# $\mathbb{Q}_p$ not isomorphic to $\mathbb{Q}_q$ for $p\neq q$

I have to show that for $p, q$ prime and $p \neq q$ the fields $\mathbb{Q}_p$ and $\mathbb{Q}_q$ are not isomorphic. I have tried a proof but it seems far too complicated. What's the best way to approach this problem?

Ok, my idea was to assume $p < q$ and choose $u \in \{1, \ldots, p-1\}$ so that $u$ is not a square in $\mathbb{Q}_p$. The equation \begin{equation*} u X^2 + p Y^2 = Z^2 \end{equation*} has no solution since the Hilbert symbol $\left( \frac{u, p}{p} \right) = \left( \frac{u}{v} \right) = -1$.

All field isomorphisms "map the $1$ to the $1$", so suppose there is an isomorphism $\phi\colon \mathbb{Q}_p \rightarrow \mathbb{Q}_q$, we must have $\phi(u) = \phi(\underbrace{1+\cdots+1}_{u-\text{times}})= \phi(1) + \cdots + \phi(1) =1 + \cdots + 1=u$. Similarly for $p$.

Applying the isomorphism to the equation $u X^2 + p Y^2 = Z^2$ in $\mathbb{Q}_p$: $$\phi(u X^2 + p Y^2) = \phi(Z^2) \Longrightarrow u \,\phi(X)^2 + p \,\phi(Y)^2 = \phi(Z)^2$$ gives us an equation in $\mathbb{Q}_q$: $$u \,\tilde{X}^2 + p \,\tilde{Y}^2 = \tilde{Z}^2\, ,$$ but this equation is solvable in $\mathbb{Q}_q$, the Hilbert symbol $\left(\frac{u, p}{q} \right)= +1$, because $u, p\in\mathbb{Z}^\times_p$, which is a contradiction.

• One idea: Clearly $\mathbb{F}_p$ and $\mathbb{F}_q$ are different. Use Hensel's lemma to somehow "lift" the "differentness" of the residue fields. Mar 31, 2015 at 5:13

For any two primes $p \neq q$ (if one of them is $2$ then it should be $p$) you can find using CRT an integer $n$, not divisible by $p$ or $q$, such that $n$ is a quadratic residue $\bmod p$ but not $\bmod q$. It follows by Hensel's lemma (if $p = 2$ then you should require that $n \equiv 1 \bmod 8$) that $n$ has a square root in $\mathbb{Q}_p$ but not in $\mathbb{Q}_q$.