Example. $\Ints[\sqrt{-5}]$ is not integrally closed, for $u=\frac{1+\sqrt{-5}}{2}\in\Rats[\sqrt{-5}]$ is integral over $\Ints[\sqrt{-5}]$ since $u^2-u-1=0$ , but $u\notin\Ints[\sqrt{-5}]$ .

Example. $R=\Ints[\sqrt{2},\sqrt{3}]$ is not integrally closed. Note that $(\sqrt{6}+\sqrt{2})/2\notin R$ , but that $$ \left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)^2=2+\sqrt{3 $$ and so $(\sqrt{6}+\sqrt{2})/2$ is integral over $\Ints$ since it satisfies the polynomial $(z^2-2)^2-3=0$ .

Example. $\Alg_K$ is integrally closed when $[K:\Rats]<\infty$ . For if $u\in K$ is integral over $\Alg_K$ , then $\Ints\subset\Alg_K\subset \Alg_K[u]$ are all integral extensions, so $u$ is integral over $\Ints$ , so $u\in\Alg_K$ by definition. In fact, $\Alg_K$ can be defined as the integral closure of $\Ints$ in $K$ .

Example. $\Complex[x,y]/(y^2-x^3)$ . This is a domain because $y^2-x^3$ is irreducible hence a prime ideal. But this quotient ring is not integrally closed. To see this, parameterize $\Complex[x,y]\rightarrow\Complex[t]$ by \begin{eqnarray*} &x&\mapsto t^2\\ &y&\mapsto t^3 \end{eqnarray*}The kernel of this map is $(y^2-x^3)$ , and its image is $\Complex[t^2,t^3]$ . Hence $$ \Complex[x,y]/(y^2-x^3)\cong\Complex[t^2,t^3 $$ and the field of fractions of the latter ring is obviously $\Complex(t)$ . Now, $t$ is integral over $\Complex[t^2,t^3]$ ($z^2-t^2$ is its polynomial), but is not in $\Complex[t^2,t^3]$ . $t$ corresponds to $\frac{y}{x}$ in the original ring $\Complex[x,y]/(y^2-x^3)$ , which is thus not integrally closed (the minimal polynomial of $\frac{y}{x}$ is $z^2-x$ since $(\frac{y}{x})^2-x=\frac{y^2}{x^2}-x=\frac{x^3}{x^2}-x=0$ ). The failure of integral closure in this coordinate ring is due to a codimension 1 singularity of $y^2-x^3$ at $0$ .

Example. $A=\Complex[x,y,z]/(z^2-xy)$ is integrally closed. For again, parameterize $A\rightarrow \Complex[u,v]$ by \begin{eqnarray*} &x&\mapsto u^2\\ &y&\mapsto v^2\\ &z&\mapsto uv \end{eqnarray*}The kernel of this map is $z^2-xy$ and its image is $B=\Complex[u^2,v^2,uv]$ . Claim $B$ is integrally closed. We prove this by showing that the integral closure of $\Complex[x,y]$ in $\Complex(x,y,\sqrt{xy})$ is $\Complex[x,y,\sqrt{xy}]$ . Choose $r+s\sqrt{xy}\in\Complex(x,y,\sqrt{xy}), r,s\in\Complex(x,y)$ such that $r+s\sqrt{xy}$ is integral over $\Complex[x,y]$ . Then $r-s\sqrt{xy}$ is also integral over $\Complex[x,y]$ , so their sum is. Hence $2r$ is integral over $\Complex[x,y]$ . But $\Complex[x,y]$ is a UFD, hence integrally closed, so $2r\in\Complex[x,y]$ and thus $r\in\Complex[x,y]$ . Similarly, $s\sqrt{xy}$ is integral over $\Complex[x,y]$ , hence $s^2xy\in\Complex[x,y], s\in\Complex(x,y)$ . Clearly, then, $s$ can have no denominator, so $s\in\Complex[x,y]$ . Hence $r+s\sqrt{xy}\in\Complex[x,y,\sqrt{xy}]$ .