Let $p$ be a prime congruent to $1\mod4$. Then I know $x^2\equiv -1 \mod p$ has a solution. How do I show that $x^2=-1$ has a solution in $\mathbb{Q}_p$?

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    $\begingroup$ Do you know Hensel's Lemma ? $\endgroup$ – quid Jan 11 '16 at 19:53
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    $\begingroup$ en.wikipedia.org/wiki/Hensel%27s_lemma $\endgroup$ – Thomas Andrews Jan 11 '16 at 19:53
  • $\begingroup$ @quid ah, thanks $\endgroup$ – usr0192 Jan 11 '16 at 19:54
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    $\begingroup$ You can construct solutions modulo $p^n$ recursively, using Hensel, which is the $p$-adic version of Newton's method for solving an equation. $\endgroup$ – Thomas Andrews Jan 11 '16 at 19:54

Here's the answer, given in the comments by quid and Thomas Andrews, I'm writing it up for my own sake:

Let $f(x)=x^2+1$. We prove by induction on $n$ that $f(x)\equiv 0 \mod p^n$ has a solution $x_n \in \mathbb{Z}$ such that $x_n\equiv x_{n-1} \mod p^{n-1}$.

Base case: We know from elementary number theory that $f(x)=0$ has a solution $x_1 \in \mathbb{Z}$.

Inductive step: Assume for some $n\geq 1$ we have $x_n \in p^n$ such that $f(x_n)\equiv 0 \mod p^n$. Then $x_n \not\equiv 0 \mod p$ (otherwise $f(x_n)=x_n^2+1 \equiv 1 \mod p$) so $f'(x_n)=2x_n \not\equiv 0 \mod p$. Hence Hensel's lemma allows us to conclude there exists $x_{n+1} \in \mathbb{Z}$ such that $x_{n+1}=x_n \mod p^n$ and such that $f(x_{n+1})=0 \mod p^{n+1}$. Explicitly, we can take $$x_{n+1}=x_n + tp^n, \quad \text{where } t=-\frac{f(x_n)}{p^n}\cdot(f'(x_n))^{-1} $$ where the expression for $t$ is computed in $\mathbb{Z}/p$. (See the wikipedia link to hensel's lemma given in the comments by Thomas Andrews). Finally, the $\alpha=\{x_n\}$ forms an element in $\mathbb{Q}_p=\lim Z/p^n $ such that $\alpha^2+1 =0 $.


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