I want to prove that $-1$ has no square root in $\mathbb Z_3$ (3-adic integers).
By the definition of inverse limit, we know that there is a ring homomorphism $\phi$ from $\mathbb Z_3$ to $\mathbb Z/3 \mathbb Z$. So, if $\alpha$ is the square root of $-1$ in $\mathbb Z_3$ then $\phi(\alpha)$ must be the square root of $-1$ in $\mathbb Z/3 \mathbb Z$. Since, there is no square root of $-1$ in $\mathbb Z/3 \mathbb Z$ this should imply $-1$ has no square root in $\mathbb Z_3$ (3-adic integers).
Is the above argument correct? or am I missing something?