# definition of conductor of abelian extension

I study the book Silverman`s Advanced Topics in the Arithmetic of Elliptic curves. I have some questions about the notion "conductor" of an abelian extension and Hilbert class field.

Proposition 3.1(Artin reciprocity) Let $L/K$ be a finite abelian extension of number fields. There exists 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 $\mathfrak{c})$.

I try to show that if the proposition is true for any two ideals $\mathfrak{c}_1,\mathfrak{c}_2$ then also true for $\mathfrak{c}_1+\mathfrak{c}_2$. Since $\mathfrak{c}_1,\mathfrak{c}_2$ are divisible by all ramified primes then it follows that $\mathfrak{c}_1+\mathfrak{c}_2$ is divisible by all ramified primes. But I coud not show that $$for \ all \ \alpha\in K^* \ with \ \alpha\equiv 1 \ mod \ (\mathfrak{c}_1+\mathfrak{c}_2).$$ Then the book defines the "conductor" of the extension $L/K$ as the largest ideal for which proposition is true and denote it by $\mathfrak{c}_{L/K}$.

My second question: Let $H$ be the Hilbert class field(max. unramified abelian ext.) of a number field $K$. Then how to show the conductor $\mathfrak{c}_{H/K}$ of $H/K$ is equal to $(1)$. According to the definition of conductor, do we assume that the ring $(1)=\mathcal{O}_K$ is divisible by all ramified primes, I am a bit confused.

Third question: Let H Hilbert class field of a number field K. The book defines $I(\mathfrak{c})$ := group of fractional ideals of $K$ which are "relatively prime" to $\mathfrak{c}$ and $P(\mathfrak{c}):=\{ (\alpha): \alpha\in K^*,\ \alpha\equiv 1 \ mod \mathfrak{c}\}$. What does it mean "relatively prime" for fractional ideals. If it means $$I(\mathfrak{c})=\{\mathfrak{a}\in I_K : \mathfrak{a}+\mathfrak{c}=\mathcal{O}_K \}$$ then since $\mathfrak{c}$ is integral, it follows that all elements of $I(\mathfrak{c})$ are "integral" ideals. Similarly, since $\mathfrak{c}$ is an integral ideal and $1\in\mathcal{O}_K$, all elements of $P(\mathfrak{c})$ are "integral" principle ideals. But these then do not need to be groups, isnt it? According to the book, if we set $\mathfrak{c}=(1)$ we would get $I((1))=I_K$= group of non-zero fractional ideals, and $P((1))$ = group of principle ideals. As a result I could not get ideal class group $Cl(K)$ so I could not infer the important fact that Artin map induces isomorphism between ideal class group $Cl(K)$ and Galois group of $H/K$.

• @user1952009 in the definition of $P(\mathfrak{c})$, the condition $\alpha\equiv 1$(mod$\mathfrak{c}$) does not permit non integral fractional ideals. I still dont understand why does $P((1))$ consists of all fractional ideals of $K$. – ersin May 22 '17 at 10:11

For the third question: two fractional ideals $\mathfrak{a}$ and $\mathfrak{b}$ are relatively prime if in their factorization in integral primes (wich may include negative powers) they don't share any factors, i.e, $\text{min}(\text{ord}_{\mathfrak{p}}(\mathfrak{a}),\text{ord}_{\mathfrak{p}}(\mathfrak{b}))=0$, for all primes $\mathfrak{p}$.

The conductor of $H$ is $(1)$ because by definition of the Hilbert class field $\mathfrak{c}=(1)$ satisfies the proposition of the Artin reciprocity, since there can't be any larger ideal with the same property ($(1)$ is the largest of all integral ideals) this must be the conductor.

if the prop. is true for any two ideals $\mathfrak{c}_1,\mathfrak{c}_2$ then also true for $\mathfrak{c}_1+\mathfrak{c}_2$.

This is equivalent to the existence of the conductor, I guess the author doesn't include a proof because is a brief review of class field theory. Here is a proof based on Januz Algebraic Number fields (pp.201 Lemma 6.2)

Let $\mathfrak{c}=\mathfrak{c}_1+\mathfrak{c}_2$ and suposse $\alpha \equiv 1 \bmod \mathfrak{c}$, the key step is to find $\beta \in K^*$ with the following conditions: $\beta \equiv \alpha\bmod \mathfrak{c}_2$, $\beta \equiv 1\bmod \mathfrak{c}_1$ and $(\beta)$ is prime to $\mathfrak{c}$.

To find this $\beta$ we use the chinesse remainder theorem. Let $\mathfrak{p}$ be a prime and $\mathfrak{p}^{a_1}$, $\mathfrak{p}^{a_2}$ the exact powers that divide $\mathfrak{c}_1$ and $\mathfrak{c}_2$, if $a_2\leq a_1$ we require $\beta \equiv 1\bmod \mathfrak{p}^{a_1}$ and if $a_1\leq a_2$ we require $\beta \equiv \alpha \bmod \mathfrak{p}^{a_2}$, because $\alpha \equiv 1 \bmod \mathfrak{c}$ the $\beta$ constructed in this way satisfies what we want.

Now since the proposition is true for $\mathfrak{c}_2$ we have $(\alpha,L/K)=(\beta,L/K)$ and since is true for $\mathfrak{c}_1$ we also have $(\beta,L/K)=1$, thus $(\alpha,L/K)=1$.

• Yes, but why you include $J$ ? It's enough to ask $I$ and $(a)$ to be relatively prime to $\mathfrak{c}$. – CRKarl May 21 '17 at 22:47
• Let $H$ be the Hilbert class field of a number field $K$. Now I try to show the ideal $(1)$ is indeed the conductor of $H/K$ Let $\alpha \equiv 1 (mod (1))$ and $\alpha\mathcal{O}_K=\mathfrak{p}_1...\mathfrak{p}_n$. Then I need to show $((\alpha),H/K)=1\in Gal(H/K)$ i.e. $\sigma_{\mathfrak{p}_1}...\sigma_{\mathfrak{p}_n}=1$, where $\sigma_\mathfrak{p}$ is Frob element at $\mathfrak{p}$. Maybe if all primes would completely split then all $\sigma_{\mathfrak{p}_i}$ would be identity so we were done. So I dont know how to proceed. – ersin May 21 '17 at 22:51
• Well, What I had in mind is the usual definition of $H$ through the existence theorem which tell us that for each $\mathfrak{c}$ and group $R$ such that $P(\mathfrak{c}) \subset R \subset I(\mathfrak{c})$ there is a unique extention $L/K$ ramifed only in the primes that divide $\mathfrak{c}$ such that the kernel of the artin map $( \cdot,L/K)$ is exactly $R$. Then one defines $H$ to be the extention for $\mathfrak{c}=(1)$ such that the kernel is $P((1))$ and from this we see that $H/K$ the maximal unramified extention of $K$, but from the definition $((\alpha),H/K)=1$ for any $\alpha$. – CRKarl May 21 '17 at 23:05