Computer calculations show that the class group of $F = \mathbb{Q}[x]/(x^2 - x + 18) \simeq \mathbb{Q}(\sqrt{-71})$ is $C_F = \mathbb{Z}/7\mathbb{Z}$. Here is the table.

The ideal class group is defined as the fractional ideals modulo the principal (fractional) ideals. It measures the extent to which unique factorization fails. It would be instructive to see the prime ideals which factor in this manner.

In the article on fractional ideals there's an exact sequence: $$ 0 \to \mathcal{O}_K^\times \to F^\times \to I_F \to C_F \to 0 $$ so if I enter the values that I know so far: $$ 0 \to \Big[\mathbb{Z}[\sqrt{-71}]^\times \simeq \{ +1, -1\}^\times \Big] \to \mathbb{Q}(\sqrt{-71})^\times \to I_F \to \mathbb{Z}/7\mathbb{Z} \to 0 $$ From the abstract algebra point of view, I don't know how a cycle group emerges here. $\mathbb{Q}(\sqrt{-71})^\times$ is an infinitely generated abelian group under the operation (and also $\mathbb{Q}^\times$). I'd really like to see how to use the exact sequence.

The word "ideal" is used here differently:

  • a fractional ideal of $\mathcal{O}_F$ is an $\mathcal{O}_F$-module $I \subseteq F$ (of "fractions") with $rI \subseteq \mathcal{O}_F$ for some $r \in F$.

Another question about the "ring theory" involved here. Using different example:

  • 1
    $\begingroup$ $\mathcal{O}_K \ne \mathbb{Z}[\sqrt{-71}]$ because $-71 \equiv 1 \bmod 4$ $\endgroup$
    – lhf
    May 17, 2021 at 20:00
  • 1
    $\begingroup$ Class number calculations are taxing for me. But I predict that the class of the ideal $$\mathfrak{p}=(2,\frac{1+\sqrt{-71}}2)$$ is a generator. With more than a bit of luck its seventh power is one the principal ideals $(\frac{\pm 21+\sqrt{-71}}2)$. After all $$N(\frac{21+\sqrt{-71}}2)=\frac{21^2+71}4=2^7.$$ Compare with this. Hope you can reuse the arguments given there. $\endgroup$ May 17, 2021 at 20:15
  • $\begingroup$ A good alternative option is to wait for Will Jagy to show up. Anyway, I don't know of a way of using that exact sequence. I have mostly thought of that as one way of writing the definition rather than as a way of calculating the class group. $\endgroup$ May 17, 2021 at 20:19
  • $\begingroup$ Actually this is likely more difficult than the linked example. That's because this time both $3$ and $5$ split, and the Minkowski bound is $\approx 5.36$, so we need to deal with those as well. Identifying their classes as powers of $\mathfrak{p}$ is not strightforward for me at this hour. $\endgroup$ May 17, 2021 at 20:30
  • $\begingroup$ Hmm. The principal ideals generated by $(1\pm\sqrt{-71})/2$ and $(3\pm\sqrt{-71})/2$ have respective norms $18=2\cdot3^2$ and $20=2^2\cdot5$. Undoubtedly they can be used to write the classes of the primes above $3$ and $5$ in terms of powers of $\mathfrak{p}$ (or its conjugate). Also $N(46\pm\sqrt{-71})=3^7$. $\endgroup$ May 17, 2021 at 20:55

3 Answers 3


The mappings between binary quadratic forms and quadratic field ideals are described briefly in BLAH. One source is H. Cohen, A Course in Computational Algebraic Number Theory

In order to display the composition operation with the fewest tears, we may take the seven forms $ax^2 + bxy + c y^2 $ as $$ (1,21,128) ,(2,21,64) , (4,21,32) , (8,21,16) , (16,21,8), (32,21,4), (64,21,2) , $$ where Dirichlet's recipe for Gauss composition amounts to multiplying the left coefficients, with $(128,21,1)$ being the identity element once again.

