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.

## 2 Answers

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.

This is an algebraic and affine version of what Karl wrote. I could not produce a concrete example, but here is how one can try to do it.

Take a normal non-CM domain $B$ over some field of characteristic $0$ (see this question for some concrete examples). Find a Noether normalization $A \subset B$ (Macaulay 2 can do it for you).

Next, use the primitive element theorem to find $z \in B$ such that $R=A[z]$ has the same quotient field as $B$. Clearly then $\bar R = B$, and $R$ is a hypersurface since $A$ is a polynomial ring. The equation is likely to be messy, though.