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Let $p$ be a prime, let $X=\operatorname{Spec}\mathbb{Z}_{(p)}$ and let $V=\operatorname{Spec}\mathbb{Z}_{(p)}[i]\to X $ where $i^2+1=0$. Let $U=\operatorname{Spec}\mathbb{Q}\to X$. Then $\{U,V\}$ is an etale cover of $X$ for odd $p$.

My question comes from this exercise. I'm having trouble showing that the map $\operatorname{Spec}\mathbb{Q}\to\operatorname{Spec}\mathbb{Z}_{(p)}$ is etale. A map $f:X\to Y$ is etale iff it is flat and unramified. How to show that $\operatorname{Spec}\mathbb{Q}\to \operatorname{Spec}\mathbb{Z}_{(p)}$ is locally of finite type?

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The two schemes here are affine, so we don't need to bother with affine coverings. We just use the schemes themselves.

$\Bbb Z_{(p)}$ is the subring of $\Bbb Q$ where we allow only denominators which are not divisible by $p$. So we get $$ \Bbb Q\cong \Bbb Z_{(p)}[1/p]\cong \Bbb Z_{(p)}[x]/(px-1) $$ showing that the morphism is indeed (locally) of finite type.

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