I'm trying to prove that the quadric surfaces $Q_1:xy-zw=0$ and $Q_2:z^2-xy=0$ are normal (exercise 3.17(b) from Hartshorne's Algebraic Geometry).
According to the exercise, $Y$ 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 plan,now I'm stuck because I can't think of anything else.
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 $\alpha\in R$.