Confusion in definition of associate, unit, monic associate and their use in definition of "irreducible" for polynomial rings I am having some confusion over the concepts of "associate", and "unit" in the context of polynomial rings. This is partly due to the fact that the concept of "irreducible" of polynomials are defined using different terms but end up being equivalent in meaning.  Also, what how to find associates and monic associates in different polynomial rings over $Z_p$ for different $p$.
The concept of "associate", in the case of commutative rings, it is defined as follows:

Let $F$ denote a field.  
An element $a$ in a commutative ring with identity $R$ is said to be an $\textit{associate}$ of an element $g$ of $R$ if $a=bu$ for some unit $u$.  In this case $b$ is also an associate of $a$ because $u^{-1}$ is a unit and $b=au^{-1}$. 

Question 1:  Am I right to understand that in the case of finite field $Z_p$ where $p$ is not prime, then the associate only comes from units in $Z_p$.  In the case of $p=4, 6, 8$, their associates are respectively, $\{1,3\}$, $\{1,5\}$ and $\{1,3,5,7\}$
In polynomial rings, a polynomial $f(x)\in R[x]$ is a unit in $R[x]$ if and only if $f(x)$ is a constant polynomial that is a unit in $R$. If $F$ is a field,, then the units in $F[x]$ are the nonzero constants in $F$.  The definition of "associate" for polynomial is defiend as: 

$f(x)$ is an associate of $g(x)$ in $F[x]$ if and only if $f(x)=cg(x)$ for some nonzero $c\in F$

Question 2:  For polynomials rings, associates is defined in relation to the polynomial's coefficients, it still make sense to define associates over $Z_p[x]$ where $p$ is not prime.  I asked this because from the following question 
"Every nonzero $f(x)\in F[x]$ has a unique monic associate in $F[x]$"
If this statement holds, then finding an associate of a polynomial $f(x)$ can work in every finite field $Z_p$ for all value of $p$.  But for the case of monic associate, it only make sense to talk about a polynomial's monic associate over $Z_p$ only when $p$ is prime.  The reason monic associate is different is because if I am given a polynomial $f(x)=\sum_{k=0}^{n}a_{k}x^{k}\in F[x]$, $f$'s monic associate is $c\sum_{k=0}^{n}a_{k}x^{k} = \sum_{k=0}^{n}ca_{k}x^{k}\equiv x^{k} + \{\text{sum of terms}\}$ , where $c$ and $a_{n}$ are inverse of each other.
As exercises:
monic associate for
$3x^{5}-4x^{2}+1$ in $Z_{7}[x]$ and in $Z_{12}[x]$ 
$5(3x^{5}-4x^{2}+1)=15x^{5}-20x^{2}+5 \equiv x^{5}-6x^{2}+1$ in $Z_7[x]$ Also, since every element in $Z_7[x]$ are units, we can multiply by any number $c\in Z_p$ to get a different monic associate. 
But for $Z_{12}[x]$, the units are $\{1,5,7,11\}$, so there does not exist any monic associate for $3x^{5}-4x^{2}+1$. however if the leading coefficient of a polynomial are $\{1,5,7,11\}$, then it is possible to find a monic associate because they are their own inverses.
For the case of finding associate in $Z_p[x]$, they can be found in cases for all values of $p$.   As examples: 
for $3x^{5}-4x^{2}+1$ in $Z_{5}[x]$, we can multiply this polynomial by any non zero constant congruent some other numbers for its coefficients.  $c(3x^{5}-4x^{2}+1)=c3x^{5}-c4x^{2}+c \equiv a_{5}x^{5}-a_{2}x^{2}+c$ in $Z_7[x]$. Same reasoning applies to the case of $Z_{12}[x]$
Finally for the different definitions of irreducibility.  I got two different definitions.  In the second one below, the term "units' are used.  How is that the same as "associate".  I think there a way to prove that units are also associates.  I am not sure what the correct statement is.

Let $F$ be a field.  A nonconstant polynomial $p(x)\in F(x)$ is said to be irreducible if its only divisors are its associates and the nonzero constant polynomials (units).  A nonconstant polynomial that is not irreducible is said to be reducible.


Let $D$ be an integral domain.  A polynomial $f(x)$ from $D[x]$ that is neither the zero polynomial nor a unit in $D[x]$ is said to be $\textit{irreducible over $D$}$ if, whenever $f(x)$ is expressed as a product $f(x)=g(x)h(x)$, with $g(x)$ and $h(x)$ from $D[x]$, then $g(x)$ or $h(x)$ is a unit in $D[x]$.  A nonzero non-unit element of $D[x]$ that is not irreducible over $D$ is called $\textit{reducible over $D$}$.

Thank you in advance
 A: Q1: I don't quite understand what you've said, but if $F$ is a field then every nonzero element is associate to every other nonzero element because they're all units. So the equivalence classes under the equivalence relation "$x$ is associate to $y$" are $0$ and everything else.
Q2: I don't quite understand what you've said here either. There's no question mark so I don't know what your question is. If $F$ is a field then every polynomial in $F[x]$ has a unique monic associate; this condition in fact characterizes fields. As you've correctly noticed, $\mathbb{Z}/12$ is not a field.
Q3: The two definitions are equivalent. I don't know what you mean by "units are also associates." The grammar of the term "associate" is that it takes as input two elements; it's "$x$ is associate to $y$," so it doesn't mean anything to say that $x$ is associate without specifying a $y$ that it's associate to.
To prove $\Rightarrow$ (the first definition implies the second), suppose $f$ is irreducible according to the first definition. If $f(x) = g(x) h(x)$ then $g, h$ are both divisors of $f$, so by hypothesis $g, h$ must be either associates of $f$ or units. If $g$ is an associate of $f$ then $h$ is a unit and if $g$ is a unit then $h$ is an associate of $f$; moreover the same is true with $g$ and $h$ swapped. So at least (in fact exactly) one of $g$ and $h$ is a unit, which means the second definition holds.
To prove $\Leftarrow$ (the second definition implies the second), suppose $f$ is irreducible according to the second definition (and that $D$ is a field). If $g$ is a divisor of $f$, we want to show that it's either a unit or an associate of $f$. We have $f = gh$ for some polynomial $h$, and by hypothesis either $g$ or $h$ is a unit. If $g$ is a unit then we're done and if $h$ is a unit then $g$ is an associate (just like in the previous proof). So the first definition holds.
