then $f$ has the form $x^2 -2ax + a^2$.

Although the hint is pretty much straight forward I failed with getting the solution.

I'll appreciate any help.

Thanks.

Answer 1

The way the map is described in the hint is a little awkward. The idea is the following: Given two elements $a$, $b$ in $\mathbb F$, we know that $$(x-a)(x-b) = x^2 -(a+b)x + ab$$ is a polynomial of degree $2$ with roots $a$ and $b$. So we consider the map \begin{align} F\colon \mathbb F \times \mathbb F &\longrightarrow \{\, f\in \mathbb F[x] : \deg(f)=2\,\}\\ (a,b) &\longmapsto x^2 -(a+b)x + ab, \end{align} that maps $(a,b)$ to a polynomial with roots $(a,b)$. Now given any polynomial of degree $2$ in $\mathbb F$ we can make it into a monic polynomial (one with leading coefficient $1$) with the same roots. So we can assume $f=x^2+px+q$. If $f$ was to factor in monic linear factors $f=(x-a)(x-b)$, this factorization is unique up to swapping $a$ and $b$. Thus, any polynomial that factors into linear factors will appear in the image of $F$. Since $(a,b)\mapsto(-(a+b),ab)$ is not surjective, there is a polynomial $f=x^2+px+q$ that doesn't factor into linear factors.

Answer 2

Clearly, if $a\neq b$, then $(a,b)$ and $(b,a)$ are different and have the same image. Then, the map is not an injection. Since $\Bbb F\times\Bbb F$ is finite, the map is not a surjection. So there is a pair $(c_0,d_0)$ that is not the image of any pair $(a,b)$. Therefore, $x^2+c_0x+d_0$ has no roots. Since its degree is $2$, this is equivalent to be irreducible.

Remark: If $A$ and $B$ are finite sets and $f:A\to B$ is a surjection, then there is at least one element of $A$ for each element of $b\in B$, namely a preimage of $b$. But if $f$ is not an injection, then there is some $b$ which has more than one preimage. Then there are more elements in $A$ than in $B$. Of course, this situation is impossible if $A=B$.
To sum up: if $A$ is a finite set and a map $f:A\to A$ is not an injection, then it is not a surjection either.