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

Question

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?

Answer 1

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 $2\mathbb Z/4\mathbb Z$.

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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 $x\in \sqrt{I}$.