Understanding the proof of: $p$ is prime and $a \in \mathbb{Z}_p \implies x^p – a \in \mathbb{Z}_p$ is reducible.

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I am trying to understand the proof of: $p$ is prime and $a \in \mathbb{Z}_p \implies x^p – a \in \mathbb{Z}_p$ is reducible. In the proof, it is shown that $a$ is a zero of $x^p – a$ which implies that $(x-a)$ is a factor of $x^p – a$. I follow the proof completely till this point. However, then the proof claims that $(x-a)$ being a factor of $x^p – a$ implies that $x^p – a$ is reducible over $\mathbb{Z}_p$. I don't understand how this is true; I can see why this would be true, intuitively, but I am trying to develop a formal proof for a more general statement. To be specific, inspired from this question, I am trying to prove:

If $F$ is a field s.t. $f(x) \in F[x]$ and $f(x)$ has a factor $(x-a) \implies$ $f(x)$ is reducible over $F$.

Proof. Suppose $F$ is a field s.t. $f(x) \in F[x]$ and that $f(x)$ has a factor $(x-a)$. Then, I thought of using the division algorithm, but the division algorithm is valid only for finite polynomials, not for any arbitrary polynomial in $F[x]$. So, I am having trouble completing this proof.

My question: Is what I am trying to prove even true? How can I complete the proof above?

Edit: The book that I am reading only has a definition of irreducible polynomials, which is as follows:

A nonconstant polynomial $f(x) \in F[x]$ is irreducible over $F$ if $f(x)$ cannot be expressed as a product $g(x)h(x)$ of two polynomials $g(x)$ and $h(x)$ in $F[x]$ both of lower degree than the degree of $f(x)$.

Based on this, I think it would be valid to deduce the following definition:

A nonconstant polynomial $f(x) \in F[x]$ is reducible over $F$ if $f(x)$ can be expressed as a product $g(x)h(x)$ of two polynomials $g(x)$ and $h(x)$ in $F[x]$ both of lower degree than the degree of $f(x)$.

Best Answer

Then, I thought of using the division algorithm, but the division algorithm is valid only for finite polynomials, not for any arbitrary polynomial in $F[x]$.

I'm not sure what you mean by "finite polynomials". In what way is $x-a$ not finite?

We shall assume $\deg f \ge 2$.

Since you have seen that $x - a$ is a factor, we must have that $f(x) = (x - a)g(x)$ for some $g(x) \in F[x]$. (By definition of "factor".)

Note that $\deg g = \deg f - 1$ and thus, both $g$ and $x-a$ have a degree strictly less than that of $f$ showing that $f$ is reducible.

Additional note: In $\Bbb Z/p\Bbb Z$, we have the identity that $a^p = a$, so you can explicitly factor $x^p - a$ as $(x - a)^p$.

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