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.

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    $\begingroup$ Note that $i$ is in ${\bf Q}_5$. $\endgroup$ Dec 6, 2017 at 5:32
  • $\begingroup$ Pardon, So what happens $\endgroup$
    – Math123
    Dec 6, 2017 at 5:35
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    $\begingroup$ @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. $\endgroup$ Dec 6, 2017 at 6:19
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    $\begingroup$ $(1+i) = (1+i,2)= (1-i,2) = (1-i)$ 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 $\widehat{\mathbf{Q}(i)_{(2+i)}}$. $\endgroup$
    – reuns
    Dec 6, 2017 at 10:02
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    $\begingroup$ 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. $\endgroup$ Dec 7, 2017 at 7:37

1 Answer 1


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 $\mathbf Q_2(i)$.


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