Showing that $\mathbb Q(\sqrt{17})$ has class number 1

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?

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

• 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? :/ May 5, 2013 at 14:28
• @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 $p_1$. May 5, 2013 at 14:32
• In other words, try to prove that one of the ideals is generated by $(3+\sqrt{17})/2$ and the other by $(3-\sqrt{17})/2$. May 5, 2013 at 14:44
• @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? May 5, 2013 at 15:21
• @Thomas: Correct! 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.

• +1 Surely you mean $(5-\sqrt{17})/2$ in your last sentence? 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.

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

$$\left(2,\dfrac{1\pm\sqrt{17}}{4}\right)$$

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

$$|x^2-xy-4y^2|=2.$$

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

• Where did you get the condition involving $x,y$ from? May 2, 2023 at 13:25
• 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. May 2, 2023 at 13:40