2
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I tried to calculate the class number with help of the Minkowski bound of $M \approx 5$. So if an ideal has norm $1$, it is the ring of integers. If it has norm $2$, it is $(2, 1+\sqrt{-17})$, which is not principal. Norm 3 gives us the ideals $(3, 1+\sqrt{-17})$ and $(3, 2+\sqrt{-17})$, which are both not principal. The only ideal of norm 4 is the ideal (2) and there arent any of norm 5.

I don't know how to find the class number given this information. Can you guys please help?

Thanks in advance!

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4
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From your calculation, you know the following facts about the class number $ h = h_{\mathbb{Q}(\sqrt{-17})} $ already

  1. $ h > 1 $, because $ \mathfrak{p}_2 = (2, 1 + \sqrt{-17}) $ is a non-principal ideal

  2. $ h \leq 5 $, because you have found 5 ideals with norm less than the Minkowski bound.

We have $ h = 2, 3, 4, \text{ or } 5 $. But to pin down $ h $, you need to do more work...

Step 1: From your computation of $ \mathfrak{p}_2 = (2, 1 + \sqrt{-17}) $ being an ideal of norm 2, you probably have found that $ (2) = \mathfrak{p}_2^2 $ since $ -17 \equiv 3 \pmod{4} $. Or just directly check this by computing the product $ \mathfrak{p}_2^2 $.

Because $ (2) $ is principal, this shows that in the class group, $ [\mathfrak{p}_2] $ is an element of order 2. But what do you know about the order of an element in a finite group? It must divide the order of the group. So which possibilities for $ h $ can we now eliminate?

Step 2: We can ask a similar question about $ \mathfrak{p}_3 = (3, 1 + \sqrt{-17}) $: is $ \mathfrak{p}_3^2 $ principal? If it was, then it would be an ideal of norm $ 3 \times 3 = 9 $. So it would be generated by an element $ \alpha $ with norm $ \pm 9 $. But solving $ N(\alpha) = x^2 + 17y^2 = 9 $ shows $ \alpha = \pm 3 $. But when you have factored $ (3) $ to find $ (3) = \mathfrak{p}_3 \widetilde{\mathfrak{p}_3} $, you can also say $ \mathfrak{p}_3 \neq \widetilde{\mathfrak{p}_3} $. This means that by the unique factorisation of ideals in number rings, $ \mathfrak{p}_3^2 \neq (3) $, so it cannot be principal.

(Alternatively: compute that $ \mathfrak{p}_3^2 = (9, 1 + \sqrt{-17}) $. If this is going to equal $ (3) $, then we must have $ 1 + \sqrt{-17} \in (3) $, but clearly $ 3 \nmid 1 + \sqrt{-17} $, so this is not possible. Again conclude $ \mathfrak{p}_3^2 $ is not principal.)

This means that in the class group $ [\mathfrak{p}_3] $ has order $ > 2 $. This shows that $ h > 2 $ because the order of an element must divide the order of the group.

Conclusions: You have enough information now to conclude that $ h = 4 $, agreeing with Will Jagy's answer. Step 1 shows that $ h = 2, 4 $, and step 2 shows that $ h = 4 $, so we're done.

In fact we also know the structure of the class group $ \mathcal{C}(\mathbb{Q}(\sqrt{-17})) $. It is a group of order $ h = 4 $, so either $ \mathcal{C}(\mathbb{Q}(\sqrt{-17})) \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \text{ or } \mathbb{Z}_4 $. But since $ \mathfrak{p}_3 $ has order $ > 2 $, it must have order 4. This shows that $ \mathcal{C}(\mathbb{Q}(\sqrt{-17}) \cong \mathbb{Z}_4 $.

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1
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I get $4.$ Maybe someone will describe that in language you find suitable. For positive forms (imaginary fields), the number of classes of forms, same as number of reduced forms of the discriminant $-68,$ agrees with your calculation.

 jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./classGroup
Absolute value of discriminant? 
68
Discr  -68 = 2^2 * 17  class  number  4

 all  
      68:  < 1, 0, 17>    Square        68:  < 1, 0, 17>
      68:  < 2, 2, 9>    Square        68:  < 1, 0, 17>
      68:  < 3, -2, 6>    Square        68:  < 2, 2, 9>
      68:  < 3, 2, 6>    Square        68:  < 2, 2, 9>
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