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Let $L =\Bbb{Q}(\sqrt{2},\sqrt{-3})$ and $ K = \Bbb{Q}(\sqrt{2})$. Now the prime $2$ is totally ramified in two of the three subfields of $L$ and from ramification theory we know that $2$ splits as $P^2$ in $\mathcal{O}_L$. How can I determine the exact prime decomposition of $2\mathcal{O}_L$? I do this because I want to prove that every prime factor of $2\mathcal{O}_L$ is principal, thus showing that $\mathcal{O}_L$ is a PID (The minkowski bound is less than 3).

However my problem now is that the usual way to do this via factoring a certain polynomial mod $p$ does not work. To simplify this, I tried to see if I could factor the ideal $(\sqrt{2})$ of $\mathcal{O}_K$ in the bigger ring $\mathcal{O}_L$. However this is not so simple: we have

$$\operatorname{disc} \mathcal{O}_L = 16 \cdot 36$$

and $\operatorname{disc} \mathcal{O}_K[\sqrt{-3}] = (-96)^2 $ (this latter calculation was done here) and thus the index $$\left[\mathcal{O}_L : \mathcal{O}_K[\sqrt{-3}] \right] = 4.$$

Since $2|4$ I can't apply the usual procedure even to factoring $\sqrt{2}$ in $\mathcal{O}_L$.

My question is: How can I find the exact factorisation of $2\mathcal{O}_L$? More generally how can I find the factorisation of prime ideals when the usual process does not work?


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There is an algorithm of Buchmann and H.W. Lenstra. It always works. It is described in Cohen's book. But it's no fun to do it by hand. – Hans Giebenrath Jan 27 '13 at 17:11
@HansGiebenrath If I had SAGE I could do it, but I would like a computation by hand. – user38268 Jan 27 '13 at 17:27
2 is not ramified in the subfield $\mathbb{Q}(\sqrt{-3})$. – Ted Jan 27 '13 at 17:42
May I suggest you edit the title of your question? The current title is a question that's different from the one you want an answer to. – Daan Michiels Jan 27 '13 at 19:20
@DaanMichiels I have edited my title. – user38268 Jan 28 '13 at 2:49
up vote 1 down vote accepted

Let $e$, $f$, and $g$ be the usual prime splitting parameters for 2 in the extension $L$. Looking at how 2 splits in ${\mathbf Q}(\sqrt{2})$ shows $2|e$. Looking at how 2 splits in ${\mathbf Q}(\sqrt{-3})$ shows $2|f$. Since $efg = 4$, we must have $e = 2, f = 2$, and $g = 1$. Thus $2{\mathcal O}_L = {\mathfrak p}_4^2$, where ${\mathfrak p}_4$ is a prime ideal in ${\mathcal O}_L$ of norm 4. But also $2{\mathcal O}_L = (\sqrt{2}{\mathcal O}_L)^2$, so $(\sqrt{2}{\mathcal O}_L)^2 = {\mathfrak p}_4^2$. By unique factorization, ${\mathfrak p}_4 = \sqrt{2}{\mathcal O}_L$ is principal.

This is simply saying that the prime over 2 in ${\mathbf Q}(\sqrt{2})$ stays prime in ${\mathbf Q}(\sqrt{2},\sqrt{-3}) = {\mathbf Q}(\sqrt{2},\zeta_3)$, which in retrospect is clear since the residue field at $(\sqrt{2})$ in ${\mathbf Q}(\sqrt{2})$ is ${\mathbf F}_2$ and $x^2 + x + 1$ - the minimal polynomial of $\zeta_3$ over ${\mathbf Q}(\sqrt{2})$ - is irreducible over ${\mathbf F}_2$. That is, if you'd think of $L$ as ${\mathbf Q}(\sqrt{2})(\zeta_3) = K(\zeta_3)$, a quadratic extension of $K$, and ${\mathcal O}_L$ as the integral closure of ${\mathcal O}_K$ in $L$, then you can use the same methods to factor $\sqrt{2}{\mathcal O}_K$ in ${\mathcal O}_L$ as you'd use to factor a rational prime in a quadratic field. After all, $K$ has class number 1, just like ${\mathbf Q}$. When you write that the "usual" methods don't work, it's not true! You need to think of number fields in a relative way, one number field as an extension of another, instead of everything as an extension of ${\mathbf Q}$.

