Originally Posted by

**ymar** Let $\displaystyle R$ be a commutative ring. Let $\displaystyle Nil(R)$ and $\displaystyle Z(R)$ denote the nilradical and the singular ideal respectively. I can prove that $\displaystyle Nil(R)\subseteq Z(R).$ In Lam's second book there is an exercise in which the reader is told to show that $\displaystyle Nil(R)\subseteq_e Z(R)$ and to give a counterexample for the equality.

I can't do either. I'm quite sure the first one must be easy, but not for me, as usual. I'll try to think more about it and maybe I'll get it finally. So if it's very, very easy, please don't tell me how to do it. But if you think some hint would be helpful, then please post it.

For the second, I think I know too little to find the example. I know that the example can't be a Noetherian ring. Actually, it can't have the ACC property on annihilators of elements. All non-Noetherian commutative rings I know are polynomial rings in infinitely many variables and power series in infinitely many variables. But I know nothing about them really. In the polynomial ring in infinitely many variables, there is the ideal chain

$\displaystyle (X_1)\subseteq (X_1,X_2)\subseteq ...$

which doesn't terminate. But these ideals aren't annihilators of elements for any ring, because $\displaystyle X_i$ doesn't annihilate anything non-zero. So this is not what I want. Polynomial rings would be nice though because I know what nilpotents look like there...