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I am trying to factor the ideal $(8)$ into a product of prime ideals in $\mathbb{Q}(\sqrt{-7})$.

I am not exactly sure how to go about doing this, and I feel I am missing some theory in the matter. What I have noted is that we have $$8=(1+\sqrt{-7})(1-\sqrt{-7})=2^3$$ and so I surmise the factorization will involve these numbers somehow. I also noted that $N(8)=64$, so the prime ideal factors will have norms that multiply out to $64.$ Unfortunately I cannot proceed further than this. Any hints or references are appreciated.

Martin Sleziak
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Coffee_Table
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2 Answers2

8

Let $K=\mathbb{Q}(\sqrt{-7})$, so that $\mathcal{O}_K=\mathbb{Z}[\frac{1+\sqrt{-7}}{2}]$ because $-7\equiv 1\bmod 4$. Let $(8)$ denote the ideal generated by $8$ in $\mathcal{O}_K$.

Because $(8)=(2)^3$, it will suffice to determine the factorization of $(2)$ in $\mathcal{O}_K$, and then the factorization of $(8)$ will be the same with the exponents multiplied by $3$.

Note that $$(\tfrac{1+\sqrt{-7}}{2})(\tfrac{1-\sqrt{-7}}{2})=(2).$$ The norm of $\frac{1+\sqrt{-7}}{2}$ is $$N\mathopen{\big(}\tfrac{1+\sqrt{-7}}{2}\mathclose{\big)}=\mathopen{\big(}\tfrac{1}{2}\mathclose{\big)}^2+7\mathopen{\big(}\tfrac{1}{2}\mathclose{\big)}^2=2,$$ and similarly with $\frac{1-\sqrt{-7}}{2}$, so that $\mathcal{O}_K/\mathopen{\big(}\frac{1+\sqrt{-7}}{2}\mathclose{\big)}$ and $\mathcal{O}_K/\mathopen{\big(}\frac{1-\sqrt{-7}}{2}\mathclose{\big)}$ have cardinality $2$, and they are therefore the field $\mathbb{F}_2$.

Thus the ideals $\mathopen{\big(}\frac{1+\sqrt{-7}}{2}\mathclose{\big)}$ and $\mathopen{\big(}\frac{1-\sqrt{-7}}{2}\mathclose{\big)}$ are prime, so that $$(8)=\mathopen{\big(}\tfrac{1+\sqrt{-7}}{2}\mathclose{\big)}^3\mathopen{\big(}\tfrac{1-\sqrt{-7}}{2}\mathclose{\big)}^3.$$

Zev Chonoles
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  • To clarify: Would this argument be equally valid if one were to drop all mention of ideals and consider the problem in thr context of $\mathcal{O}_K$ itself, with the element 8? – Coffee_Table Apr 08 '13 at 03:14
  • @Coffee_Table Might I ask what you are referring to? Maybe you talk about idèles and adèles? Thanks in advance. – awllower Apr 09 '13 at 00:30
  • @awllower : I am referring to neither, just plain old ideals. Assuming you are referring to my comment right above (and not my original question), I mean to say that we could find the prime factorization of the number 8 in the ring $\mathcal{O}_K$ in an almost identical manner as the one described in the answer above, rather than finding the prime factorization of the ideal $(8)$. – Coffee_Table Apr 09 '13 at 01:22
  • @Coffee_Table I see.Thanks for the clarification then. – awllower Apr 09 '13 at 03:52
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As you noted, $N(8) = 64$. Also note that $N(1 \pm \sqrt{-7}) = 8$ and $N(2) = 4$. Since the ring of algebraic integers of $\textbf Q(\sqrt{-7})$ is said to be a unique factorization domain and, as you noticed, $(1 - \sqrt{-7})(1 + \sqrt{-7}) = 2^3 = 8$, this must mean that the two apparently distinct factorizations are in fact incomplete factorizations, just as $4^3$ would be an incomplete factorization of $64$ in $\textbf Z$.

As it turns out, $$\frac{1 + \sqrt{-7}}{2}$$ is an algebraic integer and it belongs in this domain, since its minimal polynomial is $x^2 - x + 2$. Therefore, as numbers, we have $$\left( \frac{1}{2} - \frac{\sqrt{-7}}{2} \right)^3 \left( \frac{1}{2} + \frac{\sqrt{-7}}{2} \right)^3 = 8.$$

However, for ideals, we need to verify that $$\left\langle \frac{1}{2} - \frac{\sqrt{-7}}{2} \right\rangle \not \subseteq \left\langle \frac{1}{2} + \frac{\sqrt{-7}}{2} \right\rangle$$ nor vice-versa. A couple of divisions will quickly confirm that $\langle 2 \rangle$ is a splitting, not ramifying ideal. Therefore, $$\langle 8 \rangle = \left\langle \frac{1}{2} - \frac{\sqrt{-7}}{2} \right\rangle^3 \left\langle \frac{1}{2} + \frac{\sqrt{-7}}{2} \right\rangle^3.$$

David R.
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