Let $X\subseteq\mathbb{P}^{N}$ be a quasiprojective variety of dimension $N-1$, 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](http://stacks.math.columbia.edu/download/properties.pdf) 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?



If $X$ has non-singular normalization $\dim (\mathrm{Sing(X)})=\dim (X)-1$?
**Proposition:** *If $X$ is S2, then the normalization map $\nu : X^{n} \to X$ is an isomorphism outside a set of pure codimension 1 (in either $X$ or $X^{n}$).*


*Proof:* Let $Z \subseteq X$ be the locus where $\nu$ is not an isomorphism and let $W \subseteq X^{n}$ be the (scheme-theoretic) pre-image of $W$. For a contradiction, by localizing at the generic point of a component of $Z$, one may assume that $Z$ is of codimension $\geq 2$. Let $i : U = X \setminus Z \hookrightarrow X$ and $j : V = X^{n} \setminus W \hookrightarrow X^{n}$. Then since $X$ is S2, $$i\_\* O\_U = O\_X \text{ and } j\_\* O\_V = O\_{X^n}.$$ 
But $O\_U = O\_V$ and so $$\nu\_\* O\_{X^n} = \nu\_\* j\_\* O\_V = i\_\* O\_U = O\_X.$$ Hence $X$ is normal. 


### In your case...


Now, since hypersurfaces are S2 (even Cohen-Macaulay), the locus where the normalization map is not an isomorphism pure codimension 1 as you desired.


