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At Barbara's request, I am posting this as an answer (with a correction).

Assume that $X$ is Cohen-Macaulay. If $D_{k}(f)$ has the expected dimension, then it is Cohen-Macaulay. This is, of course, far from being non-singular, but limits how bad the singularities can be (e.g., no embedded points and all irreducible components have the same dimension).

I learned about this result from Chapter 4 of Geometry of algebraic curves. I do not have the book handy, but I think this result was originally proven in [Hochster-Eagon, "A class of perfect determinantal ideals"].

Chapter 4 has a general discussion of determinantal varieties that might be helpful. I seem to remember that they prove, if $D_{k}(f)$ has expected dimension, then the singular locus of $D_{k}(f)$ is contained in $D_{k-1}(f)$ (part of the theorem cited by Francesco Polizzi, but without genericity), but I could be misremembering.
I was misremembering.

show/hide this revision's text 2 added 336 characters in body

At Barbara's request, I am posting this as an answer (with a correction).

Assume that $X$ is Cohen-Macaulay. If $D_{k}(f)$ has the expected dimension, then it is Cohen-Macaulay. This is, of course, far from being non-singular, but limits how bad the singularities can be (e.g., no embedded points and all irreducible components have the same dimension).

I learned about this result from Chapter 4 of Geometry of algebraic curves. I do not have the book handy, but I think this result was originally proven in [Hochster-Eagon, "A class of perfect determinantal ideals"].

Chapter 4 has a general discussion of determinantal varieties that might be helpful. I seem to remember that they prove, if $D_{k}(f)$ has expected dimension, then the singular locus of $D_{k}(f)$ is contained in $D_{k-1}(f)$ (part of the theorem cited by Francesco Polizzi, but without genericity), but I could be misremembering.

show/hide this revision's text 1

At Barbara's request, I am posting this as an answer (with a correction).

Assume that $X$ is Cohen-Macaulay. If $D_{k}(f)$ has the expected dimension, then it is Cohen-Macaulay. This is, of course, far from being non-singular, but limits how bad the singularities can be (e.g., no embedded points and all irreducible components have the same dimension).

I learned about this result from Chapter 4 of Geometry of algebraic curves. I do not have the book handy, but I think this result was originally proven in [Hochster-Eagon, "A class of perfect determinantal ideals"].