$R/\operatorname{rad}(0)$ is Cohen-Macaulay

Question

If $(R,m)$ is a local Noetherian ring, is it true that $\bar R=R/\operatorname{rad} (0)$ is a Cohen-Macaulay ring?

I think that we should take a maximal chain of prime ideals of length $d$ under $\bar m$ and try to find an $\bar R$-sequence of length $d$ in $\bar m$ to get grade of $\bar m$ at least height of $\bar m$.

Any help would be appreciated!

Answer 1

If this would be true, then any reduced local ring would be Cohen-Macaulay. But this is clearly false, see this answer for example.

That the ring is local is not relevant, since Cohen-Macaulay is a local property anyway.