Intro
This is the final part of the problem 9.4.20 (e), (d) from Dummit and Foote's Abstract Algebra. They formulate it as "Determine all the factorizations of $f(x)=x$ in $\Bbb{Z}/n\Bbb{Z}[x]$ where $n$ is the product of $k$ distinct primes" and ask to show that $x=(10x+21)(15x+16)(6x+25)$ in $\Bbb{Z}/30\Bbb{Z}[x]$.
$\Bbb{Z}/n\Bbb{Z}[x]$ is not an integral domain when $k>1$.
The simplest case when $n=6=2\cdot3$, so that $x$ is factorized in $\Bbb{Z}/6\Bbb{Z}[x]$ was previously considered here, but the proposed solution "$1⋅x,\ 5(5x),\ (3+4x)(3x+4),\ (3x+2)(3+2x)$" is not entirely correct.
${1,5}$ are units in $\Bbb{Z}/6\Bbb{Z}[x]$.
If we are to list all the factorizations, then also $5(3x+4)(2x+3),\ 5(3x+2)(4x+3)$ should be added to the given list.
If we are to list only the factorizations unique up to multiplication by units, then together with an obvious $5(5x)$ (which the author did mention) one of the other two factorizations, say $(3x+2)(3+2x)$, must be excluded from the given list, because $(3+4x)(3x+4)=1\cdot(3+4x)(3x+4)=5\cdot5\cdot(3+4x)(3x+4)=(5(3+4x))(5(3x+4))=(15+20x)(15x+20)=(3+2x)(3x+2)$ in $\Bbb{Z}/6\Bbb{Z}[x]$.
⦁ I constrain the argument to factorization with maximal number of nonconstant factors unique up to multiplication by units (up to associates) since other factorizations (with less number of nonconstant factors) can be obtained from it. I decided to first explore the general case $\Bbb{Z}/n\Bbb{Z}[x]$ and then apply its results to $\Bbb{Z}/30\Bbb{Z}[x]$. By "associates" below I mean their usual definition.
Part 1
Let $n=p_a p_b…p_j p_k$ with $p_i$-s being distinct primes in $\Bbb{Z}$ numbered alphabetically. According to Chinese Remainder Theorem (CRT), $$\Bbb{Z}/n\Bbb{Z}[x]≅\Bbb{Z}/p_a \Bbb{Z}[x]×\Bbb{Z}/p_b \Bbb{Z}[x]×…×\Bbb{Z}/p_k \Bbb{Z}[x]$$(this is almost trivial to show so I omit the proof). Denote this isomorphism as $π$.
The mapping of $x=(1\ mod\ n)x∈\Bbb{Z}/n\Bbb{Z}[x]$ under $π$ is $(1\ mod \ n)x↔((1\ mod\ p_a )x,\ (1\ mod\ p_b )x,\ …,\ (1\ mod\ p_k )x)$ or $$x↔(x,x,…,x)$$
hence the image $\overline{f(x)}$ of $f(x)=x$ in all the quotient rings $\Bbb{Z}/p_i\Bbb{Z}[x]$ is $\overline{f(x)}=x$.
Claim 1: $f(x)=x$ has no factor of degree $≥2$ in $\Bbb{Z}/n\Bbb{Z}[x]$.
Suppose the contrary: $f(x)=x$ has a factor of degree $≥2$. Then the leading coefficient $K$ of this factor is nonzero in $\Bbb{Z}/n\Bbb{Z}$, hence there is at least one prime $p_i$ such that $p_i∤K$ in $\Bbb{Z}$, therefore $K∉(p_i)$. Then in the field $\Bbb{Z}/(p_i)=\Bbb{Z}/p_i\Bbb{Z}$ we have that $K≠0$, hence (because there are no zero divisors in a field) $deg\ \overline{f(x)}>1$ in $\Bbb{Z}/p_i\Bbb{Z}[x]$, hence $\overline{f(x)}≠x$, a contradiction. ◼
So, the factorization of $x$ in $\Bbb{Z}/n\Bbb{Z}[x]$ can be written as:
$$x=C(a_1 x+a_0 )(b_1 x+b_0 )(c_1 x+c_0 )…$$
where $C$ is some constant polynomial, i.e., $C∈\Bbb{Z}/n\Bbb{Z}$.
Claim 2: The constant factor $C$ in the factorization of $x$ in $\Bbb{Z}/n\Bbb{Z}[x]$ is a unit in $\Bbb{Z}/n\Bbb{Z}$.
