What is the class group of $\Bbb{Q}(\sqrt{-41})$?

Question

What is the class group of $\Bbb{Q}(\sqrt{-41})$? I've found that it's generated by $P_2, P_3, P_5, P_7$ as per Dedekind's theorem, but I'm having a bit of trouble finding the relations between the elements. I've shown $P_5$ is inert and $P_2 = P_3$, but can't work out any other relations. Any help would be great, thanks.

Answer 1

The following argument is correct but is based on the false premises that $[P_2]=[P_3]$ and $5$ is inert and is hence invalid:

$P_2$ is ramified (as $-41 \equiv 3 \pmod 4$), so it has order 1 or 2 in the class group. $P_2=(2,1+\sqrt{-41})$ which is non-principal so it has order 2. Let $K=\mathbb Q(\sqrt{-41})$ and let $[P_i]$ denote the class containing $P_i$ in the class group. Then using the Minkowski bound $\lambda(-41) \approx 8.15,$ I consider the possible integral ideals of norm at most 8: $$Cl(K)\subseteq\{[\mathcal O_K], [P_2], [P_3],[P_2]^2,[P_2][P_3], [P_7], [P_2]^3\}.$$ Assuming your calculations are correct, $[P_2]=[P_3]$, $P_2^2=(2) \in [\mathcal O_K]$, so $[P_2]^2=[\mathcal O_K]$ and $[P_2]^3=[P_2]$ and $[P_2P_3]=[P_2]^2=[\mathcal O_K]$ and so $$Cl(K)\subseteq\{[\mathcal O_K], [P_2], [P_7]\}.$$ Then this is a group of order $\leq 3$ with an element of order 2, so it must be cyclic order 2, generated by $[P_2]$.

Answer 2

So the Minkowski bound is indeed generated by the prime ideals dividing $(2),(3),(5),(7)$. Then the minimal polynomial of $\alpha = \sqrt{-41}$ is $x^2 + 41$ which reduces modulo $p$ as \begin{align} x^2 + 41 &\cong (x+1)^2 \pmod{2} \\ x^2 + 41 &\cong (x+1)(x-1) \pmod{3} \\ x^2 + 41 &\cong (x-2)(x-3) \pmod{5} \\ x^2 + 41 &\cong (x+1)(x-1) \pmod{7} \end{align}

Then let $P_2 = (2, \alpha + 1)$, $P_3 = (3, \alpha + 1)$, $P_5 = (5, \alpha +2)$, $P_7 = (7, \alpha +1)$. Then the equivalence classes of these ideals generate the ideal class group $C_K$.

Now I'm guessing you got the hint to factorise $I_1 = (1 + \alpha)$, $I_2 = (2 + \alpha)$, $I_3 = (3+ \alpha)$.

We can see that the norms of these ideals are respectively \begin{align} N(I_1) &= 42 = 2\times3\times7, \\ N(I_2) &= 45 = 3\times3\times5, \\ N(I_3) &= 50 = 2\times5\times5. \end{align} Then this suggests we should try multiplying the prime ideals to get ideals of the proscribed norms. By a few long calculations which I'm not going to go into we see that \begin{align} P_2P_3P_7 =& I_1, \\ P_3^2P_5 =& I_2, \\ P_2P_5^2 =& I_3. \\ \end{align} Thus the ideal class $[P_7] = [P_2][P_3]^{-1}$, so the ideal class group is generated by $[P_2],[P_3]$ and $[P_5]$, then $[P_3]^2 = [P_5]^{-1}$ so the ideal class group is generated by $[P_2]$ and $[P_3]$ and lastly $[P_2]^{-1} = [P_2] = [P_5]^2 = [P_3]^4$, so the ideal class is generated by $[P_3]$, which is of order $8$. So the ideal class group is isomorphic to the cyclic group of order $8$, and note that $8$ is coprime to $3$.