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If $R$ is a commutative ring with ideal $I\subseteq R$, show that $a^n \in I$ implies $-a^n\in I$ for some $n\in\mathbb N$.

This is an intermediate step in attempting to show that the set $$\operatorname{rad} I :=\{r\in R:r^n\in I\text{ for some }n\in\mathbb N\}$$ is closed under subtraction (which is an intermediate step in showing that it is an ideal). I have successfully shown that for any $x,y\in\operatorname{rad}I$, we have $m,n\in\mathbb N$ such that $x^m,y^n\in I$, which implies $$(x+y)^{m+n}=\sum_{k=0}^{m+n}{m+n\choose k}x^ky^{m+n-k};$$ in other words, $\operatorname{rad}I$ is closed under addition. However, it is not entirely clear how I could get subtraction from this. If $R$ has a multiplicative identity $1$, then it is simple: For any $r\in R$, $$(ry)^n = r^ny^n \in I,$$ since $I$ is an ideal and $r^n\in R$. In particular, choose $r=-1\in R$. Then the result is obvious. But what could you do if $R$ has no multiplicative identity $1$? I thought about it: If $n$ is even, then you get exactly what you want: $$(-y)^n = \underbrace{(-y)(-y)\cdots(-y)}_{n\text{ times}} = \underbrace{yy\cdots y}_{n\text{ times}} = y^n.$$ But when $n$ is odd, things are slightly more difficult to interpret: $$\underbrace{(-y)(-y)\cdots(-y)}_{n\text{ times}} = -y^n.$$ Is it necessarily true that $-y^n\in I$ as well? If so, could someone provide some insight or a hint as to how we can accomplish this based on the fact that $R$ is solely a commutative ring and $I\subseteq \operatorname{rad} I \subseteq R$ is an ideal?

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    $\begingroup$ If $x\in I$, $-x\in I$ too, since an ideal is a subgroup for the additi law.) $\endgroup$ – Bernard May 8 '17 at 1:27
  • $\begingroup$ @Bernard, thanks! Someone posted this much earlier, but they deleted it. I thank both of you for helping me see the connection. $\endgroup$ – Decaf-Math May 8 '17 at 1:57

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