# Pencil of singular affine hypersurfaces differing only in the constant term

Question--quick version: Does there exist a (nonconstant) polynomial $f \in \mathbb C[x,y,z]$ such that for all $c \in \mathbb C$, the affine hypersurface cut out by $f + c$ is singular?

Motivated version: Suppose you have a singular hypersurface $\mathbf V(f)$ in $\mathbb C^3$. Since a generic hypersurface is smooth, you can deform the hypersurface to be smooth by adding a bit of noise to every coefficient of $f$ (i.e., all coefficients of degree $\leq$ the degree of $f$). Can you achieve the same effect by adding noise only to the constant term? I would be surprised if there were no counterexamples, but I can't think of any.

Thoughts: You can't get an example just by putting e.g. a triple point in the base locus, because the pencil in question has empty affine base locus. (Perhaps this would give you singularities at infinity, but I only care about affine singularities.) The other approach that occurs to me is to make the pencil the image of a singular variety under a pencil of affine transformations, but I don't think it's ever possible to do this by shifting the constant term.

• This is Sard's Theorem, aka, generic smoothness. The critical locus of the morphism $f:\mathbb{A}^3\to \mathbb{A}^1$ is a proper closed subset of $\mathbb{A}^1$. – Jason Starr Jan 18 '16 at 2:02

## 1 Answer

As you observe, there is no base locus on the affine part. This allows you to use the Bertini theorem as stated in Hartshorne Corollary III.10.9 and Remark III.10.9.2. The latter says that the projectivity assumption can be dropped if the linear system is finite-dimensional, which is certainly the case in your question.

Edit: As Jason points out, the affine version with a single function can be directly inferred from "Sard's theorem". This is essentially what the argument that I gave uses: see Corollary III.10.7.