Group Isomorphism Question

Question

In this pdf (https://kconrad.math.uconn.edu/blurbs/gradnumthy/classgroupKronecker.pdf) the author claims that the ideal class group of the ring of integers of $\mathbb{Q[\sqrt{-199}]}$ is the cyclic group $\mathbb{Z_9}$. I wish to prove this for some practice identifying ideal class groups. Here is a summary of some of the progress I have made:

(I) Since $-199 \equiv 1 mod 4$, we see that the algebraic integers are $\mathbb{Z[\frac{1 + \sqrt{-199}}{2}]} \cong \mathbb{Z[x]}/(h(x))$ where $h(x) = x^2 - x + 50$

(II) Calculating the Minkowski bound, I need to check whether $2,3,5,7$ split in $\mathbb{Z[\frac{1 + \sqrt{-199}}{2}]}$. I found that all but 3 split.

(III) Here, $(2) = PP^{*}$, $(5) = QQ^{*}$ and $(7) = SS^{*}$, where $P = (2, \frac{1 + \sqrt{-199}}{2})$, $Q = (5, \frac{1 + \sqrt{-199}}{2})$, and $S = (7, 3 - \frac{1 + \sqrt{-199}}{2})$. I determined that these prime ideals do not equal their conjugate prime ideals.

(IV) I determined that the order of $[S]$ in the ideal class group is 3. Indeed, $(373) = (12 + \sqrt{-199})*(12 - \sqrt{-199}) = (7)^3 $= $S^3S^{*3}$ and $(7)$ does not divide the ideals $(12 + \sqrt{-199}),(12 - \sqrt{-199})$

(V)This is where I am stuck!!

$(70) = (4 + \frac{1 + \sqrt{-199}}{2})(4 + \frac{1 - \sqrt{-199}}{2}) = (2)(5)(7) = PP^*QQ^*SS^*$. I determined that the principle ideals $(4 + \frac{1 + \sqrt{-199}}{2})$ , $(4 + \frac{1 - \sqrt{-199}}{2})$ are not divisible by the ideals $(2) = PP^*, (5)= QQ^*, (7) = SS^*$. Hence:

$(4 + \frac{1 + \sqrt{-199}}{2}) = (P or P^*)*(QorQ^*)*(SorS^*)$. WLOG lets assume its $PQS$. This $[P][Q][S] = identity$. Hence $[Q]^2 = [P]^{-2}[S]$.

Now what do I do?; I cant seem to prove that $[P]$ or $[Q]$ has order 9. How do I show that one of these have order 9? Any help will be appreciated.

An Update I determined the order of $[P]$ is 9 and so $[P]^9 = [S]^3$

Answer 1

Well, he is correct, there is an isomorphism with the group of equivalence classes of positive binary forms of discriminant $-199,$ under Gauss composition...

The easiest way is to find equivalent forms which all have the same middle term, so that Dirichlet's description of Gauss composition is evident. In this case, forcing the middle coefficient to be $43$ works nicely. Each triple $\langle a,b,c \rangle$ refers to the binary form $$ f(x,y) = a x^2 + b xy + c y^2. $$

From Henri Cohen, A Course in Computational Number Theory, especially pages 225-229: we have Theorem 5.2.8, when $D<0$ is congruent to $0$ or $1 \pmod 4,$ we have a mapping from the positive forms (well, eqivalence classes) of that discriminant $$ \langle a,b,c \rangle \; \; \mapsto \; \; \; a \mathbb Z + \frac{-b + \sqrt D}{2} \mathbb Z $$ Lehman's notation is different. Also, for real fields and indefinite forms the mapping is usually two to one; it's a long story.

The first form is the group identity, the second is a group generator. $$ \langle 1, 43, 512 \rangle $$ $$ \langle 2, 43, 256 \rangle $$ $$ \langle 4, 43, 128 \rangle $$ $$ \langle 8, 43, 64 \rangle $$ $$ \langle 16, 43, 32 \rangle $$ $$ \langle 32, 43, 16 \rangle $$ $$ \langle 64, 43, 8 \rangle $$ $$ \langle 128, 43, 4 \rangle $$ $$ \langle 256, 43, 2 \rangle $$

Indeed, Dirichlet's method gives $$ \langle 2, 43, 2^8 \rangle \circ \langle 2^k, 43, 2^{9-k} \rangle = \langle 2^{k+1}, 43, 2^{8-k} \rangle $$

There is a recent book by Lehman, where the entire book takes binary forms and quadratic fields side by side, illustrating the isomorphism over and over, as a lead-in to later studies in algebraic number theory

 199:  < 1, 1, 50>    Square       199:  < 1, 1, 50>
 199:  < 2, -1, 25>    Square       199:  < 4, 3, 13>
 199:  < 2, 1, 25>    Square       199:  < 4, -3, 13>
 199:  < 4, -3, 13>    Square       199:  < 5, 1, 10>
 199:  < 4, 3, 13>    Square       199:  < 5, -1, 10>
 199:  < 5, -1, 10>    Square       199:  < 2, 1, 25>
 199:  < 5, 1, 10>    Square       199:  < 2, -1, 25>
 199:  < 7, -5, 8>    Square       199:  < 7, 5, 8>
 199:  < 7, 5, 8>    Square       199:  < 7, -5, 8>