This is a subtask of Problem 226 from Fernando Gouvea's book $p$-adic numbers - An Introduction. In this book, the author mentions that the ramification index $e = e(\mathbb{Q}_5(\sqrt{2})/\mathbb{Q}_5)$ is exactly $1$ and I want to understand why this is true.
- As the ramification index is a divisor of $[\mathbb{Q}_5(\sqrt{2}):\mathbb{Q}_5]=2$, we must have $e \in \{1,2\}$.
Let us assume $e = 2$. If we can find a contradiction, then $e$ must be $1$ and we are done.
- The ramification index $e$ is defined to be the unique positive integer such that $v(\mathbb{Q}_5^\times) = \frac{1}{e}\mathbb{Z}$ where $v$ denotes the valuation on $\mathbb{Q}_5(\sqrt{2})$. If $e$ is $2$, then there must exist an $x=a+b\sqrt{2}$ with $a,b \in \mathbb{Q}_5$ so that $v(x) = \frac{1}{2}$.
- Furthermore, the coefficient $b$ must be non-zero, otherwise $v(x)$ would be an integer which is a contradiction to $v(x) = \frac{1}{2}$.
- Calculating yields us $$\frac{1}{2} = v_5(x) = (a+b\sqrt{2})(a-b\sqrt{2})= a^2-2b^2$$ and would like to show that there can not exist such coefficients $a,b \in \mathbb{Q}_5$ in the first place. I can see that $a,b$ can not be both integers. But I was not able to extend my solution further.
Could someone help me with this problem? Thank you!