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L.S.,

In an exercise for my algebraic number theory homework I came across the following problem:

I would like to factor ideals $(2)$ and $(7)$ in $K = \mathbb{Q}(i, \sqrt{14})$. I managed to show that $\alpha = \frac{\sqrt{14} + \sqrt{-14}}{2}$ is a primitive element with minimal polynomial $f = X^4 + 49$. However, when I compute disc$(\mathbb{Z}[\alpha]) = N(f'(\alpha)) = 2^87^6$ there is problem. I already computed the discrimimant $\Delta(K) = 2^87^2$. So for $p = 2,7$, I cannot conclude that $p$ doesn't divide $[\mathcal{O}_K:\mathbb{Z}[\alpha]]$. (Or can I maybe for $p = 2$ because both disc$(\mathbb{Z}[\alpha])$ and $\Delta_K$ have the same power of 2?) Unfortunately, that is needed to use $X^4 - 49$ to factor ideals $(2)$ and $(7)$. So then my thinking was as follows: when I (might) find some other primitive element $\beta$, for which disc$(\mathbb{Z}[\beta]$) doesn't contain factors 2 and 7, I just use the minimal polynomial for $\beta$ for the ideal factorization. But now I wonder: how best to look for such element? For instance $\beta = \sqrt{14} + i$ is primitive, but disc$(\mathbb{Z}[\beta])$ also has factors 2 and factors 7, I believe..

Many thanks!

Willem

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    $\begingroup$ Note that $X^4-49=(X^2-7)(X^2+7)$, so it cannot be the minimal polynomial of $\alpha$ if $\Bbb{Q}(\alpha)=K$. Perhaps you meant $X^4+49$? $\endgroup$ – Servaes Dec 5 '15 at 17:47
  • $\begingroup$ I am sorry, that is what I meant! I'll edit right now, thank you $\endgroup$ – Willem Beek Dec 5 '15 at 19:16
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First note that if $\beta\in\mathcal{O}_K$ then $\Bbb{Z}[\beta]\subset\mathcal{O}_K$, and if $\operatorname{rank}\Bbb{Z}[\beta]=\operatorname{rank}\mathcal{O}_K$ then $$\operatorname{disc}\Bbb{Z}[\beta]=[\mathcal{O}_K:\Bbb{Z}[\beta]]^2\cdot\Delta(K),$$ so certainly $2^8\times7^2$ will divide $\operatorname{disc}\Bbb{Z}[\beta]$, meaning you won't find the sort of $\beta$ you are looking for. But you've made a good start already, so let me recap and guide you in the right direction:

The quartic number field $K:=\Bbb{Q}(i,\sqrt{14})$ indeed has a primitive element $$\alpha:=\frac{\sqrt{14}+\sqrt{-14}}{2}=(1+i)\sqrt{\tfrac72},$$ with minimal polynomial $f:=X^4+49$ over $\Bbb{Q}$. Some computations show that $$\operatorname{disc}\Bbb{Z}[\alpha]=\Delta f=2^8\times7^6\qquad\text{ and }\qquad\Delta(K)=2^8\times7^2,$$ so that $[\mathcal{O}_K:\Bbb{Z}[\alpha]]=7^2$. Now you can conclude that $\Bbb{Z}[\alpha]\supset\Bbb{Z}$ is regular at all primes except $7$, so you should be able to factor the ideal $(2)\subset\mathcal{O}_K$.

The ideal $(7)\subset\Bbb{Z}[\alpha]$ is singular, by Kummer-Dedekind for example, but from the minimal polynomial of $\alpha$ it is easy to see that $(7)=(\alpha)^2$ holds in $\mathcal{O}_K$, so it remains to factor $(\alpha)\subset\mathcal{O}_K$. Can you take it from here?

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    $\begingroup$ Thank you very much! I wouldn't know how to factor $(\alpha)$, but what I do know is that I can find a primitive element $\beta = \sqrt{14} + i$ with disc$(\mathbb{Z}[\beta] = 7^2m$, where $m$ has no factors 7. So then, if not mistaken(?), by your argument I think can use the minimal polynomial for $\beta$ to factor $7$, which yields $(7) = (7,\beta^2 + 1)^2$. $\endgroup$ – Willem Beek Dec 7 '15 at 10:15
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    $\begingroup$ Yes, that certainly works! $\endgroup$ – Servaes Dec 7 '15 at 13:50
  • $\begingroup$ Dear @servaes, Could I maybe ask one more question related to this exercise? I would like to show that all prime ideals of norm 5 are generated by one prime ideal of norm 5. I already factored $(5) = (5, \alpha + 1)(5, \alpha - 1)(5,\alpha + 2)(5,\alpha - 2)$, so those are the only 4 prime ideals of norm 5. My thinking was to look for elements of norm 25, so then some product of two prime ideals of norm 5 should be principal. I found the elements $\alpha + 1 + i$ and $\alpha - 1 - i$. But how now to proceed? many thanks! $\endgroup$ – Willem Beek Dec 9 '15 at 14:52
  • $\begingroup$ I don't understand what you are asking. What does it mean for "all prime ideals of norm $5$ to be generated by one prime ideal of norm $5$"? How does a prime ideal generate other prime ideals? $\endgroup$ – Servaes Dec 9 '15 at 15:01
  • $\begingroup$ I'm sorry for not writing it down properly. The question was about the class group.I meant to say that all classes of prime ideals of norm 5 are generated by one class of a prime ideal of norm 5. I tried a long time now, but I'm afraid that even if I can show that $(\alpha + i + 1) = \mathfrak{p}_5\mathfrak{q}_5$ and $(\alpha - i - 1) = \mathfrak{p}'_5\mathfrak{q}'_5$ with $\mathfrak{p}_5,\mathfrak{q}_5,\mathfrak{p}'_5,\mathfrak{q}'_5$ all different prime ideals of norm 5, I cannot conclude that all classes of prime ideals of norm 5 are generated by one such class. Could you maybe give a hint? $\endgroup$ – Willem Beek Dec 9 '15 at 15:12

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