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I wish to show that if $N$ is an ideal of a commutitive ring $R$, then $$\sqrt{N} = \{a \in R \mid \exists k \in \mathbb{N}^{*} \textrm{s.t. }a^k \in N\}$$ is an ideal of $R$.

Is this sufficient: " If r is an element of $R$ and $a$ an element of √N with $a^n \in N$ then $(ra)^n = r^na^n$ is in $N$, so $ra$ is in $√N$. "?

Can someone show an example where ideals of a commutative ring, $\sqrt{N}$ may not be equal to $N$?

rschwieb
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Ozera
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You're right, but it is not sufficient, as you also have to check that $x,y\in\sqrt{N}$ implies that $x+y\in \sqrt{N}$. For this, see my answer here :

$Rad(I)$ is an ideal of $R$

(Note that the $R$ there is your $A$ here.)

Now as you saw, $N$ is always contained in $\sqrt{N}$, but the inclusion can be strict, for example : the radical of $4\mathbf{Z}$ in $\mathbf{Z}$ is $2\mathbf{Z}$. Note $J$ the radical of $I = 4\mathbf{Z}$ and let's show $J = 2\mathbf{Z}$. Take $x\in J$ so that there is an $n\in\mathbf{N}^*$ such that $x^n \in 4\mathbf{Z}$. Then $4$ divides $x^n$, so that $2$ divides $x^n$, and as $2$ is prime, this implies that $2$ divides $x$, that is, that $x\in 2\mathbf{Z}$. We have shown that $J\subseteq 2\mathbf{Z}$. Inversely take $x= 2y\in 2\mathbf{Z}$. Then $x^2 = 4 y^2 \in I$, so that $x\in\sqrt{I} = J$, and we are done.

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Olórin
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  • Ah, I see now. Thank you. I'll continue reading your post in the other thread. – Ozera Mar 02 '15 at 13:20
  • I would by the way appreciate people to avoid correcting right things to replace them by wrong ones : "you're right" means "you are right", correcting it by "your right" makes no sense. – Olórin Mar 02 '15 at 13:20
  • @Ozera Np ! If your happy, just validate the answer, so that the case is closed. ;-) – Olórin Mar 02 '15 at 13:20
  • There is a time limit in accepting answers. The question will be closed in < 2min – Ozera Mar 02 '15 at 13:22
  • Oh really, why will it be closed ? (I mean, I may contribute here, by I'm not hyper aware of all habits here) – Olórin Mar 02 '15 at 13:23
  • Ah ok, sorry, it's my fault ! By "case is closed" I meant : the answer is accepted so people know that an answer has been accepted. – Olórin Mar 02 '15 at 13:24
  • I edited my answer. – Olórin Mar 02 '15 at 13:48
  • Robert, what is $N^$. Do you mean $\mathbb{N}^$? – Ozera Mar 02 '15 at 14:07
  • $\mathbf{N} = \mathbb{N} = {0,1,2,\ldots}$. $\mathbf{N}^* = \mathbb{N}^* = {1,2,3,\ldots}$. – Olórin Mar 02 '15 at 14:08
  • Aha. Okay, thank you. – Ozera Mar 02 '15 at 14:08
  • Usually, I mean historically, people wrote $\mathbb{N}$ on a blackboard because it was quicker to wrote that writing $\mathbf{N}$. But the notation passed to typesetting and books. ;-) – Olórin Mar 02 '15 at 14:10
  • I had thought you were just keeping to your typo faith as in the previous thread. Neat information! – Ozera Mar 02 '15 at 14:16