# Codimension of non-flat locus

Let $$X$$, $$Y$$ be integral separated schemes of finite type over $$\mathbb{C}$$, $$Y$$ be normal, $$f:X\rightarrow Y$$ be a surjective morphism of schemes. Can the non-flat locus of $$f$$ be non-empty and have codimension $$\geq 2$$ in $$X$$?

Let $$Y$$ be a cone over a smooth projective variety which is a product and projectively normal (so $$Y$$ is normal). For example let $$Y$$ be the cone over $$\mathbb P^1\times \mathbb P^1\subseteq \mathbb P^3$$. In general, say $$Y$$ is a cone over $$V\times W$$.

Next let $$H\subseteq W$$ be an effective Cartier divisor (In the $$\mathbb P^1\times \mathbb P^1$$ example, $$H$$ is simply a point) and let $$H'=V\times H$$. Finally, let $$f:X\to Y$$ be the blow up of $$Y$$ along its subscheme $$Z$$ which is the cone over $$H'$$. Then $$f$$ is an isomorphism (and hence flat) outside $$Z$$. Since $$Z$$ is a Weil divisor, which is Cartier except at the vertex, $$f$$ is a small morphism, so it is an isom outside a codimension $$2$$ subset.

Addendum: Here is an example with $$Y$$ smooth. Start with the above example with $$V=\mathbb P^1$$ and $$W=\mathbb P^n$$ with $$n>1$$ and $$V\times W$$ embedded with the Segre embedding. Construct the same (and for the sake of avoiding confusion, let's denote it differently) $$f_0:X_0\to Y_0$$ and let $$v\in Y_0$$ denote the vertex of the cone. Then a relatively simple computation shows that then $$f^{-1}(v)\simeq V=\mathbb P^1$$ (for the actual computation see Prop 3.3 of this paper. Note that this means that the exceptional set of $$f$$ is $$1$$-dimensional. Another simple calculation shows that $$X_0$$ is smooth (this is where the choice of $$V$$, $$W$$ and the embedding matters).

Now let $$f:X\colon =X_0\times _{Y_0}X_0\to Y\colon=X_0$$. Then $$f:X\to Y$$ is birational (because of the dimensions there is only one component) and it's exceptional set (on $$X$$) is $$2$$-dimensional. From the construction and by the assumption that $$n>1$$ we obtain that $$\dim X=\dim X_0= n+2>3$$. Hence $$f$$ is an isomorphism (in particular, flat) outside a codimension $$2$$ subset of $$X$$ (but it is obviously not flat along the exceptional set).

• what happens if $Y$ is smooth?
– user137767
Apr 5, 2019 at 18:03
• @StepanBanach: I added an example where $Y$ is smooth Apr 7, 2019 at 2:15
• The second example ($X$ and $Y$ smooth, $f$ iso away from a curve in $X$) seems to contradict van den Waerden's theorem. What am I missing? Apr 7, 2019 at 3:00
• @PiotrAchinger: I didn't claim that $X$ was smooth. Apr 7, 2019 at 3:36
• if I understand correctly, for a surjective birational morphism between connected smooth schemes of finite type over $\mathbb{C}$, the non-flat locus can not be non-empty of codimension $\geq 2$. Is it true that the non-flat locus of a surjective morphism between connected smooth schemes of finite type over $\mathbb{C}$ can not be non-empty of codimension $\geq 2$?
– user74900
Apr 7, 2019 at 8:44

Let $$n \ge 2$$, $$X = \mathbb{A}^n$$, $$Y = \mathbb{A}^n/\{\pm 1\}$$ and $$f \colon X \to Y$$ the quotient morphism. The non-flat locus of $$f$$ is the point $$f(0) \in Y$$.

Yes. Let $$Y$$ be $$\mathrm{Spec} \ k[w,x,y,z]/(wz-xy)$$. Let $$X$$ be $$\mathrm{Proj} \ k[w,x,y,z, s,t]/(wz-xy,\ wt-xs,\ yt-zs)$$ where $$w$$, $$x$$, $$y$$ and $$z$$ are in degree $$0$$ and $$s$$ and $$t$$ are in degree $$1$$. Then $$Y$$ is three dimensional with a singularity at $$w=x=y=z=0$$. The map $$X \to Y$$ is a resolution of this singularity; it is an ismorphism (and hence flat) away from the singularity and the fiber over the singularity is $$\mathbb{P}^1$$, so the nonflat locus is $$1$$-dimensional inside the $$3$$-fold $$X$$.

This is a special case Sándor Kovács example, namely, it is what happens when you blow up the cone on $$\mathbb{P}^1 \times (\mathrm{point})$$ inside the cone on $$\mathbb{P}^1 \times \mathbb{P}^1$$.

Try $$Y=\mathrm{Spec}\,\mathbb{C}[x^4, x^3y, xy^3, y^4]$$ and $$X$$ its normalization.

• isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
– user137767
Apr 5, 2019 at 17:21
• @StepanBanach You did not mention normal in your question. Apr 5, 2019 at 18:28