How does the prime (3) ramify in the extension?
问题内容
I extend $\mathbb Q$ to first adjoin the roots of $x^2-1=9$ and then the roots of $(x^2-1)^2-1=9$, i.e. $x^4-2x^2-9=0$. In the second extension, I have to determine whether the prime ideal $(3)$ ramifies. I use the Newton Polygon to only deduce that $1$ determines the ramification index, which does not say much. The factorization $x^2(x^2-2)$ modulo $3$ seems to indicate there might be ramification but I want to consult you guys about what can I look for further.
回答 (1)
The roots of $x^4-2x^2-9 = (x^2-1)^2 - 10$ are $\pm\sqrt{1\pm\sqrt{10}}$.
We have $\sqrt{10}=\sqrt{1+9}\in\mathbb Q_3$, given by the binomial series $\sqrt{1+9} = 1 + \frac12 9 + O(3^3) = 1 - 9 + O(3^3)$. This implies that also $\pm\sqrt{1-\sqrt{10}} = \pm\sqrt{1 - (1 - 9 +O(3^3))} = \pm3 + O(3^2)\in\mathbb Q_3$.
The other two roots are not in $\mathbb Q_3$, but we can still find an "easy" extension of $\mathbb Q_3$ containing them: We have $\sqrt{1+\sqrt{10}} = \sqrt{2 + 9 + O(3^3)} = \sqrt2 \sqrt{1+ O(9)}$. That last square root is in $\mathbb Q_3$, so $\mathbb Q_3(\sqrt{2})$ contains those remaining roots $\pm\sqrt{1+\sqrt{10}}$. Since $\mathbb Q_3(\sqrt2)\supset\mathbb Q_3$ is unramified, we finally get that $(3)$ is unramified in the original extension of $\mathbb Q$.
We actually got more info, namely that the factorization of $(3)$ is $P_1 P_2 Q$ with $P_1$ and $P_2$ of norm $3$ and $Q$ of norm $9$.