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A field extension $L/K$ (of number fields) admits a relative power integral basis if $\mathcal{O}_L = \mathcal{O}_K[\alpha]$ for some $\alpha \in \mathcal{O}_L$. I'm looking for a simple example in which this occurs and $L/K$ is unramified.

I've found references that imply that this is possible, e.g. in `A Note on Integral Bases of Unramified Cyclic Extensions of Prime Degree' (H. Ichimura), but the difficulty in that series of papers is to construct extensions with power integral basis that do not have normal integral bases. As such, I'd like to think that simpler examples exist.

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up vote 2 down vote accepted

EDIT The former example has to be modified because it gave the trivial extension.

Let $K$ be such that the equation $a^2-4b=-1$ (changed 1 to -1) has a solution with $a,b\in O_K$. Then $O_K[X]/(X^2+aX+b)$ is unramified over $O_K$ because it has invertible discriminant. So if $L=K[\alpha]$ with $\alpha$ a zero of $X^2+aX+b$, then $O_L=O_K[\alpha]$ (the right hand side is unramified over $O_K$, hence integrally closed) and $L/K$ is unramified.

As a concrete example, take $K=\mathbb Q[\sqrt{3}]$, $\alpha$ a zero of $X^2+\sqrt{3}X+1\in K[X]$ and $L=K[\alpha]$.

Add One can also say that the discriminant of the familly $\{1, \alpha\}$ is $-1$, so the discriminant of $O_L/O_K$ is $-1$. Which implies that $L/K$ is unramified.

To get a familly of examples, one can take for $b$ any positive integer.

For any $d\ge 3$, using a geometric interpretation on the polynomials $X^d+a_{d-1}X^{d-1}+\dots+a_0$, with indeterminates $a_i$, one can show that there exists $L/K$ unramified of degree $d$ such that $O_L$ is monogeneous over $O_K$ (i.e. $O_L=O_K[\alpha]$).

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Though I suppose the trivial field extension does actually satisfy all of the desired criteria... :) – Cam McLeman Feb 29 '12 at 11:51
@CamMcLeman: sure :). – user18119 Feb 29 '12 at 12:10

Here is another example of a quadratic extension:

Let $K=\mathbb{Q}(\sqrt{-5})$ and let $L=K(i)$, where $i^2=-1$. As it turns out, $L$ is the Hilbert Class Field of $K$, and therefore $L/K$ is everywhere unramified.

Exercise 2.42 (pages 51 and 52), in five parts, determines the ring of integers of $\mathbb{Q}[\sqrt{m},\sqrt{n}]$ for $m$ and $n$ distinct squarefree integers different from $1$. (See this question and response.) In particular, $L=\mathbb{Q}(\sqrt{5},\sqrt{-5})$ and therefore $\mathbb{O}_L$ has an integral basis $$B=\left\{1,\frac{1+\sqrt{5}}{2},\sqrt{-5},\frac{\sqrt{-5}+i}{2}\right\}.$$ Let $\alpha = \frac{\sqrt{-5}+i}{2}$. Then I claim that $\mathcal{O}_K[\alpha] = \mathcal{O}_L$. Clearly, $\mathcal{O}_K[\alpha]\subseteq \mathcal{O}_L$, so it suffices to show that $B\subseteq \mathcal{O}_K[\alpha]$.

We know that $\mathcal{O}_K = \mathbb{Z}[\sqrt{-5}]$. Thus, $1,\sqrt{-5}$, and $\alpha$ belong to $\mathcal{O}_K[\alpha]$. Also, $i=2\alpha-\sqrt{-5}\in \mathcal{O}_K[\alpha]$. It remains to show that $\frac{1+\sqrt{5}}{2}\in \mathcal{O}_K[\alpha]$, but $$\frac{1+\sqrt{5}}{2} = 1+i\cdot\frac{i+\sqrt{-5}}{2}=1+i\cdot \alpha\in \mathcal{O}_K[\alpha].$$ Hence, $\mathcal{O}_K[\alpha]=\mathcal{O}_L$.

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