Please help me to prove that every subring of $\mathbb Z[x]$ has Ascending Chain Condition on Principal ideals.
Every Subring of Z[x] has Ascending Chain Condition on Principal Ideals
2 Answers
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.
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0Why writing up a direct proof as a contradiction? (I know that this happens all the time ...) – 2013-12-31
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.
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0@Arthur I may be wrong but to apply ACCP we need prime ideals of this subring, it is not necessary for the subring to be ideal itself – 2012-11-22