Is the normalization of a Cohen-Macaulay domain necessarily Cohen-Macaulay? I suspect that the answer is no, but I don't have a counterexample. I am most interested in "geometric" situations, so one can place assumptions like excellence on the domain if it's relevant.



Cohen-Macaulay domain with non-Cohen-Macaulay normalization?
No, take a normal non-Cohen-Macaulay $d$-dimensional projective variety $X \subseteq \mathbb{P}^n$ and do a series of generic projections from points not on the variety so that its image is a hypersurface $Y \subseteq \mathbb{P}^{d+1}$. 


The generic hypersurface projection $Y$ will be Cohen-Macaulay (its a hypersurface), but the normalization is $X$, which is not Cohen-Macaulay. 


