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Let $\alpha:=\mathbb{Q}(\sqrt[3]{17})$ and $K:=\mathbb{Q}(\alpha)$. We know that $$\mathcal{O}_K=\left\{\frac{a+b\alpha+c\alpha^2}{3}:a\equiv c\equiv -b\pmod{3}\right\}.$$ I have to show that $K$ has class number $1$, i.e. $\mathcal{O}_K$ is a PID. The Minkowski bound $\lambda <9$, so we should consider the primes $2, 3, 5, 7$. It's easy to show that

  • $2\mathcal{O}_K=\mathfrak{p}_1\mathfrak{p}_2$, with $\mathfrak{p}_1=(2, \alpha+1)$ and $\mathfrak{p}_2=(2, \alpha^2+\alpha+1)$
  • $3\mathcal{O}_K=\mathfrak{p}_3^2\mathfrak{p}_4$ (I can't compute these primes explicitly)
  • $5\mathcal{O}_K=\mathfrak{p}_5\mathfrak{p}_6$, with $\mathfrak{p}_5=(5, \alpha+2)$ and $\mathfrak{p}_6=(5, \alpha^2+3\alpha-1)$
  • $7\mathcal{O}_K=\mathfrak{p}_7$

Now, how can I show that, for example, $\mathfrak{p}_1$ and $\mathfrak{p}_5$ are principal ideals? (I can't find elements with norm $2$ or $5$). The situation for the prime $3$ is more complicated: the book suggests to find elements in $\mathcal{O}_K$ with norm $3$ that are coprime (this implies that $\mathfrak{p}_3$ and $\mathfrak{p}_4$ are principal), but I can't find these elements.

Note that $N_{K/\mathbb{Q}}(a+b\alpha+c\alpha^2)=a^3+17b^3+17^2c^3-3\cdot 17 abc$.

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    $\begingroup$ The standard way of proceeding is collecting many elements of small norm and forming quotients; if you have elements of norm 2 and 6 you can in general find an integer with norm 3. $\endgroup$ – franz lemmermeyer Jan 11 '16 at 16:25
  • $\begingroup$ @franzlemmermeyer: can you give an example? Thanks. $\endgroup$ – user72870 Jan 11 '16 at 17:16
  • $\begingroup$ I have barely studied cubic rings, so I could be way off on this, but doesn't $K$ here not have any numbers of the form $$\frac{a + b \alpha + c \alpha^2}{3}$$ with none of $a, b, c$ divisible by $3$? Like if $a = b = c = 1$, the minimal polynomial has a leading coefficient of $27$, not $1$. $\endgroup$ – Mr. Brooks Jan 11 '16 at 22:14
  • $\begingroup$ @Mr.Brooks: your counterexample is not valid, since I've written $a\equiv c \equiv -b \pmod{3}$. $\endgroup$ – user72870 Jan 11 '16 at 22:27
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    $\begingroup$ @user72870 You're absolutely right, I neglected to notice that important little detail. I tried $b = 2$ and that gives me $x^3 - x^2 - 11x - 12$. $\endgroup$ – Mr. Brooks Jan 11 '16 at 22:40
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The first thing to do is get a clean basis without annoying congruence conditions : put $\beta=\frac{\alpha^2-\alpha+1}{3}$. Then $[1,\alpha,\beta]$ is a $\mathbb Z$-basis of $\mathcal{O}_K$.

We apply Franz Lemmermeyer's method and look for elements of the form $x+y\alpha+z\beta$ with interesting norms and $x,y,z$ small.

A little inspection shows that

$$ N(2+\alpha+\beta)=N(2-\beta)=2, N(1+\alpha+\beta)=N(1-\alpha+\beta)=3, N(3-\alpha)=2 \times 5 $$

A few additional checks and computations from here then reveals that

$$ \begin{array}{lclcl} \mathfrak{p}_1 &=& (2+\alpha+\beta) &=& \Bigg( \frac{\alpha^2+2\alpha+7}{3}\Bigg) \\ \mathfrak{p}_2 &=& (2-\beta) &=& \Bigg( \frac{-\alpha^2+\alpha+5}{3}\Bigg) \\ \mathfrak{p}_3 &=& (1+\alpha+\beta) &=& \Bigg( \frac{\alpha^2+2\alpha+4}{3}\Bigg) \\ \mathfrak{p}_4 &=& (1-\alpha+\beta) &=& \Bigg( \frac{\alpha^2-4\alpha+4}{3}\Bigg) \\ \mathfrak{p}_5 &=& (\frac{3-\alpha}{2+\alpha+\beta}) &=& \Bigg( \frac{-2\alpha^2-\alpha+16}{3}\Bigg) \\ \mathfrak{p}_6 &=& (\frac{5(2+\alpha+\beta)}{3-\alpha}) &=& \Bigg( \frac{11\alpha^2+28\alpha+74}{3}\Bigg) \\ \end{array} $$

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