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It is well documented that there are nine imaginary quadratic number fields with class number 1 and hence they have narrow class number 1 as well. However, are there imaginary quadratic number fields with class number > 1 and narrow class number 1? If so, are there lots of them?

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if you tell us what you understand the equivalence relations defining the class group and the narrow class group to be, I think you will find the answer to this question. –  KCd Apr 20 '12 at 22:16
Note, by the way, that the real quadratic field ${\mathbf Q}(\sqrt{3})$ has class number 1 and narrow class number 2. Think about what you're asking with this example in mind, even though it is not imaginary quadratic. –  KCd Apr 20 '12 at 22:22
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1 Answer

up vote 6 down vote accepted

The narrow class number of a number field $K$ is just the cardinality of the corresponding narrow class group $\operatorname{Cl^+}(K) = I(K)/P^+(K)$ where $P^+(K)$ is the group of principal fractional ideals $(\alpha) = \alpha \mathcal{O}_K$ whose generator $\alpha \in K$ is totally positive.

Now, this is basically the important part here, because $\alpha$ is totally positive if for every embedding $\sigma: K \hookrightarrow \mathbb{R}$, the image of $\alpha$ under the embedding $\sigma(\alpha) > 0$.

Then in the case you're interested in, say when $K = \mathbb{Q}(\sqrt{d})$ is an imaginary quadratic field, there are no embeddings $\sigma: K \hookrightarrow \mathbb{R}$, thus the condition for a principal fractional ideal $(\alpha)$ to belong to $P^+(K)$ is trivially satisfied, therefore $P^+(K) = P(K)$, the whole group of principal fractional ideals.

Thus in the case when $K$ is an imaginary quadratic number field, or more generally a number field with no real embeddings, the narrow class group and the class group are actually the same and thus the class number and the narrow class number are also the same.

Added: Matt E's comment

As Matt E comments, the ideal class group $\operatorname{Cl}(K)$ is a quotient of the narrow class group by virtue of the third isomorphism theorem for groups, that is

$$ \operatorname{Cl}(K) = I(K)/P(K) \cong \frac{\frac{I(K)}{P^+(K)}}{\frac{P(K)}{P^+(K)}} = \frac{\operatorname{Cl^+}(K)}{\frac{P(K)}{P^+(K)}} $$

and therefore as he observes in his comment below, if the narrow class group $\operatorname{Cl^+}(K)$ has order $1$ then as a consequence the class group $\operatorname{Cl}(K)$ also has order $1$ and thus

$$\text{narrow class number} = 1 \implies \text{class number} = 1$$

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Dear Adrian, This is a nice and succinct explanation. You could also make explicit the fact that the class group is a quotient of the narrow class group, and hence that if the latter is trivial (has order $1$), so is the former. So, essentially for reasons of definition, class number $> 1$ but narrow class number $1$ is impossible for any number field. Regards, –  Matt E Apr 21 '12 at 1:02
@MattE Dear Matt, thank you very much for your comment. I have added that approach to my answer as you suggested. –  Adrián Barquero Apr 21 '12 at 3:17
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