TITLE: If $X$ has non-singular normalization $\dim (\mathrm{Sing(X)})=\dim (X)-1$?
QUESTION [5 upvotes]: Let $X\subseteq\mathbb{P}^{N}$ be an algebraic variety, and let 
$$
\nu:X^{\nu}\rightarrow X
$$
be its normalization. Let us suppose that $X^{\nu}(\neq X)$ is smooth. I wonder if in this case
$$
\dim (\mathrm{Sing(X)})=\dim (X)-1.
$$
I think Serre's Normality Criterion (See Lemma 12.5 of this text) has something to do with it, but I can't see how. 
Maybe
$$
X^{\nu} \text{ smooth }\Rightarrow
$$
$$
\Rightarrow
 X\text{ satisfies the property $(S_{2})$ (i.e. $\forall$ $x\in X, \mathrm{depth}(\mathcal{O}_{X,x})\geq \min\{2,\dim(\mathcal{O}_{X,x})\}$)},
$$
in which case the equality follows from the fact that $X$ is reduced, not normal, and would satisfy $(R_{1})$ if 
$$
\dim (\mathrm{Sing(X)})<\dim (X)-1.
$$
Is that implication true? At least, is the equality true?

REPLY [3 votes]: You are on the right track. If the singularity has codimension greater than one, it satisfies $R_1$, so to make it non-normal, it must fail $S_2$. Further you want the normalization to be smooth. So, here is an example. Let $X$ be the spectrum of $k[x^2,xy,y^2,x^3,y^3]$. It is not normal, its fraction field is $k(x,y)$ and its integral closure is $k[x,y]$. It also satisfies $R_1$.