Let $F$ be a field and $f(x) \in F[x]$ be a polynomial of degree $> 1$. If $f(a) = 0$ for some $a \in F$, then $f(x)$ is reducible over $F$.

Question

My Attempt at this is as follows:

Let $f(x) = a_0+a_1x+a_2x^2+...+a_mx^m$ $\forall$ $a_i$ $\in$ $F[X]$.

Given $f(a)=0$

$\Rightarrow$ $$f(x) = (x-a)t(x) \label{a}\tag{1}$$ such that $\deg(t(x))\geq1$,

and since $\deg(x-\alpha)=1$, we can express $f(x)$ as a product of two functions such that their degree is not zero, hence $f(x)$ is reducible.

So for proving the above result is this approach allowed? Or should I use some other way or do I have to prove the equation which I used in $\ref{a}$?

Answer 1

You should use polynomial division, otherwise you're simply stating what you want to prove. The division algorithm states that if $f(x)$ and $g(x)$ are polynomials, with $g\ne0$, then there exist unique polynomials $t(x)$ and $r(x)$ such that

1. $f(x)=g(x)t(x)+r(x)$;
2. $\deg r(x)<\deg g(x)$ (or $r=0$).

In this case we can take $g(x)=x-\alpha$, so $$ f(x)=(x-\alpha)t(x)+r(x) $$ where $r=0$ or has degree less than $\deg(x-\alpha)=1$. Therefore $r(x)=c$ is constant.

Evaluate at $\alpha$: $$ 0=f(\alpha)=(\alpha-\alpha)t(\alpha)+c=c $$ Hence $f(x)=(x-\alpha)t(x)$. Since $\deg f(x)>1$, $\deg t(x)>0$, so $f$ is reducible.

Answer 2

Your proof is correct. I would not write “s.t. $\deg t(x)\geqslant1$”. It's more specific than that (although what you wrote is correct): $\deg t(x)=m-1$. And $f(x)$ doesn't became reducible; it is reducible.

And the equality $(1)$ doesn't have to be proved (as part of this proof), since it is a standard theorem about polynomials.