Let $L/K$ be an unramified abelian extension. I want to show that its conductor is (1). To do this, I need to show the following two facts hold:
- (1) is divisible by all primes of $K$ that ramify in $L$.
- For all $\alpha\in K^*$ with $\alpha \equiv 1$ mod $(1)$, $((\alpha),L/K)=1$ where $(.,L/K)$ is the Artin map.
Ok, since there is no ramified primes, first condition is satisfied. But, I have problem at the second condition. How can we ensures that $((\alpha),L/K)=1$. My attempt: Write $(\alpha)=\prod_\mathfrak{p} \mathfrak{p}^{n_\mathfrak{p}}$. Then $((\alpha),L/K)=\prod_\mathfrak{p} \sigma_\mathfrak{p}^{n_\mathfrak{p}}$ where $\sigma_\mathfrak{p}$`s are Frobenius element at $\mathfrak{p}$. Now I expect the Frobenius elements to be identity. I would be grateful, if anyone enlighten me.