Every Subring of Z[x] has Ascending Chain Condition on Principal Ideals 
Please help me to prove that every subring of $\mathbb Z[x]$ has Ascending Chain Condition on Principal ideals.

 A: Let $A\subset\Bbb Z[x]$ be a subring and suppose that $A$ doesn't satisfy ACCP, so there exists $f_1A\subset f_2A\subset\cdots\subset f_nA\subset\cdots$ a strictly ascending chain of principal ideals in $A$. We get $\cdots f_n\mid\cdots\mid f_2\mid f_1$ (in $A$) and therefore $\deg f_1\ge \deg f_2\ge\cdots\ge \deg f_n\ge\cdots$. Then there exists an $m\ge 1$ such that $\deg f_m=\deg f_{m+1}=\cdots$. Since $f_{m+1}\mid f_m$ there is $a_{m+1}\in\Bbb Z$, $a_{m+1}\ne -1,0,1$, such that $f_m=a_{m+1}f_{m+1}$, and this shows that $a_{m+1}$ divides the leading coefficient of $f_m$. Continuing this way we find an infinite sequence $(a_i)_{i\ge m+1}$ of integral numbers different from $-1,0,1$ such that $a_{m+1}\cdots a_{m+j}$ divides the leading coefficient of $f_m$ for all $j\ge 1$, a contradiction. 
A: start with a principal ideal $I_0$ generated by a polynomial $p\in\mathbb{Z}[x]$. 
Say $p$ has degree n. 
Now consider an ascending chain of principal ideals 
$$
I_0\subset I_1\subset I_2\subset\ldots
$$
and their corresponding elements. Prove that after there exists $j\in\mathbb{N}$ such that the generator of $I_j$ has degree which is strictly smaller than $n$.
Now proceed by induction.
