how to prove that an algebraic variety is normal but not smooth? Let $X=\{(x_1,x_2,x_3) \in \mathbb C^3 : x_1^2=x_2^2+x_3^2\}$, an algebraic variety. How do i prove that $X$ is normal, but not smooth? I guess the non-smoothness appears at the point $(0,0,0)$, but I don't know how to put that rigorously.
 A: Here is a nice criterion by Serre for normality. A variety is normal if and only if it satisfies the Serre's conditions $R_1$ and $S_2$. Any hypersurface satisfies $S_2$. To check $R_1$, suffices to say that the singular locus is codimension at least two. In your example, the origin is the only singular point (as pointed out by Rene Schipperus, by Jacobian criterion) and thus has codimension of singular locus is 2. You can find a proof of the above theorem in Serre's Local Algebra.
A: In this case, we can also show normality just by noting that the coordinate ring is integrally closed, by the following:
Lemma (Hartshorne, Exc. II.6.4).
Let $f \in k[x_1,\ldots,x_n]$ be a square-free nonconstant polynomial over a field of characteristic $\ne 2$. Then, $k[x_1,\ldots,x_n,z]/(z^2-f)$ is integrally closed.
Proof Sketch. First, you can show the quotient field $K = \operatorname{Frac}A$ is equal to $k(x_1,\ldots,x_n)[z]/(z^2-f)$, which is a degree $2$ extension over $k(x_1,\ldots,x_n)$, hence Galois with Galois group $\mathbf{Z}/2$ generated by $z \mapsto -z$. Now let $\alpha = g + hz \in K$, where $g,h \in k(x_1,\ldots,x_n)$. Then, $\alpha$ has minimal polynomial $X^2 - 2gX + (g^2-h^2f)$, for
  \begin{equation*}
    (g+hz)^2 - 2g(g+hz) + (g^2 - h^2f) = g^2 + 2ghz + h^2z^2 - 2g^2 - 2ghz + g^2 - h^2f = 0.
  \end{equation*}
  Thus, $\alpha$ is integral over $k[x_1,\ldots,x_n]$ if and only if
  $2g$ and $g^2-h^2f \in k[x_1,\ldots,x_n]$, which is true if and only if $g, h^2f \in
  k[x_1,\ldots,x_n]$. Now if $\alpha$ is integral, and $h$ had nontrivial
  denominator, then $h^2 \notin k[x_1,\ldots,x_n]$ since $f$ is square-free;
  hence $h \in k[x_1,\ldots,x_n]$ so $\alpha \in A$. Thus, $A$ is integrally
  closed in $K$. $\blacksquare$
