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My professor wrote down that $\sqrt{\mathrm{ann}_R(R/I)} = \sqrt{I}$ for $I$ a proper ideal of $R$. Clearly if $R$ is unital ring we have that $\mathrm{ann}_R(R/I) = I$. So the radical relation is true. However, is this true for when $R$ is not unital?

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  • $\begingroup$ I added the "ring-theory" tag to your post. $\endgroup$ – Robert Lewis Nov 29 '17 at 17:14
  • $\begingroup$ Remember to include some extra tag besides abstract algebra, since the later is very general. $\endgroup$ – Xam Nov 29 '17 at 17:49
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Since $RI\subseteq I$, you always have that $I\subseteq \mathrm{ann}_R(R/I)$, unity or not. What identity buys for you is that $1\cdot \mathrm{ann}_R(R/I)\subseteq I$.

Stringing those two together, you'd get $I=\mathrm{ann}_R(R/I)$.

As an example of how this can fail, consider the ring $(x)$ inside $F[x]$ containing the ideal $(x^2)$. You can see that $\mathrm{ann}_{(x)}\big(\frac{(x)}{(x^2)}\big)=(x)$.

Or, for that matter, the ring $2\mathbb Z$ in $\mathbb Z$, and the quotient $2\mathbb Z/4\mathbb Z$.


From the containment $I\subseteq \mathrm{ann}_R(R/I)$ you immediately get $\sqrt{I}\subseteq \sqrt{\mathrm{ann}_R(R/I)}$.

Now suppose that $x\in \sqrt{\mathrm{ann}_R(R/I)}$. Then $x^n\in \mathrm{ann}_R(R/I)$, so that $x^nR\subseteq I$. Normally you'd say that $x^n\cdot 1\in I$, and conclude $x\in\sqrt{I}$, but you can't do that here. But (surprise! 🎉) $x^nx\in I$, and indeed $x\in \sqrt{I}$.

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    $\begingroup$ @伽罗瓦 I don't know for sure immediately, but I'm thinking about it. It seems obvious that $\sqrt{I}\subseteq \sqrt{ann(R/I)}$, but I don't know about the reverse. $\endgroup$ – rschwieb Nov 29 '17 at 17:30
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    $\begingroup$ @伽罗瓦 I suddenly saw why it is true and added it to the solution. Who is teaching you these things with rings without identity, by the way, if you don't mind mentioning? $\endgroup$ – rschwieb Nov 29 '17 at 17:39
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    $\begingroup$ @rschwieb In the non-unital case I think you have to consider $\mathrm{ann}_{(x)}\big(\frac{(x)}{(x^2)}\big)$. $\endgroup$ – user26857 Feb 25 at 17:02
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    $\begingroup$ @user26857 Yes, that was the intention, and unfortunately the suggested edit missed this :) Thanks for pointing it out. $\endgroup$ – rschwieb Feb 25 at 17:15
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    $\begingroup$ @stressedout Well, that goes without saying. You didn't have to justify your improvement: it is just fine! It's unfortunate though that "ann" does not have a simple counterpart like "sin" and "cos" do. That was probably my main reason for not bothering with it in the first place. $\endgroup$ – rschwieb Feb 25 at 17:22

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