Equality of $p$-adic fields How can I prove that $\mathbb{Q}_3(\sqrt{2})=\mathbb{Q}_3(\sqrt{5})$. I only did modulo reduction, but how to prove it directly?
 A: One way of seeing it is to write $5 = 2 \cdot (1 + \frac{3}{2})$ and use the power series
$$(1+x)^{1/2} = 1 + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{1}{16} x^3 + \cdots$$
which converges $3$-adically for $x = \frac{3}{2}$ (since the coefficients are in $\mathbb{Z}[\frac{1}{2}]$, hence $\mathbb{Z}_3$ and since $\big|\frac{3}{2}\big|_3<1$).  So $1 + \frac{3}{2}$ has a square root in $\mathbb{Q}_3$ so $5$ has a square root in $\mathbb{Q}_3(\sqrt{2})$.
Equivalently, you can apply Hensel's lemma, which is more general but less explicit than the above convergent series (although it amounts to the same: Hensel's lemma is proved using $p$-adic approximation of this kind, typically with a kind of Newton's method).
Finally, a more sophisticated way of putting things (essentially Hensel's lemma) is that the finite unramified extensions of $\mathbb{Q}_p$ (meaning, those which leave the value group unchanged) naturally correspond to the finite extensions of $\mathbb{F}_p$ (and can be described explicitly as Witt vectors thereof): since $\mathbb{F}_3(\sqrt{2})$ and $\mathbb{F}_3(\sqrt{5})$ evidently coincide, and since $\mathbb{Q}_3(\sqrt{2})$ and $\mathbb{Q}_3(\sqrt{5})$ are unramified, they coincide.
A: The point is that $\frac{\sqrt{2}}{\sqrt{5}}$ is in $\mathbb{Q}_3$. To see this, just apply Hensel's lemma to the polynomial equation $5x^2-2 = 0.$ 
