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For example, suppose $K=\mathbb{Q}(\sqrt{10})$, $I:=(2, \sqrt{10})$, $I$ is an ideal of $O_K$, is $I$ principal?

To solve this problem I read this topic: How to show an ideal is principal but I still have some questions:

$1$. If we want to calculate the norm of $I$, we need to find a basis of $O_K$ and $I$, a basis of I is $\{2, \sqrt{10}\}$, how to find a basis of $O_K$? I just know $O_K$ is the integer ring of $K$, but I do not know other proposition of structure about $O_K$.

$2$. If $I=(i)$ is principal, then we have $N_K(I)=\vert{N_{K/\mathbb{Q}}(i)}\vert$, does $\vert{N_{K/\mathbb{Q}}(i)}\vert$ mean the norm of $i$?

Appreciating for any idea and suggestion.

Bernard
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Deemo
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    The class number of $\mathcal{O}_K$ is $2$, so that not every ideal is principal (the ring is not factorial, too). I found this post, which has a little bit more details. Note that $10\equiv 2\bmod 4$, so we know a basis of $\mathcal{O}_K$ (google for ring of integers of quadratic number fields on this site). – Dietrich Burde Mar 25 '20 at 11:50
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    See also this post, and see also here for the ideal $(2,\sqrt{10})$. – Dietrich Burde Mar 25 '20 at 12:02
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    For general $K$ and $I$ this problem is difficult, and algorithms for determining such bases and generators are quite involved. – Servaes Mar 25 '20 at 12:06
  • @DietrichBurde Thank you for your comment, I read these posts but I do not find how to find a basis of $O_K$. We can know there exist a basis but how to find it? – Deemo Mar 25 '20 at 12:22
  • @Servaes Thank you for your comment, can we have some ways to calculate this example which I mentioned in this question? – Deemo Mar 25 '20 at 12:25
  • @DietrichBurde Yes I googled this topic. Maybe I need to check more – Deemo Mar 25 '20 at 12:28
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    Here is a help. The basis is given, for example, here for the general case $\Bbb Q(\sqrt{D})$. – Dietrich Burde Mar 25 '20 at 12:31
  • @DietrichBurde Thank you so much. – Deemo Mar 25 '20 at 12:40

1 Answers1

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Here is a way to proceed without too much algebraic number theory. Of course, I assume you know that $O_K=\mathbb{Z}[\sqrt{10}]$ (see all the comments)

Of course, the unlerlying idea is that $I$ has norm $2$, and that there is no element in $O_K$ having norm $\pm 2$, but we don't need that.

In what follows, $N=N_{K/\mathbb{Q}}$.

Assume that $I=(\alpha)$ for some $\alpha\in O_K=\mathbb{Z}[\sqrt{10}]$. So $\alpha\mid 2$ and $\alpha\mid \sqrt{10}$ in $O_K$. Since norms are multiplicative, taking the norms yields $N(\alpha)\mid 4$ and $N(\alpha)\mid -10$ in $\mathbb{Z}$, so $N(\alpha)\mid 2$. It follows that $N(\alpha)=\pm 1,\pm 2$. The case $N(\alpha)=\pm 1$ implies that $\alpha$ is a unit, and that $I=O_K$. I let you check this is not the case. So $N(\alpha)=\pm 2$. Writing $\alpha=a+b\sqrt{10}$, we have to solve $a^2-10 b^2=\pm 2$. In particular, $\pm 2$ is a square in $\mathbb{Z}/10\mathbb{Z}$. You can check this is not the case. Hence $I$ is not principal.

Side comment: if you want to show that $N(I)=2$, then you have several ways to proceed, depending what you know in algebraic number theory.

First method (very elaborate): since $2\in I$, you can try to factor $(2)$ as a product of prime ideals and see what happens. Since $O_K=\mathbb{Z}[\sqrt{10}]$, the factorisation a prime number $p$ is reflected by the factorisation of $X^2-10$ mod $p$. Here, $X^2-10=X^2$ mod $p$, and then $(2)=(2,\sqrt{10})^2=I^2$ by a famous theorem of Dedekind. In particular, $N(I^2)=N(I)^2= N(2O_K)=\vert N(2)\vert =4$, so $N(I)=2$.

Second method(elementary): Well, you can check that $I=\{a+b\sqrt{10}, a\in 2\mathbb{Z},b\in\mathbb{Z}\}$. This is not totally obvious, and does not comes directly from the definition of $I$, since a generic element of $I$ has the form $2z_1+\sqrt{10}z_2,$ with $z_1,z_2\in O_K $(and not $z_1,z_2\in\mathbb{Z}$), so this requires a bit of easy computation. Hence, a $\mathbb{Z}$-basis of $O_K$ is $(1,\sqrt{10})$, while a basis of $I$ is $(2,\sqrt{10})$. Thus, as an abelian group, we get $O_K/I\simeq \mathbb{Z}/2\mathbb{Z}$ and $N(I)=\vert O_K/I\vert =2$.

GreginGre
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