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

Example. $\mathbb{C}[x,y]/(y^{2}-x^{3})$. 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 $\mathbb{C}[x,y]\rightarrow\mathbb{C}[t]$ by

 $\displaystyle x$ $\displaystyle\mapsto t^{2}$ $\displaystyle y$ $\displaystyle\mapsto t^{3}$

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

 $\mathbb{C}[x,y]/(y^{2}-x^{3})\cong\mathbb{C}[t^{2},t^{3}]$

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

 $\displaystyle x$ $\displaystyle\mapsto u^{2}$ $\displaystyle y$ $\displaystyle\mapsto v^{2}$ $\displaystyle z$ $\displaystyle\mapsto uv$

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

Title examples of integrally closed extensions ExamplesOfIntegrallyClosedExtensions 2013-03-22 17:01:32 2013-03-22 17:01:32 rm50 (10146) rm50 (10146) 9 rm50 (10146) Example msc 13B22 msc 11R04