TITLE: Is the normalization map bijective?
QUESTION [8 upvotes]: Let $X$ be an integral scheme of finite type over a field. Then there is a surjective finite map $\tilde{X} \to X$ from the normalization $\tilde{X}$ of $X$. 
Is this going to be bijective?
In the simplest non-normal case, namely the spectrum of $k[x^2, x^3] = k[t,u]/(t^3 -u^2)$, the map is bijective, because the curve is geometrically just a cubic curve (the set of all $(v^2, v^3)$ in affine 2-space, geometrically). I can't find it asserted anywhere that the map is necessarily bijective, though.

REPLY [11 votes]: If $X$ is reduced and finite type over an algebraically closed field of characteristic zero, then there is a largest finite map $Y \to X$ which is bijective (where $Y$ is also reduced).  This map is the seminormalization, $Y = X^{SN}$.  In fact, it always factors the normalization map $X^N \to X^{SN} \to X$ where $X^N$ is the normalization.