Is the nilpotent exponent of the Jacobson radical of algebra a Morita equivalent invariant？

Question

I have known that two rings A and B are said to be Morita equivalent if the left module categories mod A and mod B are equivalent. Many properties of module category are Morita equivalent invariant, and we know ideal of A and B are one to one corespondents. I want to know is the nilpotent exponent of Jacobson radical of ring a Morita equivalent invariant？I think it is right for left Artin rings. Firstly, we have an equivalence functor F:mod A $\to$ mod B, for an A-module M, we have $rad^i M=\mathfrak{r}^i M$, where $\mathfrak{r}$ is the Jacobson radical of $A$. Secondly, we note that the nilpotent exponent of Jacobson radical is the maximum of radical length of $M$, for each A-module $M$. Finally, I think $F$ preserve the length of radical sequence of $M$, i.e. $rl_A M=rl_B F(M)$ . Therefore, the nilpotent exponent of Jacobson radical of left artin rings a Morita equivalent invariant. But I’m not sure it is right, and I want to know is it true for arbitrary ring.

Answer 1

Yes. The natural correspondence between the ideals of two Morita equivalent rings $R$ and $S$ preserves products of ideals, takes $\operatorname{rad}R$ to $\operatorname{rad}S$, and of course takes the zero ideal to the zero ideal.

So $\operatorname{rad}^nR=0$ if and only if $\operatorname{rad}^nS=0$.

You can find proofs of the necessary results in, for example, Proposition 18.44 and Corollary 18.50 of

Lam, T. Y., Lectures on modules and rings, Graduate Texts in Mathematics. 189. New York, NY: Springer. xxiii, 557 p. (1999). ZBL0911.16001.