Let $K=\mathbb Q(\sqrt{d})$ with $d=17$. The Minkowski-Bound is $\frac{1}{2}\sqrt{17}<\frac{1}{2}\frac{9}{2}=2.25<3$.
The ideal $(2)$ splits, since $d\equiv 1$ mod $8$. So we get $(2)=(2,\frac{1+\sqrt{d}}{2})(2,\frac{1-\sqrt{d}}{2})$ and $(2,\frac{1\pm\sqrt{d}}{2})$ are two ideals of norm $2$.

Now if we can show that $(2,\frac{1\pm\sqrt{d}}{2})$ are principal ideals, then we know that every ideal class contains a principal ideal, which shows that the class number is $1$.

But how can we show that $(2,\frac{1\pm\sqrt{d}}{2})$ are principal?


5 Answers 5


Hint: $$ \left(\frac{3+\sqrt{17}}2\right)\left(\frac{3-\sqrt{17}}2\right)=\frac{9-17}4=-2. $$

  • $\begingroup$ Basically I tried multiplying one of the generators $2$ and $\frac{1+\sqrt{d}}{2}$ with something in $\mathcal O_K$ in order to obtain the other, but no luck. How does your identity help me? :/ $\endgroup$
    – Phil-ZXX
    May 5, 2013 at 14:28
  • 1
    $\begingroup$ @Thomas: $$\frac{3+\sqrt{17}}2=2-\frac{1-\sqrt{17}}2.$$ The identity shows you how to get $2$ as a multiple of $p_1=(3+\sqrt{17})/2$, and this sum then tells how you get $(1-\sqrt{17})/2$ as a multiple of $p_1$. $\endgroup$ May 5, 2013 at 14:32
  • $\begingroup$ In other words, try to prove that one of the ideals is generated by $(3+\sqrt{17})/2$ and the other by $(3-\sqrt{17})/2$. $\endgroup$ May 5, 2013 at 14:44
  • $\begingroup$ @JyrkiLahtonen: So $2$ is multiple of $p_1=\frac{3+\sqrt{17}}{2}$, hence $2\in (p_1)$, and $p_1-2=-\frac{1-\sqrt{17}}{2}$, hence $\frac{1-\sqrt{17}}{2}\in (p_1)$. And vice versa $p_1=2-\frac{1-\sqrt{17}}{2}\in (2,\frac{1-\sqrt{17}}{2})$. This shows $(2,\frac{1-\sqrt{17}}{2})=(p_1)$ principal? $\endgroup$
    – Phil-ZXX
    May 5, 2013 at 15:21
  • $\begingroup$ @Thomas: Correct! $\endgroup$ May 5, 2013 at 15:48

We show that $$\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle = \left\langle \frac{5+\sqrt{17}}{2}\right\rangle.$$ Since, $$\frac{5+\sqrt{17}}{2}=2+\frac{1+\sqrt{17}}{2},$$ we have $$\frac{5+\sqrt{17}}{2}\in\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle,$$ thus $$\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle \subseteq \left\langle \frac{5+\sqrt{17}}{2}\right\rangle.$$ Now these two ideals have the same Norm--namely, $2$. Therefore, $$\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle = \left\langle \frac{5+\sqrt{17}}{2}\right\rangle.$$ The proof of the principality of $\frac{5-\sqrt{17}}{2}$ is similar.

  • 3
    $\begingroup$ +1 Surely you mean $(5-\sqrt{17})/2$ in your last sentence? $\endgroup$ Oct 26, 2013 at 19:00

Here because of a duplicate. Though it does seem like the original asker does pop in from time to time.

I'm not sure if the duplicate asker is aware of algebraic integers such as $$\theta = \frac{1}{2} + \frac{\sqrt{17}}{2},$$ which is a solution to $x^2 - x - 4$, though the duplicate asker is aware of the Minkowski bound.

