Find polynomial to use for prime ideal factorization 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
 A: 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?
