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The Minkowski bound for $\mathbb{Q}(\sqrt{-199})$ (with ring of integers $\mathbb{Z}(\frac{1+\sqrt{-199}}{2})$) is $\frac{2}{\pi}\sqrt{199} < 11$, so we need to consider $2$, $3$, $5$, $7$. Since $|\mathbb{Z}(\frac{1+\sqrt{-199}}{2}):\mathbb{Z}(\sqrt{-199})|=2$, we have to apply Dedekind on $(2)$ with the minimal polynomial $x^2 + x + 50$ (instead of $x^2 + 199$). The latter factorizes mod 2 as $x(x+1)$, so $(2) = (2, \frac{1+\sqrt{-199}}{2})(2, 1 + \frac{1+\sqrt{-199}}{2})$, neither of the factors being principal (and hence both of order $2$), as both have norm $2$ and $2 = (x+\frac{y}{2})^2 + \frac{199y^2}{4}$ is insoluble in $\mathbb{Z}$.

However, http://www.lmfdb.org/NumberField/2.0.199.1 claims that the class group is $C_9$, so there can't be elements of order $2$, which seemingly contradicts the above. Where is the mistake in my approach?

Any help appreciated!

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  • $\begingroup$ I think it was in Stewart & Tall that I read that class number 2 means not UFD but you can at least count on the distinct factorizations each having the same number of factors, e.g., the two factorizations of 6 in $\mathbb Z[\sqrt{-5}]$ both have two factors each. But $$50 = 2 \times 5^2 = \left(\frac{1}{2} - \frac{\sqrt{-199}}{2}\right)\left(\frac{1}{2} + \frac{\sqrt{-199}}{2}\right).$$ $\endgroup$ – Robert Soupe Jun 1 at 16:37
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The latter factorizes mod 2 as $x(x+1)$, so $(2)=\left(2,\frac{1+\sqrt{−199}}{2}\right) \left(2,\frac{3+\sqrt{−199}}{2}\right)$, neither of the factors being principal (and hence both of order 2)

Why do you think that these prime ideals being nonprincipal implies that they have order 2?

Labeling the prime ideals, say $(2) = \mathfrak{p}\mathfrak{p}'$. If the order of $\mathfrak{p}$ is 2, then necessarily $\mathfrak{p} = \mathfrak{p}'$ (hint: you can use that they have the same norm and are equal in the class group). Can you show that isn't the case?

In fact, $\mathfrak{p}^k$ isn't principal for $k = 1,2,\dots,8$, but $\mathfrak{p}^9 = \left(\frac{43\pm\sqrt{-199}}{2}\right)$ (sign depending on choice of $\mathfrak{p}$). Hence the order of $\mathfrak{p}$ in the class group is 9.

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  • $\begingroup$ "If the order of $\mathfrak{p}$ is 2, then necessarily $\mathfrak{p} = \mathfrak{p}'$" is not a true statement for general quadratic fields. $\endgroup$ – user670344 May 30 at 3:09
  • $\begingroup$ It confuses me why you would edit your answer to make a typographical change and yet not correct a mathematical mistake. You imply that two ideals $\mathfrak{p}$ and $\mathfrak{p}'$ of norm $2$ which are equal in the class group are actually equal which is transparently false. $\endgroup$ – user670344 May 30 at 16:50
  • $\begingroup$ @user670344: And it confuses me why you continue to make this point. The order of $\mathfrak{p}$ isn't 2, so any statement that follows is vacuously true. One can show in this case (not in a general quadratic field or anything like that) that if the order of $\mathfrak{p}$ was 2, then necessarily $\mathfrak{p} = \mathfrak{p}'$. $\endgroup$ – Brandon Carter May 30 at 22:08
  • $\begingroup$ Because I don't like to see misleading statements. Yes, I understand that it is vacuously true, but you are claiming that there is a natural argument to imply that $\mathfrak{p} = \mathfrak{p}'$ using the given hint. What is that argument exactly? $\endgroup$ – user670344 May 30 at 22:27
  • $\begingroup$ No response, as expected. It's ok to make mistakes. But refusing to acknowledge them is just intellectually dishonest. $\endgroup$ – user670344 Jun 1 at 22:33

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