Note that $C∉(p_i)$ for each prime $p_i$ since otherwise there would be $\overline{C}=\overline{0}$ in the field $\Bbb{Z}/(p_i)=\Bbb{Z}/p_i\Bbb{Z}$ and hence $x=\overline{0}(\overline{a_1}x+\overline{a_0})(\overline{b_1}x+\overline{b_0})…=\overline{0}$ in $\Bbb{Z}/p_i\Bbb{Z}[x]$, a contradiction. Therefore, $p_i∤C$ in the ring $\Bbb{Z}$ for all primes $p_i$ so that $g.c.d.(C,n)=1$ in the ring $\Bbb{Z}$, hence $C∈(\Bbb{Z}/n\Bbb{Z})^×$, i.e., $C$ is a unit in $\Bbb{Z}/n\Bbb{Z}$. ◼
Therefore, taking any factor $f_1x+f_0$ in the factorization of $x$ we can write $x=C(a_1 x+a_0 )(b_1 x+b_0 )...(f_1 x+c_f )…=(a_1 x+a_0 )(b_1 x+b_0 )...(C(f_1 x+f_0 ))…=(a_1 x+a_0 )(b_1 x+b_0 )...(Cf_1 x+Cf_0)...=(a_1 x+a_0 )(b_1 x+b_0 )...$ with $Cf_1 x+Cf_0$ being associate to $f_1 x+f_0$. So, up to associates of its factors $x$ in $\Bbb{Z}/n\Bbb{Z}[x]$ can be written as $$x=(a_1 x+a_0 )(b_1 x+b_0 )(c_1 x+c_0 )…$$,
i.e., with the constant factor being $1$.
Claim 3: $x$ has at most $k$ nonconstant factors in $\Bbb{Z}/n\Bbb{Z}[x]$.
The isomorphism $π$ maps $x$ to the $k$-tuple $(x,x,…,x)$ with each $i$-th coordinate $x∈\Bbb{Z}/p_i\Bbb{Z}[x]$. This $k$-tuple can be written as a product of at most $k$ nonconstant tuples-factors in $\Bbb{Z}/p_a \Bbb{Z}[x]×\Bbb{Z}/p_b \Bbb{Z}[x]×…×\Bbb{Z}/p_k\Bbb{Z}[x]$, since otherwise (because all $\Bbb{Z}/p_i\Bbb{Z}$ are fields) at least one coordinate of our $k$-tuple $(x,x,…,x)$ would be of degree $>1$ hence $≠x$, a contradiction. So, $(x,x,…,x)$ has at most $k$ nonconstant factors in $\Bbb{Z}/p_a \Bbb{Z}[x]×\Bbb{Z}/p_b \Bbb{Z}[x]×…×\Bbb{Z}/p_k\Bbb{Z}[x]$, hence by the isomorphism $π$ follows that $x$ has at most $k$ nonconstant factors in $\Bbb{Z}/n\Bbb{Z}[x]$. ◼
So, up to associates a factorization of $x$ in $\Bbb{Z}/n\Bbb{Z}[x]$ can be written as: $$x=(a_1 x+a_0)(b_1 x+b_0)…(k_1 x+k_0)$$
Let $(x,x,…,x)$ be a product of maximal possible number $k$ of nonconstant factors in $\Bbb{Z}/p_a \Bbb{Z}[x]×\Bbb{Z}/p_b \Bbb{Z}[x]×…×\Bbb{Z}/p_k\Bbb{Z}[x]$, i.e., $(x,x,…,x)=q_a (x) q_b (x)…q_k (x)$, where $q_i(x)$ is a $k$-tuple. Since $\Bbb{Z}/p_i\Bbb{Z}$ are fields and so contain no zero divisors, we necessarily have $q_i(x)=(C_a,C_b,…,C_h,C_ix,C_j,…,C_k)$ [after renumbering $q_i(x)$-s if necessary so as to have exactly $i$-th coordinate nonconstant in each $q_i(x)$], that is, only one coordinate is nonconstant and for any pair $q_i (x),q_j (x)$ the non-constant coordinates $i,j$ are distinct; furthermore, this nonconstant coordinate is necessarily of degree 1 and has no constant term, i.e., is exactly of the form $C_i x$.
Thus,
$$(x,x,…,x)=(C_{1a} x, C_{1b},…,C_{1k})(C_{2a}, C_{2b}x,…,C_{2k})…(C_{sa},C_{sb},…,C_{sk} x) ⇔ \\ (x,x,…,x)=(\prod C_{ia},\prod C_{ib},...,\prod C_{ik})(x, 1, …, 1)(1, x, 1,…,1)…(1, 1, …, 1, x)$$
where $C_{if}$ is a unit in $\Bbb{Z}/p_f\Bbb{Z}$ (because $\Bbb{Z}/p_f\Bbb{Z}$ is a field), so that $(\prod C_{ia},\prod C_{ib},...,\prod C_{ik})=u$ is a unit in $\Bbb{Z}/p_a \Bbb{Z}[x]×\Bbb{Z}/p_b \Bbb{Z}[x]×…×\Bbb{Z}/p_k\Bbb{Z}[x]$.