So this tells us $N(\theta) = -4$, while obviously $N(2) = 4$. This suggests that $$¿ N\left(\left\langle 2, \frac{1}{2} + \frac{\sqrt{17}}{2} \right\rangle\right) = 4 ?$$ However, if $\mathcal O_{\mathbb Q(\sqrt{17})}$ does indeed have class number 1, that would mean that 16 has one distinct factorization (ignoring multiplication by units) and so $$16 = 2^4 = \left(\frac{1}{2} - \frac{\sqrt{17}}{2}\right)^2 \left(\frac{1}{2} + \frac{\sqrt{17}}{2}\right)^2$$ represents incomplete factorizations, just like $16 = 4^2 = 2 \times 8$ in $\mathbb Z$.

It's not a given that this is a Euclidean domain even if it does have class number 1. However, it wouldn't hurt to try. And so we find by the Euclidean algorithm that $$\gcd\left(2, \frac{1}{2} + \frac{\sqrt{17}}{2}\right) = \frac{5}{2} + \frac{\sqrt{17}}{2},$$ and indeed $$2 + \frac{1}{2} + \frac{\sqrt{17}}{2} = \frac{5}{2} + \frac{\sqrt{17}}{2}.$$

Furthermore, since $$\frac{5}{2} - \frac{\sqrt{17}}{2} \in \left\langle \frac{5}{2} + \frac{\sqrt{17}}{2} \right\rangle,$$ it follows that $$\langle 2 \rangle = \left\langle \frac{5}{2} + \frac{\sqrt{17}}{2} \right\rangle^2.$$ That's a principal ideal after all.

Since $$\left(\frac{17}{3}\right) = -1$$ (that's the Legendre symbol), 3 is prime in this ring. But it's well over the Minkowski bound anyway, so we're done.

EDIT: Jyrki Lahtonen points out a mistake I made regarding $\langle 2 \rangle$. The correct factorization is $$\langle 2 \rangle = \left\langle \frac{5}{2} - \frac{\sqrt{17}}{2} \right\rangle \left\langle \frac{5}{2} + \frac{\sqrt{17}}{2} \right\rangle.$$ This does not detract from the point that these are all principal ideals.

  • $\begingroup$ @ Robert Soupe Thanks for your answer . The last part I do not understand . Why is 3 a prime ? $\endgroup$ Jul 24, 2019 at 6:47
  • $\begingroup$ @Jyrki Maybe it was just a silly temporary confusion with $\mathbb Z[\sqrt{-17}]$. Thanks for pointing it out. $\endgroup$ Jul 25, 2019 at 2:41
  • $\begingroup$ 3 is prime because there are no solutions to $x - 17y = 3$, much less $x^2 - 17y^2 = 3$. $\endgroup$ Jul 25, 2019 at 2:49

Let $x=\frac{1+\sqrt{17}}{2}$ be the generator of the ring of integers $R=\Bbb Z[x]$, of $\Bbb Q[x]$. Then $x^2=x+4$.

OP asked ten years ago whether $(2,x)$ is principal ideal, that is, whether there is $\alpha\in R$ such that $(2,x)=(\alpha).$ First canditate coming to mind for $\alpha$ is $x+2$.

Indeed, if $I=(x+2)$ then $(x+2)^2=x^2+4x+4=5x+8\in I$, also $5(x+2)=5x+10\in I$ so that $5x+10-(5x+8)=2\in I$ and $x+2-2=x\in I.$ Hence, $(x,2)\subset I.$ But, the other inclusion, $I\subset (2,x)$, is trivial. Hence, $(x,2)=I=(x+2)$ is principal.


The quadratic integer $n=(1\pm\sqrt{17})/2$ has minimal polynomial $n^2-n-4$, so the ideals


will be principal if natural integers $x,y$ exist such that


As this condition holds when $x=2,y=1$ (also $x=1,y=-1$) we are done.

  • $\begingroup$ Where did you get the condition involving $x,y$ from? $\endgroup$ May 2, 2023 at 13:25
  • $\begingroup$ Match the norm of the quadratic form with that of the ideal ($2$). The existence of a solution $x=2,y=1$ goes along with $2+\psi:\psi=(1+\sqrt{17})/4$ being a prime factor of $2$ which would be needed to get a UFD/class 1. $\endgroup$ May 2, 2023 at 13:40

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