Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm a little stuck with the proof of a theorem I'm trying to understand. The theorem is as follows:

"For odd prime $p$, suppose for $\alpha \in Q_{p}$ (the p-adic rationals) that $|\alpha|_p=1$. Then $\exists\beta\in Q_p$ such that $\alpha=\beta^2\iff \exists\gamma\in Z/pZ$ such that $|\alpha-\gamma^2|_p<1$."

The proof is:

Suppose $\exists\gamma\in Q_p$ such that $|\alpha -\gamma^2|_p<1$, (i.e. $\beta^2\equiv\alpha(modp)$ is soluble). Now we construct a sequence $(\beta_n)$ by letting $\beta_1=\gamma$ and defining $\beta_n$ to satisfy:

$|\beta_n^2-\alpha|_p<\frac{1}{p^n}$ and $|\beta_{n+1}-\beta_n|<\frac{1}{p^n}$

If we take $\beta_n$ as given, then we take $\beta_{n+1}=\beta_n+\delta_n$, so that $\beta_{n+1}^2=\beta_n^2+2\beta_n\delta_n+\delta_n^2$, and it is sufficient to take $\delta_n=\frac{\alpha-\beta_n^2}{2\beta_n}$.

Conversely, the necessity is obvious if we choose $\gamma=\beta^2$. $\square$

I feel that I'm missing something which is stopping me from understand this. $|\beta_{n+1}-\beta_n|_p<\frac{1}{p^n}\implies p^n$ divides $(\beta_{n+1} - \beta_n)$, and if $\beta_{n+1}=\beta_n+\delta_n$ then this must mean $\delta_n$ is divisible by $p^n$.

If we take $\delta_n$ as $\delta_n=\frac{\alpha-\beta_n^2}{2\beta_n}$, then this gives us $\beta_{n+1}^2=\alpha+\delta_n^2 \implies \beta_{n+1}^2-\alpha=\delta_n^2$. Then we have shown that $\beta_{n+1}-\alpha$ is divisible by $p^{n+1}$, i.e. that $|\beta_{n+1}^2-\alpha|_p<\frac{1}{p^{n+1}}$.

Was this the aim of the proof? An inductive argument on the terms of $(\beta_n)$? If not what is it that I have misunderstood in this theorem? Many thanks in advance for any replies, I would love to understand this.

share|improve this question
2  
The idea is that the sequence $(\beta_n)$ is Cauchy (w.r.t the $p$-adic metric), so converges to an element $$\beta=\lim_{n\to\infty}\beta_n$$ in the complete field $\mathbb{Q}_p$. Also, simultaneously you prove that the sequence $(\beta_n^2)$ converges to $\alpha$, so at the limit you have $\beta^2=\alpha$. Your notation is a bit funny at some points. For example, surely $\gamma$ must come from the ring of $p$-adic integers $\mathbb{Z}_p$ as opposed to the quotient ring $\mathbb{Z}/p\mathbb{Z}$. True, the test still amounts to checking that $\alpha$ reduced modulo $p$ is a quadratic residue. –  Jyrki Lahtonen Jan 3 '13 at 14:41
    
Thanks for your reply. I see now that $(\beta_n)$ is Cauchy (as defined by $|\beta_{n+1}-\beta_n|_p \rightarrow 0$ I believe?). I think I'm starting to understand it, I see what you mean about having $\beta^2=\alpha$ at the limit. Is what I said towards the end of my post enough to show this, that if we assume $|\beta_n^2-\alpha|_p<\frac{1}{p^n}$ then $|\beta_{n+1}^2-\alpha|_p<\frac{1}{p^{n+1}}$? –  Traxter Jan 3 '13 at 14:53
    
Correct. That shows that $$\lim_{n\to\infty}\beta_n^2=\alpha.$$ Therefore (by continuity of the squaring map) $$\beta^2=(\lim_{n\to\infty}\beta_n)^2=\lim_{n\to\infty}\beta_n^2=\alpha.$$ BTW, this is also what Matt E is saying in his answer (+1). –  Jyrki Lahtonen Jan 3 '13 at 17:40
    
Ah great, thanks! –  Traxter Jan 3 '13 at 22:27
add comment

1 Answer

up vote 3 down vote accepted

The argument is constructing a Cauchy sequence of $p$-adic numbers $\beta_n$ whose limit is equal to a square-root of $\alpha$. (This is the standard way to exhibit a $p$-adic number with some property, since the $p$-adics are defined as a certain completion of $\mathbb Q$.)

share|improve this answer
    
Thanks for the reply Matt :) –  Traxter Jan 3 '13 at 22:27
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.