So the mapping $x↔(x,x,…,x)$ by the isomorphism $π$ can be written as: $$1(a_1 x+a_0)(b_1 x+b_0)…(k_1 x+k_0)=u(x, 1, …, 1)(1, x, 1,…,1)…(1, 1, …, 1, x)$$
Since isomorphism maps units to units, $1↔u$. Since isomorphism maps identity to identity, it can only be that $1↔(1,…,1)$, so that $u=(1,…,1)$.
⦁ Thus, when considering factorizations of $x$ in $\Bbb{Z}/n\Bbb{Z}[x]$ unique up to associates, the mapping $x↔(x,x,…,x)$ performed by the isomorphism $π$ can be considered as: $$(a_1 x+a_0 )(b_1 x+b_0 )…(k_1 x+k_0 )↔(x,1,…,1)(1,x,1,…,1)…(1,…,1,x)\ [map]$$ where each factor $f_1 x+f_0$ on the left is mapped bijectively (because of the isomorphism) to some $k$-tuple on the right. Therefore, if $n=p_a p_b…p_j p_k$, there is only one unique up to associates factorization of $x$ in $\Bbb{Z}/n\Bbb{Z}[x]$ with maximal possible number $k$ of factors. [result]
Now our goal is to find these factors. Consider any factor $f_1x+f_0$. The isomorphism $π$ maps it as $$f_1x+f_0↔(\ (f_1+ (p_a) )x+f_0+ (p_a),… ,(f_1+ (p_f) )x+f_0+ (p_f) ,… ,(f_1+ (p_k) )x+f_0+(p_k)\ ) ⇔ \\ f_1x+f_0↔(\ (f_1\ mod\ p_a )x+f_0\ mod\ p_a,… ,(f_1\ mod\ p_f\ )x+f_0\ mod\ p_f,… ,(f_1\ mod\ p_k )x+f_0\ mod\ p_k)\ )$$
where the ideals $(p_i)$ are ideals in $\Bbb{Z}/n\Bbb{Z}$.
Since each $k$-tuple on the right of the $[map]$ above has exactly one nonconstant coordinate $x$ and all the other coordinates equal to $1$, there is exactly one prime, say $p_f$ (after renaming if necessary), such that $f_1∉(p_f)$, furthermore, $f_1≡1\ (mod\ p_f)$, whereas $f_1∈(p_a), (p_b),…,(p_e), (p_g),…,(p_k)$, so that $f_1∈(p_a)⋂ (p_b)⋂…⋂(p_e)⋂(p_g)⋂…⋂(p_k)$, or, since the ideals $(p_a), (p_b),…,(p_e), (p_g),…,(p_k)$ are pairwise comaximal in the commutative ring $\Bbb{Z}/n\Bbb{Z}$, $f_1∈(p_a)(p_b)…(p_e)(p_g)…(p_k)=(p_ap_b…p_ep_g…p_k)$. So, $f_1=p_ap_b…p_ep_g…p_kα$ for some $α∈\Bbb{Z}/n\Bbb{Z}$, hence $p_ap_b…p_ep_g…p_kα≡1\ (mod\ p_f)$, therefore, both $p_a p_b…p_e p_g…p_k,α∈(\Bbb{Z}/p_f\Bbb{Z})^×$. Since $α∈\Bbb{Z}/p_f\Bbb{Z}$, also $α∈\Bbb{Z}/n\Bbb{Z}$ and we can find it as $α=(p_a p_b…p_e p_g…p_k )^{-1}\ mod\ p_f$, the multiplicative inverse of $p_a p_b…p_e p_g…p_k$ modulo $p_f$.
For the similar reason, $f_0∈(p_f)$, whereas $f_0∉(p_a), (p_b),…,(p_e), (p_g),…,(p_k)$, furthermore, $f_0≡1\ (mod\ p_i)\ ∀i≠f$. The ideals $(p_f ),(p_a p_b…p_e p_g…p_k)$ are comaximal in $\Bbb{Z}/n\Bbb{Z}$, therefore, $(p_f )+(p_a p_b…p_e p_g…p_k)=(1)$, hence $p_f γ-p_a p_b…p_e p_g…p_k β=1$ for some $β,γ∈\Bbb{Z}/n\Bbb{Z}$, hence $p_f γ=p_a p_b…p_e p_g…p_k β+1$.Therefore, $p_f γ≡1\ (mod\ p_i )\ ∀i≠f$, so that $p_f γ$ satisfies the requirements for $f_0$. Since $f_0∈(p_f)$ and, obviously, $p_f γ∈(p_f )$ we can assume $f_0=p_f γ ⇔ f_0=p_a p_b…p_e p_g…p_k β+1$, where $β<p_f$ so as not to wrap around in $\Bbb{Z}/n\Bbb{Z}$. Such $β∈\Bbb{Z}/n\Bbb{Z}$ exists (as was just shown) and is necessarily unique (otherwise there would be two distinct factors $f_1 x+f_0$ being mapped to the same $k$-tuple by the isomorphism $π$, a contradiction).