The mapping to ideals takes $$ (A,21, \frac{128}{A}) \mapsto \left[ A, \frac{21+ \sqrt{-71}}{2} \right] $$


If the question is just how to compute the class number and give an ideal in each class, you can use the standard algorithm for imaginary quadratic fields. We need the discriminant, in our case $D=-71$, then find all the integer triples $(a,b,c)$ satisfying $b^2-4ac = D$, $|b|\leq\sqrt{|D|/3}$, $0<a\leq c$, and also $b\leq0$ if $a=c$. For each triple there corresponds a unique ideal class, and $a\mathbb Z +\frac{b+\sqrt{D}}{2}\mathbb Z$ is an ideal in that class. In our case the solutions $(a,b)$ are $(1,1)$ (the identity class), $(2,\pm 1)$, $(3,\pm 1)$, $(4,\pm 3)$, i.e. $7$ classes.

If the question is which primes decompose to ideals in which classes then wait for a number theorist (my class field theory knowledge is non-existent). If I completely missed the question, please let me know.


Nothing wrong with the other answers (+1 to all). Summarizing my "elementary" argument anyway.

  1. The algebraic integers of this field have the form $$z=\frac{a+b\sqrt{-71}}2$$ with $a,b$ rational integers of the same parity. The norm of such an element is $$N(z)=\frac{a^2+71b^2}4.$$
  2. The Minkowski bound $\in (5,6)$ meaning that the class group is generated by the prime ideals above the rational primes $2,3$ and $5$. All these primes split, and the two prime ideals above each are gotten by applying the non-trivial automorphism (=complex conjugation).
  3. The smallest exponent $\ell$ such that $a^2+71b^2=2^\ell$ has a solution with $b\neq0$ is $\ell=9$, $a=21,b=1$. Hence the principal ideals generated by $x_{\pm}=\dfrac{21\pm\sqrt{-71}}2$ have index $2^7$.
  4. If $\mathfrak{p}_2=(2,(1+\sqrt{-71})/2)$ then $\mathfrak{p}_2$ and $\overline{\mathfrak{p}_2}$ are the prime ideals lying above the rational prime $2$.
  5. Neither of the principal ideals $(x_{\pm})$ from step 3 are divisible by $(2)$. Hence (see the argument here) they must be pure powers of exactly one of the prime ideals from step 4. A look at the indices tells us that they are seventh powers. It follows that the class of $\mathfrak{p}_2$ has order $7$ in the class group. The class of $\overline{\mathfrak{p}_2}$ is obviously the inverse.
  6. Let $\mathfrak{p}_3=(3,(1+\sqrt{-71})/2))$ and $\overline{\mathfrak{p}_3}$ be the prime ideals above $(3)$. A norm calculation shows that the principal ideal $\mathfrak{a}$ generated by $(1+\sqrt{-71})/2$ has index $18$, and it is thus a product of prime ideals of respective norms $2,3,3$. It follows that $$ \mathfrak{a}=\mathfrak{q}\mathfrak{p}_3^2, $$ where $\mathfrak{q}$ is one of the prime ideals above $(2)$. In either case it follows that the class of $\mathfrak{p}_3^2$ is an element of the subgroup $G$ of the class group generated by (the class of) $\mathfrak{p}_2$. Using the elements $y_{\pm}=46\pm\sqrt{-71}$ of norm $3^7$ the same way as the numbers $x_{\pm}$ were used in step 5 we see that the order of the class of $\mathfrak{p}_3$ is also seven, so it belongs to any subgroup containing its square. That is, it is an element of $G$.
  7. Similarly, the principal ideal generated $w:=(3+\sqrt{-71})/2$ has index $20$. As $w\in\overline{\mathfrak{p}_2}$ it follows that $$(w)=\mathfrak{r}\overline{\mathfrak{p}_2}^2,$$ where $\mathfrak{r}$ is one of the prime ideals above $(5)$. As in step 6 this implies that the class of $\mathfrak{r}$ is an element of $G$.

The class group has order $h=7$, and any of the low norm prime ideals above can be used as a generator. The obvious relations from the listed principal ideals allow us to write them as powers of each other.

  • 1
    $\begingroup$ I sent you some programs, your university address $\endgroup$
    – Will Jagy
    May 19, 2021 at 21:11
  • $\begingroup$ Thanks @WillJagy. Appreciated. The problem for me is that the passage from ideal classes to quadratic forms is anything but clear. I have gotten the impression that it is standard stuff, predates the language of ideals, really. But the one intro course on ANT I have had did not get into this connection at all, which is why I default to the route why Minkowski bound. Irrespective of whether the field is quadratic or not. $\endgroup$ May 20, 2021 at 4:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .