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Let $K$ be a number field and $O_K$ be its ring of integers. Take $\alpha \in K$ and consider the principal fractional ideal $(\alpha)$. I am sure the following is true, but I couldn't quite prove it. I was wondering if someone could give me a hint on how I can find $\beta, \gamma \in O_K$ so that $\alpha = \beta/ \gamma$ and $(\beta) + (\gamma) = 1$. Thank you very much!

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  • $\begingroup$ Can you show that every element of $K$ is the quotient of two elements of $O_K$? Can you show that that implies what you want? $\endgroup$ – Greg Martin Oct 16 '15 at 21:14
  • $\begingroup$ @GregMartin Yes, I can show that every element in $K$ is a quotient of two elements of $O_K$. But I am not seeing how this implies what I want yet. Could I get a hint? Thank you! :) $\endgroup$ – Johnny T. Oct 17 '15 at 21:46
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This may not always be possible, take, for example $\mathcal{O}_k=\Bbb Z[\sqrt{-5}]$. Then consider the ideal generated by $\alpha={1+\sqrt {-5}\over 2}$. Both numerator and denominator are irreducibles, but due to lack of unique factorization we find both divide $6$, i.e.

$$\begin{cases} (1+\sqrt{-5})(1-\sqrt{-5})=6 \\ 2\cdot 3 = 6\end{cases}.$$

To see irreducibility it is sufficient to show there are no ideals of norm $2$ or $3$, which is readily verified by noting that the norm $N(a+b\sqrt{-5})=a^2+5b^2$ which clearly is never equal to $2$ or $3$ for integer $a,b$. But then it must be that $(1+\sqrt{-5})\supseteq (6)$ and therefore is contained in some maximal ideal containing $6$, but such a maximal ideal, $\mathfrak{p}$, must lie above $(2)$ or $(3)$ In fact it lies in one for each of them, $\mathfrak{p}_2=(2,1+\sqrt{-5}),\mathfrak{p}_3=(3,1+\sqrt{-5})$. This can also be seen by reducing the minimal polynomial for $1+\sqrt{-5}$ modulo $2$ and $3$. Knowing that $2$ is ramified, this means $\mathfrak{p}_2$ is the unique ideal above $2$ in $k=\Bbb Q(\sqrt{-5})$. We can see now that $(1\pm\sqrt{-5})+(2)\subseteq\mathfrak{p}_2$ so that they are not coprime. Even multiplying by conjugates we get

$${(1+\sqrt{-5})(1-\sqrt{-5})\over 2(1-\sqrt{-5})}={3\over 1-\sqrt{-5}}$$

which gives an expression where we have $(3)+(1-\sqrt{-5})=\mathfrak{p}_3'$, the other maximal ideal containing $3$ which still has them non-coprime integer ideals.

In general, say you have

$$\alpha = {\beta\over\gamma}\cdot{\delta\over\delta}$$

for some $\delta\in K$ such that $\delta\beta, \delta\gamma\in\mathcal{O}_K$. Then we ask if we may choose a $\delta$ so that $(\delta\beta)+(\delta\gamma)=\mathcal{O}_K$.

We know that $(\beta)=\mathfrak{p}\mathfrak{q}_1,(\gamma) =\mathfrak{p}\mathfrak{q}_2$. Then $\delta\in \mathfrak{p}^{-1}$ by definition, i.e.

$$(\delta)=\prod_{i=1}^r\mathfrak{p}_i^{e_i}\mathfrak{p}^{-1}$$

and, as usual, this expression is unique with all $e_i>0$. Then we have

$$(\beta\delta)+(\gamma\delta)=(\mathfrak{q}_1+\mathfrak{q}_2)\prod_{i=1}^r\mathfrak{p}_i^{e_i}.$$

Then clearly for this to be equal to $\mathcal{O}_K$ we must have $r=0$, i.e. there are no other primes, and we already know that $\mathfrak{q}_1+\mathfrak{q}_2=\mathcal{O}_K$. But then this implies $(\delta)=\mathfrak{p}^{-1}$, which would imply that $\mathfrak{p}^{-1}$ is a principal fractional ideal in the ideal class group. However we know that $\mathfrak{p}$ is a non-principal ideal, and since the ideal class group is cyclic of order $2$, we have that the ideal class for $\mathfrak{p}$ is the same as that for $\mathfrak{p}^{-1}$. However, $\mathfrak{p}^{-1}=(\delta)$ is a contradiction, since the latter is a principal ideal. Hence no such expression exists.

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  • $\begingroup$ Perhaps I am missing something... How does this show that $\alpha$ you have given me can not be expressed as $\beta / \gamma$ with $(\beta)+(\gamma) = (1)$? $\endgroup$ – Johnny T. Oct 22 '15 at 22:12
  • $\begingroup$ @JohnnyT. I have expanded my answer to include the full details. Let me know if anything else needs explanation. If you don't already know what the class groups is, that's a necessary part of the proof, so I can edit to give a quick primer enough to give you the conceptual picture as to how it implies the lack of existence in this case. Cheers. $\endgroup$ – Adam Hughes Oct 26 '15 at 23:46
  • $\begingroup$ Yes, thank you! I am just going through the details at the moment! $\endgroup$ – Johnny T. Oct 27 '15 at 18:59

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