Thus (*):
$f_1=p_a p_b…p_e p_g…p_k α$, where $α=(p_a p_b…p_e p_g…p_k )^{-1} mod\ p_f$,
$f_0=p_a p_b…p_e p_g…p_k β+1$, where $β<p_f$ such that $f_0∈(p_f )$
Part 2
Now let's apply the results obtained to factorization of $x$ in $\Bbb{Z}/30\Bbb{Z}[x]$ as an example using equations (*).
$30=2\cdot3\cdot5$, where $p_a=2,\ p_b=3,\ p_c=5$ and the number of primes $k=3$. Hence $x$ has (Claim 3) at most $3$ nonconstant linear (Claim 1) factors and so up to associates is factorized (Claim 2) as $x=(a_1 x+a_0)(b_1 x+b_0)(c_1 x+c_0)$.
$a_1=p_b p_c α$, where $α=(p_b p_c )^{-1} mod\ p_a$, hence $a_1=3\cdot5\cdotα$, where $α=(3\cdot5)^{-1} mod\ 2=(1\cdot1)^{-1}\ mod\ 2=1$, so that $a_1=3\cdot5\cdot1=15$.
$a_0=p_b p_c β+1$, where $β<p_a$ such that $a_0∈(p_a)$, hence $a_0=3\cdot5\cdotβ+1$, where $β<2$ such that $a_0=15β+1∈(2)$. Here $β=1$, so that $a_0=3\cdot5\cdot1+1=16∈(2)$.
Thus, $a_1 x+a_0=15x+16$.
$b_1=p_a p_c α$, where $α=(p_a p_c)^{-1} mod\ p_b$, hence $b_1=2\cdot5\cdotα$, where $α=(2\cdot5)^{-1} mod\ 3=(2\cdot2)^{-1} mod\ 3=1$, so that $b_1=2\cdot5\cdot1=10$.
$b_0=p_a p_c β+1$, where $β<p_b$ such that $b_0∈(p_b)$, hence $b_0=2\cdot5\cdotβ+1$, where $β<3$ such that $b_0=10β+1∈(3)$. Here $β=2$, so that $b_0=2\cdot5\cdot2+1=21∈(3)$.
Thus, $b_1 x+b_0=10x+21$.
$c_1=p_a p_b α$, where $α=(p_a p_b )^{-1} mod\ p_c$, hence $c_1=2\cdot3\cdotα$, where $α=(2\cdot3)^{-1} mod\ 5=1$, so that $c_1=2\cdot3\cdot1=6$.
$c_0=p_a p_b β+1$, where $β<p_c$ such that $c_0∈(p_c )$, hence $c_0=2\cdot3\cdotβ+1$, where $β<5$ such that $c_0=6β+1∈(5)$. Here $β=4$, so that $c_0=2\cdot3\cdot4+1=25∈(5)$.
Thus, $c_1 x+c_0=6x+25$.
So, $x=(15x+16)(10x+21)(6x+25)$ in $\Bbb{Z}/30\Bbb{Z}[x]$ — exactly the same factorization as given by Dummit and Foote.
It seems that it is the only one factorization with maximal number of $3$ nonconstant factors unique up to multiplication by units (up to associates) according to the [result] of Part 1.
As a further example using the same technique (omitting calculations):
$x=(5005x+10011)(6006x+9010)(10725x+4291)(1365x+13651)(6930x+8086)$ in $\Bbb{Z}/15015\Bbb{Z}[x]$, where $15015=3\cdot5\cdot7\cdot11\cdot13$.
It seems that it is the only one factorization with maximal number of $5$ nonconstant factors unique up to multiplication by units (up to associates) according to the [result] of Part 1.
Question
Somehow I have doubts on the [result] of Part 1. That is:
⦁ If $n$ is the product of $k$ distinct primes in $\Bbb{Z}$ is it always true that $f(x)=x$ in $\Bbb{Z}/n\Bbb{Z}[x]$ has only one factorization with maximal number of nonconstant factors unique up to multiplication by units (up to associates)?
If this is not the case, please, provide a counterexample.
P.S. By "associates" above I mean their usual definition. I know that there are several definitions of "associates" in non-integral domains, but I haven't studied this properly as of yet, so excuse me, please, if this is imprecise.