# $\mathrm{ann}(R/I) = I$ only when $R$ is unital?

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?

• I added the "ring-theory" tag to your post. – Robert Lewis Nov 29 '17 at 17:14
• Remember to include some extra tag besides abstract algebra, since the later is very general. – Xam Nov 29 '17 at 17:49

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 $$x\in \sqrt{I}$$.

• @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. – rschwieb Nov 29 '17 at 17:30
• @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? – rschwieb Nov 29 '17 at 17:39
• @rschwieb In the non-unital case I think you have to consider $\mathrm{ann}_{(x)}\big(\frac{(x)}{(x^2)}\big)$. – user26857 Feb 25 at 17:02
• @user26857 Yes, that was the intention, and unfortunately the suggested edit missed this :) Thanks for pointing it out. – rschwieb Feb 25 at 17:15
• @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. – rschwieb Feb 25 at 17:22