# 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 $$. 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. $\u2102[x,y]/({y}^{2}-{x}^{3})$. 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 $\u2102[x,y]\to \u2102[t]$ by

$x$ | $\mapsto {t}^{2}$ | |||

$y$ | $\mapsto {t}^{3}$ |

The kernel of this map is $({y}^{2}-{x}^{3})$, and its image is $\u2102[{t}^{2},{t}^{3}]$. Hence

$$\u2102[x,y]/({y}^{2}-{x}^{3})\cong \u2102[{t}^{2},{t}^{3}]$$ |

and the field of fractions^{} of the latter ring is obviously $\u2102(t)$. Now, $t$ is integral over $\u2102[{t}^{2},{t}^{3}]$ (${z}^{2}-{t}^{2}$ is its polynomial), but is not in $\u2102[{t}^{2},{t}^{3}]$. $t$ corresponds to $\frac{y}{x}$ in the original ring $\u2102[x,y]/({y}^{2}-{x}^{3})$, 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. $A=\u2102[x,y,z]/({z}^{2}-xy)$ is integrally closed. For again, parameterize $A\to \u2102[u,v]$ 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 $B=\u2102[{u}^{2},{v}^{2},uv]$. Claim $B$ is integrally closed. We prove this by showing that the integral closure of $\u2102[x,y]$ in $\u2102(x,y,\sqrt{xy})$ is $\u2102[x,y,\sqrt{xy}]$. Choose $r+s\sqrt{xy}\in \u2102(x,y,\sqrt{xy}),r,s\in \u2102(x,y)$ such that $r+s\sqrt{xy}$ is integral over $\u2102[x,y]$. Then $r-s\sqrt{xy}$ is also integral over $\u2102[x,y]$, so their sum is. Hence $2r$ is integral over $\u2102[x,y]$. But $\u2102[x,y]$ is a UFD, hence integrally closed, so $2r\in \u2102[x,y]$ and thus $r\in \u2102[x,y]$. Similarly, $s\sqrt{xy}$ is integral over $\u2102[x,y]$, hence ${s}^{2}xy\in \u2102[x,y],s\in \u2102(x,y)$. Clearly, then, $s$ can have no denominator, so $s\in \u2102[x,y]$. Hence $r+s\sqrt{xy}\in \u2102[x,y,\sqrt{xy}]$.

Title | examples of integrally closed extensions |
---|---|

Canonical name | ExamplesOfIntegrallyClosedExtensions |

Date of creation | 2013-03-22 17:01:32 |

Last modified on | 2013-03-22 17:01:32 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 9 |

Author | rm50 (10146) |

Entry type | Example |

Classification | msc 13B22 |

Classification | msc 11R04 |