Definition of a prime element from Lidl & Niederreiter Finite Fields

Question

Update: Probably, I haven't provided enough details to make my previous question clear. I will add more details to motivate my question now, and I don't think it can be considered a duplicate anymore because it is now self-contained.

Definition 1. Let $R$ be a commutative ring with a multiplicative identity. An element $a \in R$ is called a prime element if:

1. $a$ is not an invertible element.
2. If $b \mid a$, then $b$ is either an invertible element or is associated with $a$.

Suppose $F$ is a field and $F[x]$ is a polynomial ring. Then, authors define the notion of the degree of a polynomial. In particular, the degree of the zero polynomial is $-\infty$, and the degree of a constant nonzero polynomial is $0$.

Then, authors claim that prime elements in $F[x]$ play an important role, and those elements are called irreducible polynomials.

They provide an equivalent definition of an irreducible polynomial as the following: $f \in F[x]$ is called irreducible in $F[x]$ if:

1. $\deg(f) > 0$,
2. if $f = gh$, then either $g$ or $h$ is a constant polynomial.

If we stick to Definition 1 of a prime element, then I was not able to show that the definitions of irreducible polynomials are equivalent.

However, if we slightly modify the definition of a prime element, then in that case they are equivalent.

Definition 2. Let $R$ be a commutative ring with a multiplicative identity. An element $a \in R$ is called a prime element if:

1. $a$ is a nonzero element,
2. $a$ is not an invertible element,
3. if $b \mid a$, then $b$ is either an invertible element or is associated with $a$.

Proof (using Definition 2).
$\boxed{\Rightarrow}$ Suppose $f$ is a prime element of $F[x]$. Then, by Definition 2, $f \neq 0$.

Since $f$ is not a unit, $f \notin F^* := F \setminus \{0\}$. Thus, $\deg(f) > 0$.

Suppose $f = gh$. Then $g \mid f$. If $g$ is a unit, then $g$ is a constant polynomial. If $g$ is associated with $f$, then $g = af$, where $a \in F^*$. Hence, $f = afh$, which implies $f(1 - ah) = 0$. Since $f \neq 0$, it follows that $1 = ah$, and thus $h$ is a unit, implying $h$ is a constant polynomial.

$\boxed{\Leftarrow}$ Suppose that $f \in F[x]$ is an irreducible polynomial.

Since $\deg(f) > 0$, $f$ is non-constant and hence $f$ is not invertible.

If $g \mid f$, then $f = gh$. If $g$ is constant (notice that $g \neq 0$ as $f \neq 0$), then $g$ is a unit.

If $h$ is constant (notice that $h \neq 0$), then $h$ is a unit, and hence $f$ and $g$ are associated.

Is my reasoning correct?

Answer 1

If $R$ does not have zero divisors, $\{0\}$ is a prime ideal. If $R$ does have zero divisors $x,y\neq0$, that is, $xy=0$, you have such elements $x,y\neq\{0\}$ with $xy\in\{0\}$. This means that $\{0\}$ is not a prime ideal. The definition of a prime element $x$ I know is that the principal ideal $R\cdot $x is a prime ideal.

A shorter comment on your question gave this answer too, with a little less detail.