# Is This an Interesting Principal Ideal?

Let $I$ be a principal ideal in a commutative ring $R$, $x$ the generator of $I$, and $x^n\ne \sum_{i=0}^{n-1}a_ix^i,\, \forall a_i\in R\setminus I\bigcup\{0\},\, \forall n\in\mathbb N$. In other words, for this particular principal ideal, any integer power of the generator is assumed to not be expandable in its polynomial of lower orders with coefficients in the ring but not in the principal ideal. Is there a name and special interest in such an object? What are some of the interesting sufficient conditions that can achieve the aforementioned non-expandability condition?

• I think you need more conditions on $a_i$, as letting some $a_i = 0$, some $a_i = x^m$ for some $m$ will mean this is true in all rings as $x^n = 0+0+\dots + x (x^{n-1})$. – Mark Dec 30 '16 at 2:18
• @Mark: I think you may be confused about the non-expandability. It is a premise not a conclusion. I have changed the word description of the condition to make it clearer. On the other hand, I would welcome a (simple) sufficient (or even sufficient and necessary) condition that achieves the condition. – Hans Dec 30 '16 at 4:06
• This condition will needs more limitation on the $a_i$s: given any $x\in R$, $x^2=a_1x$ where $a_1=x$. – rschwieb Dec 30 '16 at 5:04
• My comment dates from when the summation started at 1 then giving for $n = 2$: $x^2 = a_1x \iff x(x-a_1) = 0$ which is impossible in an integral domain. But now we have that $x^2 = a_1x +a_0 \iff x(x-a_1) = a_0$, but the LHS is $\in I$ and the RHS $\in R \setminus I$ (and this nonzero) so the condition is valid for all rings. – Marc Bogaerts Dec 30 '16 at 20:01
• @hans Well, that's my mistake because I was actually thinking of integral elements which are closely related. Check that one out (it uses mimic polynomials). – rschwieb Jan 1 '17 at 10:48

## 1 Answer

Let $R$ be an integral domain and let $I$ be a principal ideal with $x$ s.t. $I = xR$. If $I$ does not have the property proposed in question, there exist the smallest natural number $n$ such that $a_i \in R$ and $x^n = \sum_{i=0}^{n-1}a_ix^i \iff x^n - \sum_{i=1}^{n-1}a_ix^i = x(x^{n-1} - \sum_{i=0}^{n-2}a_{i+1}x^i) = a_0$. Now $(...)\ne 0$ by $n$ being the smallest such natural number, together with $x\ne0$, we have $a_0\ne0$ as $R$ is an integral domain. But then the LHS $\in I$ and the RHS $\notin I$, a contradiction. The conclusion is that in an integral domain every principal ideal has the proposed property.

• $(...)$ could be zero, which leads to the same argument but for a lower order polynomial. So there is an inductive argument missing. Do you agree? – Hans Dec 31 '16 at 21:37
• @Hans and Marc Bogaerts: In an integral domain one can always chose $a_0\ne 0$. – user26857 Jan 1 '17 at 10:47
• @user26857: I do not understand your point. $a_0\ne 0$ would lead to, LHS not equal to RHS, a contradiction as stated in the answer. – Hans Jan 1 '17 at 22:06
• @Hans Read again my comment. – user26857 Jan 1 '17 at 22:20
• @user26857: I agree with your comment per se. What I do not understand is how it is related to the answer. Could you please state explicitly and in detail your argument? – Hans Jan 1 '17 at 22:57