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Does there exist any ring embedding from $\mathbb{Z}_p$, the ring of $p$-adic integers to $\mathbb{R}$ ?

One can realise $\mathbb{Z}_p=\varprojlim \mathbb{Z}/p^{n}\mathbb{Z}$. I can't produce any injection to $\mathbb{R}$

user371231
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    Certainly not: many negative integers are perfect squares by Hensel’s Lemma. – KCd Jul 05 '20 at 12:09
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    Also you could consider roots of unity. For $p>3$ that should resolve the question. For $p=2$ as remarked by @KCd you have $-7$ is a square and for $p=3, -2$ is a square. – sharding4 Jul 05 '20 at 12:11
  • Thanks both of you. – user371231 Jul 05 '20 at 12:26
  • @sharding4 Can you explain the $p=2$ case ? – user371231 Jul 05 '20 at 12:51
  • It's KCd's remark. Since $-7$ is a square in $\mathbb{Z}_2$, any ring embedding of $\mathbb{Z}_2$ into $\mathbb{R}$ would take a square root of $-7$ to a square root of $-7$ in $\mathbb{R}$ which is impossible. – sharding4 Jul 05 '20 at 12:55
  • @sharding4 But why $-7$ is a square in $\mathbb{Z}_2$ ? Hansel’s lemma can’t be applied right ? am I doing any mistake ? – user371231 Jul 05 '20 at 14:05
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    Hensel's Lemma is a bit of a problem in this case, but that doesn't mean $-7$ doesn't have a square root in $\mathbb{Z}_2$. See this post https://math.stackexchange.com/q/2298779/254075 for more details. The simplest answer would seem to be to calculate $\alpha=(1+\sqrt{-7},)/2$ which has minimal polynomial $\alpha^2-\alpha+2$. Then $\sqrt{-7}=2\alpha-1$ – sharding4 Jul 05 '20 at 14:51
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    @user371231 Hensel's lemma in its "full" form can absolutely be applied to $x^2 + 7$ in $\mathbf Z_2[x]$. Perhaps you have only seen the version involving lifting a simple root mod $p$ to a $p$-adic root. More general versions of Hensel's lemma exist that let you lift non-simple roots mod $p$ under suitable conditions. See section 4 of the Wikipedia page on Hensel's lemma: for $f(x) = x^2 + 7$, $|f(1)|_2 = 1/8$ and $|f'(1)|_2^ = 1/4$, so from $1/8 < 1/4$ there is $\alpha \in \mathbf Z_2$ such that $f(\alpha) = 0$. – KCd Jul 05 '20 at 17:13
  • I think that a very pleasant way of showing that $\sqrt{-7}\in\Bbb Q_2$ is to notice that $X^2+X+2$ clearly has a root in $\Bbb Q_2$, by any form of Hensel that you like. And its roots are $\frac{-1\pm\sqrt{-7}}2$. – Lubin Jul 05 '20 at 18:41

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