# Completion of $Q(i)$ with respect to prime ideal (1+i)

The completion of $Q(i)$ with respect to prime ideal (1+i) lying above 2 is $Q_{2}(i)$,. Then what about the completion of $Q(i)$ with respect to the prime ideal $(1-i)$? is it same

and since $5=(2-i)(2+i)$ what is the completion of $Q(i)$ wrt $(2-i)$ and $(2+i)$ Any help is greatly acknowledged.

• Note that $i$ is in ${\bf Q}_5$. Dec 6, 2017 at 5:32
• Pardon, So what happens Dec 6, 2017 at 5:35
• @user86925 It's unlikely you can find any material on the completions of this specific field, but advanced texts on number theory will consider the completions of arbitrary number fields. Dec 6, 2017 at 6:19
• $(1+i) = (1+i,2)= (1-i,2) = (1-i)$ thus $\widehat{\mathbf{Q}(i)_{(1+i)}}=\widehat{\mathbf{Q}(i)_{(1-i)}}$. Now $(2+i) \ne (2-i)$ thus the completions $\widehat{\mathbf{Q}(i)_{(2+i)}},\widehat{\mathbf{Q}(i)_{(2-i)}}$ are different, and are isomorphic with $a+ib \mapsto a-ib$, and hence the complex conjugaison isn't an automorphism of $\widehat{\mathbf{Q}(i)_{(2+i)}}$. Dec 6, 2017 at 10:02
• Extending slightly @reuns's nice comment - basically combining it with Gerry's comment. We have $5=(2+i)(2-i)$ in $\Bbb{Q}(i)$. And there are two square roots of $-1$ in $\Bbb{Q}_5$. Therefore the two completions $\Bbb{Q}(i)_{(2+i)}$ and $\Bbb{Q}(i)_{(2-i)}$ are both isomorphic to $\Bbb{Q}_5$. But in the former $i\equiv 3\pmod I$, and in the latter $i\equiv 2\pmod I$ (here $I$ is the maximal ideal $5\Bbb{Z}_5$). All depending on which factor of $5$ we think of as an element of $I$. Of course, the two $i$s are just negatives of each other bringing us back to reuns' last comment. Dec 7, 2017 at 7:37

Since $5=(2+i)(2-i)$, the only prime ideals above $5$ are the ideals $(2\pm i)$. But in an extension of number fields $L/K$ , you have $[L:K] =\sum [L_w:K_v]$, the sum bearing over all the $w$'s above a fixed $v$. Here $v$ is the 5-adic valuation and the global degree is 2, hence the two completions are $\mathbf Q_5$. This is actually the phenomenon called total splitting of a prime.
Things are different for 2. Although $2 =(1+i)(1-i)$, this is not a genuine prime decomposition in $\mathbf Z[i]$ because the relation $i(1-i)=1+i$ shows that the two factors $1\pm i$ differ (multiplicatively) by a unit. In fact, $2$ decomposes as $2=-i (1+i)^2$, which means that $2$ is totally ramified, with ramification index $2$, so a fortiori the local degree is $2$. It follows that the completion at the prime $2$ is $\mathbf Q_2(i)$.