Jtdcftul

Let $X$, $Y$ be integral separated schemes of finite type over $mathbb{C}$, $Y$ be normal, $f:Xrightarrow Y$ 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 $mathbb P^1times mathbb P^1subseteq mathbb P^3$. In general, say $Y$ is a cone over $Vtimes W$.

Next let $Hsubseteq W$ be an effective Cartier divisor (In the $mathbb P^1times mathbb P^1$ example, $H$ is simply a point) and let $H'=Vtimes H$. Finally, let
$f:Xto Y$ 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 $Vtimes W$ embedded with the Segre embedding. Construct the same $f:Xto Y$ and let $vin Y$ denote the vertex of the cone. Then a relatively simple computation shows that then $f^{-1}(v)simeq V$ (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 $X$ is smooth (this is where the choice of $V$, $W$ and the embedding matters).

Now let $f':X'colon =Xtimes _YXto Y'colon=X$. 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 $dim X'=dim X= n+2>3$. Hence $f'$ is an isomorphism (in particular, flat) outside a codimension $2$ subset of $X'$ (but it obviously not flat along the exceptional set).

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.