Question: Determine the $p$-adic fields which contain $$ a)\;\sqrt{-1} \qquad b)\;\sqrt{3} \qquad c)\;\sqrt{-7} \qquad d)\;\sqrt{17}$$
I have no idea on this as I am completely confused with $p$-adic fields. I mean they are field of fraction of $p$-adic ring $\mathbb{Z}_p$ and elements of $\mathbb{Z}_p$ are of the form $(\alpha_o,\alpha_1,\ldots)$ s.t. $\alpha_i \in \mathbb{Z}/p^{i+1}\mathbb{Z}$, so I have no idea how can a number like $\sqrt{7}$ be in $\mathbb{Q}_p$.
If someone can show me the working for part a) and c) it will be a huge help in understanding the concept and then I will be able to do the other two parts myself too. Thanks!
There are several ways of attacking questions like this. The most general is to transform your defining equation (such as for $\sqrt{-7}$, which I’ll use as the type example) into something to which Hensel’s Lemma applies. If you haven’t seen Hensel, I’ll show you another method further on.
Recall that Hensel says that if you have a polynomial $f(X)\in\mathbb Z_p[X]$ with the property that when you reduce its coefficients modulo $p$, you get a factorization into relatively prime factors, then you can get a corresponding factorization back in $\mathbb Z_p[X]$, in such a way that one of the “lifts” has the same degree as the factor in characteristic $p$.
Let me give an example unrelated to $\sqrt{-7}$ first. Look at the polynomial $pX^4 +X^2+X + 1$. Then when you look at this as a polynomial over $\mathbb Z/p\mathbb Z=\mathbb F_p$, you get the factorization $1\cdot(X^2+X+1)$. Hensel says that you can factor your original in the form $g\cdot h$ where $g$ has only one coefficient not divisible by $p$, and that’s its constant term; and $h$ is of degree $2$ (like $X^2+X+1$), and furthermore, all the coefficients are congruent to $1$ modulo $p$.
Now let’s look at finding a square root of $-7$ in $\mathbb Z_2$. The polynomial $f(X)=X^2+7$ is no good for Hensel, ’cause in characteristic $2$, it’s a square: your factors are not relatively prime. But if $f(X)=X^2+7$, then $g(X)=f(X+1)$ is $X^2+2X+8$, and if we can factor this over $\mathbb Z_2$, we’re good. Still, this $g$ is not good for Hensel, but look at $h(x)=\frac14g(2X)=X^2+X+2$ Now the factorization in characteristic two is $X^2+X=X\cdot(X+1)$, the factors are relatively prime, so Hensel applies, and you get an element $(\sqrt{-7}-1)/2$ in $\mathbb Z_2$, and thus a square root of $-7$.
For other primes than two, $\sqrt{-7}$ is much easier. Look at $p=11$, for example. Here, $X^2+7\equiv X^2-4\pmod{11}$, and this certainly factors into two relatively prime factors, so you see immediately that $\sqrt{-7}\in\mathbb Z_{11}$. Indeed, for primes different from $2$, $\sqrt{-7}\in\mathbb Z_p$ if and only if $-7$ is a square modulo $p$, and for a general criterion for when this happens, you call in Quadratic Reciprocity.
Now let me show you a more elementary method of finding $\sqrt{-7}$ in $\mathbb Z_2$, using the Binomial expansion of $(1+a)^{1/2}$. If you’ve ever written several terms of this expansion down, you’ve noticed that the only denominators are powers of $2$,and the exponents on the $2$ don’t grow too fast. In particular, when you write down the expansion of $(1-8)^{1/2}$, all the coefficients are integers, and the coefficients moreover are more and more highly divisible by $2$. So this series is Cauchy in $\mathbb Z_2$, and has a limit, a square root of $-7$.
Another way would be to apply Newton-Raphson to find a root of $X^2+7$, starting with the “almost-root” $1$. This is the idea of the “baby” form of Hensel’s Lemma, and it’s actually the fastest way of getting a ($p$-adic) numerical root of an equation. But I’ve already written too much.