There exist no ring homomorphism $\sigma$ between $\mathbb{Q}_3$ and $\mathbb{Q}_5$ with $\sigma(1)=1$. There exist no ring homomorphism $\sigma$ between $\mathbb{Q}_3$ and $\mathbb{Q}_5$ with $\sigma(1)=1$.
Approach: Since $\sigma(1)=1$, we can conclude that $\sigma(q) = q$ for all $q \in \mathbb{Q}$. We have that $$1+3+3^2+3^3+... \to (1-3)^{-1}$$
in $\mathbb{Q}_3$. But this does not hold in $\mathbb{Q}_5$. Therefore there exist no such ring homomorphism. Is that correct?
 A: (First, no, there is assumption of continuity, so convergence arguments are not directly relevant...)
Thinking in terms of Hensel's lemma, there will be different square roots (of rational integers) in the two $p$-adic fields. For example, exactly the rational integers $n=\pm 1\mod 5$ have square roots in $\mathbb Z/5$, and then, by Hensel's lemma, in $\mathbb Z_5\subset \mathbb Q_5$. Similarly, exactly the rational integers $n=1\mod 3$. (One should also note that a square root $\alpha$ in $\mathbb Z_p$ of a rational integer $n$ is integral over $\mathbb Z$, so is integral over $\mathbb Z_p$, so is in $\mathbb Z_p$ rather than $\mathbb Q_p$.)
Thus, for example, there is a $\sqrt{7}$ in $\mathbb Q_3$, but not in $\mathbb Q_5$. Oppositely, there is a $\sqrt{-1}$ in $\mathbb Q_5$, but not in $\mathbb Q_3$. So there can be no inclusion of either in the other (regardless of topologies).
A: No. The ring homomorphism need not be continuous with respect to the $3$- and $5$-adic (or any) topology.
Instead, look at the polynomial in $X^2 - 10$ in both fields. Use that $ℚ_5$ is the quotient field of the factorial ring $ℤ_5$ (in which $5$ is prime) and use Hensel on the polynomial in $ℚ_3$.
(Note that $ℚ_3$ is a field, so any such ring homomorphism $σ$ would be an embedding.)
