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I am trying to construct a sequence in $\mathbb Q_2$ that is formed of rational numbers and converges to $\sqrt{-7}$, to prove that $(\mathbb Q, |\cdot|_2)$ is not complete. My lecturer stated that this was an easy example but I am getting confused by my working. (Also I have just been introduced to $p$-adic numbers, so I can't use Hansel's Lemma, for example.)

So far, I said that if $\{x_n\}_{n\in \mathbb N} \to \sqrt{-7}$ then $\{x_n^2\}_{n\in \mathbb N} \to -7$ and so I must have $x_n^2 + 7$ divisible by a large power of $2$ for large $n$. Suppose I want that $2^n|x_n^2+7$ for all $n.$ I would then, for every $n$, need a root of $x^2+7$ in $\mathbb Z/2^n\mathbb Z$. But I'm unsure how to approach this kind of problem; finding quadratic residues modulo $p^n$ (here $p=2$) rather than modulo primes or products of distinct primes (where I can use the Chinese Remainder Theorem).

I would appreciate any help with my proof or an explanation of how to find quadratic residues modulo $p^n$, or both.

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Suppose you have $x_n$ such that $2^n\mid x_n^2+7$. If $2^{n+1}\mid x_n^2+7$, you can let $x_{n+1}=x_n$. If not, then $$(x_n+2^{n-1})^2+7=(x_n^2+7)+2^n x_n+ 2^{2n-2}$$ is divisible by $2^{n+1}$ (as long as $n\geq 3$) since $x_n$ must be odd. So in this case, you can define $x_{n+1}=x_n+2^{n-1}$. By induction, you can thus define a sequence $(x_n)$ such that $2^n\mid x_n^2+7$. Moreover, this sequence is Cauchy with respect to the $2$-adic norm, since by construction $|x_n-x_m|_2\leq 2^{-n+1}$ for $m\geq n$.

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  • $\begingroup$ I must be missing something, but why is the quantity divisible by $2^{n+1}?$ I would actually agree if $x_n$ was even, but as you said, it's odd. EDIT: Sorry, I didn't realise that this was the case where $2^{n+1}$ doesn't divide $x_n^2+7$. My bad. $\endgroup$ – James Jul 29 '15 at 1:45
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Just use the Binomial expansion, for $(1-8)^{1/2}$. It’s a fact, not too terribly hard to prove, that $(1+4t)^{1/2}$ is a series in $t$ with all integer coefficients. Plug in $-2$ for $t$, voilà.

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  • $\begingroup$ Thanks for your answer. I have a small question though, since I don't know much about the $p$-adics yet: How would I know whether or not the series given by the usual binomial expansion would converge to $\sqrt{-7}$ using the $2$-adic valuation, as opposed to the usual absolute value? I imagine the "4" will mean that the sequence will be $2$-adically Cauchy, but will this trick always work? Does that mean $\sqrt{-3}\in \mathbb Q_2$ also? $\endgroup$ – James Jul 30 '15 at 15:10
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    $\begingroup$ No, the series is not convergent when you plug in $-1$ for $t$; I didn’t say it was. But since the $n$-th degree (in $t$) term is divisible by $2^n$ after the substsitution, you definitely get a convergent series. It converges to $\sqrt{-7}$ because when you take the series in $t$ and square it, you get $1+4t$. This process is consistent with the procedure of substituting something small for $t$ $\endgroup$ – Lubin Jul 31 '15 at 2:30

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