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Copy from my exercise book:

Let $R$ be a commutative ring (not necessarily with identity). For an ideal $A \subseteq R$ we define the radical $\sqrt{A}$ of $A$ by $$\sqrt{A}:=\{x \in R : x^n \in A \text{ for some } n \in \mathbb{N}\}$$

Now I should show that $\sqrt{A}$ is an ideal in $R$. Everything is easy except the part where I should show that if $x \in \sqrt{A}$, then also $-x \in \sqrt{A}$. My idea was to show, that $x^n \in A$ for some $n \in \mathbb{N}$, then also $(-x)^n \in A$. Intuitively we should end up with something like $(-x)^n = \pm$ but I am not sure how to prove it. The solutions in the book are rather confusing, since they say something like since $R$ has identity...what was not assumed in the exercise.

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    See the comments [here](http://math.stackexchange.com/questions/1793633/prove-the-radical-of-an-ideal-is-an-ideal). – 2017-01-12
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    @DietrichBurde Thanks. Funny, I just proved it myself. Should wait a little longer writting a question. – 2017-01-12
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    Also, the radical of an ideal is the intersection of the prime ideals containing it, and hence an ideal itself. – 2017-01-12
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    @DietrichBurde Yes, this is the next step to prove in the exercise, thanks. I was just confused that in the solutions of this exercise in the book they assumed that $R$ has an identity element. Which is not necessary. – 2017-01-12
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    @TheGeekGreek why do you need to show that $-x\in \sqrt{A}$ ? You only need to prove that $x+y\in \sqrt{A}$ and $rx\in \sqrt{A}$ for every $x,y\in \sqrt{A}$ and $r\in R$. – 2017-01-12
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    @Xam For rings *with identity* this is clear because it covers the case $-1*x$, but for rings without identity, it is no longer clear why $A$ is an additive subgroup. (Proving that $x+y\in A$ for all $x,y\in A$ is not sufficient to prove $A$ is an additive subgroup, in general.) – 2017-01-12
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    @rschwieb ah yes, you're right. I forgot that. – 2017-01-12

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The key fact is that for any $x,y\in R$, $(-x)\cdot y=-(xy)$. Indeed, note that $$(-x)\cdot y+xy=(-x+x)\cdot y=0\cdot y=0.$$ (To prove $0\cdot y=0$, note that $0\cdot y+0\cdot y=(0+0)\cdot y=0\cdot y$.)

In particular, we find that $$(-x)^2=-(x\cdot(-x))=-(-x^2))=x^2,$$ and $$(-x)^3=(-x)\cdot(-x)^2=(-x)\cdot x^2=-x^3.$$ Continuing similarly by induction, $(-x)^n=x^n$ if $n$ is even and $(-x)^n=-x^n$ if $n$ is odd.