examples of integrally closed extensions

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

Example. $R=\mathbb{Z}[\sqrt{2},\sqrt{3}]$ 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 $\mathbb{Z}$ since it satisfies the polynomial ${(z^2-2)}^2-3=0$.

Example. ${𝒪}_K$ is integrally closed when . For if $u\in K$ is integral over ${𝒪}_K$, then $\mathbb{Z}\subset {𝒪}_K\subset {𝒪}_K[u]$ are all integral extensions, so $u$ is integral over $\mathbb{Z}$, so $u\in {𝒪}_K$ by definition. In fact, ${𝒪}_K$ can be defined as the integral closure of $\mathbb{Z}$ in $K$.

Example. $ℂ[x,y]/(y^2-x^3)$. 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 $ℂ[x,y]\to ℂ[t]$ by

$x$$\mapsto t^2$

$y$$\mapsto t^3$

The kernel of this map is $(y^2-x^3)$, and its image is $ℂ[t^2,t^3]$. Hence

$$ℂ[x,y]/(y^2-x^3)\cong ℂ[t^2,t^3]$$

and the field of fractions of the latter ring is obviously $ℂ(t)$. Now, $t$ is integral over $ℂ[t^2,t^3]$ ($z^2-t^2$ is its polynomial), but is not in $ℂ[t^2,t^3]$. $t$ corresponds to $\frac{y}{x}$ in the original ring $ℂ[x,y]/(y^2-x^3)$, 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. $A=ℂ[x,y,z]/(z^2-xy)$ is integrally closed. For again, parameterize $A\to ℂ[u,v]$ by

$x$$\mapsto u^2$

$y$$\mapsto v^2$

$z$$\mapsto uv$

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