Normal quadric surfaces in $\mathbb{P}^3$ 
I'm trying to prove that the quadric surfaces $Q_1:xy-zw=0$ and $Q_2:z^2-xy=0$ are normal (that's exercise 3.17(b) from Hartshorne's Algebraic Geometry).

The definition of normal variety given in the exercise is that $Y$ is normal when $\mathcal{O}_P$ is integrally closed for all $P\in Y$). I have a feeling that using the definition directly is not a good idea, but I'm stuck because I don't know what else to use.
 A: Using the hint Hartshorne provides for this exercise I was finally able to solve it. Hope it will be helpful for you too.
Let $P\in Q_{1}$. Since all the coordinates are symmetric, we can
assume, WLOG, that $w\neq0$. Then $\mathcal{O}_{Q_{1},P}\cong\mathcal{O}_{D\left(w\right),P}$
so it is sufficient to show that $Y=V\left(xy-z\right)\subseteq\mathbb{A}^{3}$
is normal. The morphism $\varphi:\mathbb{A}^{2}\rightarrow Y$ given
by $\left\langle x,y\right\rangle \mapsto\left\langle x,y,xy\right\rangle $
is an isomorphism (the inverse is given by $\left\langle x,y,z\right\rangle \mapsto\left\langle x,y\right\rangle $)
and thus $Y$ is normal since $A\left(\mathbb{A}^{2}\right)\cong k\left[x,y\right]$
is UFD.
Let $P\in Q_{2}$. If $x\neq0$ or $y\neq0$, $P$ has a neighborhood
isomorphic to $V\left(z^{2}-x\right)\subseteq\mathbb{A}^{3}$ which
is normal since it is isomorphic to $\mathbb{A}^{2}$. For $z\neq0$,
$P$ has a neighborhood isomorphic to $V\left(xy-1\right)\subseteq\mathbb{A}^{3}$
which is isomorphic to $\mathbb{A}^{2}\backslash V\left(x_{1}\right)$
which is an open subset of a normal variety and thus obviouslly normal.
Finally, suppose $w\neq0$, so $P$ has a neighborhood isomorphic
to $V\left(z^{2}-xy\right)\subseteq\mathbb{A}^{3}$ so we need to
show that $R=k\left[x,y,z\right]/\left(z^{2}-xy\right)$ is integrally
closed. Let $\alpha\in Quot\left(R\right)$ be integral over $R$.
$ Quot\left(R\right)/k\left(x,y\right)$ is finite field
extension of degree 2, so we can write $\alpha=f+g\cdot z$ for $f,g\in k\left(x,y\right)$.
The minimal polynomial of $\alpha$ over $k\left(x,y\right)$ is $\alpha^{2}-2f\cdot\alpha+\left(f^{2}-g^{2}xy\right)$
and since $k\left[x,y\right]$ is an UFD, by Gauss lemma, it is also
the minimal poynomial of $\alpha$ over $k\left[x,y\right]$ thus
$f\in k\left[x,y\right]$ and $f^{2}-g^{2}xy\in k\left[x,y\right]$
for which follows that $g^{2}xy\in k\left[x,y\right]$ and since $xy$
is square free, $g\in k\left[x,y\right]$ so $\alpha\in R$. 
