I am trying to learn how to generate hilbert class field of an imaginary quadratic field $K$ in the book Silvermans Advanced Topics in the Arithmetic of Elliptic curves. I dont understand some fact in the proof. Let $$(.,L/K): I^\mathfrak{m} \longrightarrow \textrm{Gal}(L/K)$$ be the Artin map for the abelian extension $L/K$ with the conductor $\mathfrak{m}$. We want to prove $L$ is the maximal unramified abelian extension (Hilbert class field) of $K$. At some point in the proof, I reached the following fact: $$((\alpha),L/K)=1 \textrm{ for all } (\alpha)\in I^\mathfrak{m}.$$ Now, somehow I need to see that $\mathfrak{m}$ has to be $(1)$. Because then I will conclude that $L/K$ is unramified. However, to show that $(1)$ is the conductor of the extension $L/K$, first I need to show that $(1)$ is divisible by all ramified primes of $K$, so I need to determine all ramified primes isn`t it? But the book without any explanation just says, the above fact implies that the conductor of $L/K$ is trivial, by the definition of conductor. Maybe I have problem at the definition of conductor. If anyone enlighten me, I would be glad.

Definition of conductor: Let $L/K$ be a finite abelian extension of number fields. Artin reciprocity guarantees the existence of an integral ideal $\mathfrak{c}$ of $K$, divisible by all the primes of $K$ that ramify in $L$, such that $((\alpha,L/K))=1$ for all $\alpha\in K$* satisfying $\alpha\equiv 1$ (mod $\mathfrak{c})$. The largest integral ideal of $K$ among such ideals is called the conductor of $L/K$.

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    $\begingroup$ You're right: the conductor for the Hilbert class field is "1": that is, that ideal class and corresponding extension is "everywhere unramified". Since the case of $\mathbb Q$ cannot illustrate this, it might be useful to do "the next" example of the Hilbert class field of $\mathbb Q(\sqrt{-5}$ and similar "small" ones. "Unramified" is often awkward to see/describe as the null case of "ramification"... $\endgroup$ – paul garrett May 22 '17 at 23:09
  • $\begingroup$ I want to show $L/K$ is unramified everywhere. To conclude this, I need to show that the its conductor is (1). This is the exact part I have problem. $\endgroup$ – ersin May 23 '17 at 11:18
  • $\begingroup$ and what is your definition of the Hilbert class field ? $\endgroup$ – mercio May 23 '17 at 17:55
  • $\begingroup$ I define hilbert class field as maximal unramified abelian extension. $\endgroup$ – ersin May 23 '17 at 20:56
  • $\begingroup$ I don't understand your question. If L has conductor m, it cannot be the Hilbert class field unless $m = (1)$. $\endgroup$ – franz lemmermeyer May 26 '17 at 9:04

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