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Thanks for your answer. When you write about field extensions relative to one another, that's what I considered in my question above. For example, I wanted to determine the splitting of $\sqrt{2}\mathcal{O}_K$ in $\mathcal{O}_L$. However the index $[ \mathcal{O}_L : \mathcal{O}_K[\sqrt{-3}] ]$ being $4$ did not allow me to do this since $2 | 4$. By the usual process, I mean I am applying Theorem 27 of Marcus' Number Fields. – user38268 Jan 28 '13 at 4:15
@BenjaLim I don't have that book, and maybe there are some special conditions in that theorem, but why are you using $\mathcal{O}_K[\sqrt{-3}]$ instead of $\mathcal{O}_K[\zeta_3]$? It seems like you should use the latter because it's the largest "obvious" set of algebraic integers in $L$. – Ted Jan 28 '13 at 4:22
@Ted I am very stupid and forgot that we can write $\Bbb{Q}(\sqrt{2},\sqrt{-3}) = \Bbb{Q}(\sqrt{2},\zeta_3)$. – user38268 Jan 28 '13 at 4:26
@Ted I have posted an answer below. – user38268 Jan 28 '13 at 4:34
@KCd I have posted an answer below clarifying what I understand of the problem. – user38268 Jan 28 '13 at 4:34

As in my question I would like to determine how $\sqrt{2}$ splits in the $\mathcal{O}_L = \Bbb{Z}[\sqrt{2},\alpha]$ where $\alpha = \frac{1 + \sqrt{-3}}{2}$. Now consider the composite

$$ \Bbb{Z}[\alpha] \to \Bbb{Z}[\sqrt{2},\alpha] \to \Bbb{Z}[\sqrt{2},\alpha]/(\sqrt{2}).$$

The kernel of this composite contains $(2)$ and now I claim it is exactly that: For the quotient $$\Bbb{Z}[\alpha]/(2) \cong \Bbb{Z}[x]/(x^2 - x + 1, 2) \cong \Bbb{F}_4$$

which is the field with four elements. Since the composite I wrote above is not the zero map, we get that the kernel is exactly $(2)$. Now $\Bbb{F}_4$ is a field and so $\sqrt{2}\mathcal{O}_L$ is maximal. In particular this means that $\sqrt{2}\mathcal{O}_K$ is inert in $\mathcal{O}_L$.

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First you should prove ${\mathcal O}_L$ equals ${\mathbf Z}[\sqrt{2},\zeta_3]$. (Why use $\alpha = \zeta_3 + 1$ instead of $\zeta_3$ in this problem?) Treating ${\mathbf Z}[\sqrt{2},\zeta_3]$ as a ${\mathbf Z}[\sqrt{2}]$-module, its ${\mathbf Z}[\sqrt{2}]$-discriminant is the ideal $(-3){\mathbf Z}[\sqrt{2}]$, which is prime (since 3 is inert), hence squarefree, so the ${\mathbf Z}[\sqrt{2}]$-index $[{\mathcal O}_L:{\mathbf Z}[\sqrt{2}][\zeta_3]]$ is the unit ideal $(1)$, and thus ${\mathcal O}_L$ is what you want. Then $x^2+x+1$ being irred. over ${\mathbf Z}[\sqrt{2}]/(\sqrt{2})$ (contd...) – KCd Jan 28 '13 at 5:10
makes $\sqrt{2}{\mathcal O}_L$ prime in ${\mathcal O}_L$. – KCd Jan 28 '13 at 5:11
@KCd Thanks for your response. I can prove that $\mathcal{O}_L$ equals $\Bbb{Z}[\sqrt{2},\zeta_3]$ using traces, but I have not seen the approach in your comment above. – user38268 Jan 28 '13 at 5:39
@BenjaLim per request I converted your answer to Wiki. But for future reference, note that you can do it yourself for answers. At the bottom of the edit box there is a tick-box that allows you to turn any of your answers into CW. It is only questions that requires moderator handling. – Willie Wong Jan 29 '13 at 9:10
@WillieWong I don't seem to see any tick-box that allows me to make my answer into CW. Do you have a screenshot/metalink of how I can do this? Thanks. – user38268 Jan 29 '13 at 9:25

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