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I have seen that you can define the usual modular arithmetic on the $p$-adic integers: For $a,b\in\mathbb{Z}_p$ and a prime $p$, $$a\equiv b\pmod{p}\iff (a-b)/p\in \mathbb{Z}_p.$$

My question is, can we also define the Legendre symbol of $p$-adic integers?

For example, for a prime $p$ and $a\in\mathbb{Z}_p$, define $$\left(\frac{a}{p}\right) = \begin{cases} 1\quad\text{if $x^2\equiv a\pmod{p}$ has a solution $x\in\mathbb{Z}_p$} \\ -1\quad\text{if $x^2\equiv a\pmod{p}$ has no solution $x\in\mathbb{Z}_p$}\\ 0\quad\text{if $a\equiv0\pmod{p}$}\end{cases}.$$

Can I use the usual properties of the Legendre symbol in this case?

Jean Marie
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the inner beauty
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    Does this answer your question? Square roots in the $p$-adics – Anne Bauval Apr 23 '23 at 07:44
  • @AnneBauval, I believe the link asks which integers are square roots in $\mathbb{Q}_p$. I was asking about the use of the usual Legendre symbol on $\mathbb{Z}_p$ and not $\mathbb{Q}_p$. Can I assume the usual properties of Legendre symbol in this case? – the inner beauty Apr 23 '23 at 08:09
  • I did not think much about it but I believe Bruno Joyal's answer answers your post. – Anne Bauval Apr 23 '23 at 08:17
  • @AnneBauval The value of $a$ (in the link, it's $n$) they were talking about is an element of $\mathbb{Z}$. In my question, $a\in \mathbb{Z}_p$. That alone, makes a difference between my question and the answer of Joyal. – the inner beauty Apr 23 '23 at 08:54
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    The answers don't assume that it is in $\Bbb{Z}$. For $p\ne 2$ on $\Bbb{Z}_p^\times$ the Legendre symbol is the same as on $\Bbb{F}_p^\times$, but on $p \Bbb{Z}_p$ it needs to take in account if the valuation is even and the image in $\Bbb{Q}_p^\times/p^\Bbb{Z}$ which makes it non-continuous (for the $p$-adic topology). – reuns Apr 23 '23 at 09:02
  • What if I assume $v_p(a)=0$, and define $$\left(\frac{a}{p}\right) = \begin{cases} 1\quad\text{if $x^2\equiv a\pmod{p}$ has a solution $x\in\mathbb{Z}_p$} \ -1\quad\text{if $x^2\equiv a\pmod{p}$ has no solution $x\in\mathbb{Z}_p$}.\end{cases}.$$

    Is the problem already fixed?

     – the inner beauty Apr 23 '23 at 09:16
  • I have added "p-adic integers" in your title in order to avoid confusion with the "ordinary" integers modulo $p$. – Jean Marie Apr 23 '23 at 11:58

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