Class Group of Ring of Integers of $\mathbb{Q}$

Let $R$ denote the ring of integers of the imaginary quadratic number field $\mathbb{Q}[\sqrt{-57}]$. I must find the ideal class group $\mathcal{C}$. Using the Minkowski Bound, I know that I need only look at the primes $2,3,5,7$. Moreover, because $x^2+57 \equiv x^2 \mod 3$, $x^2+57 \equiv x^2+2 \mod 5$, $x^2+57 \equiv x^2+1 \mod 7$, it can be seen that 2 and 3 split while 5 and 7 remain prime. We let $P\overline{P}=(2)$; it is a fact that $\langle P \rangle$ has order 2. Additionally, we let $Q\overline{Q}=(3)$.

Computing some norms, I have been unable to find anything useful; even doing a computer search did not yield any norms that had 2 and 3 as the only factors, so I could not produce any useful relations. I tried to use an intermediate prime number, but they ultimately did not tell me much.

However, I know for a fact that the answer is $\mathbb{Z}_2\times \mathbb{Z}_2$. The two facts that need to be shown for this to be true are that $\langle Q \rangle$ has order 2 and $\langle Q \rangle \neq \langle P \rangle$.

Any hint (e.g. a useful norm) would be appreciated.

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You are right in saying that $2$ and $3$ split, but you can in fact go one step further and show that they both ramify.

Then $(2) = P^2$ for some prime ideal $P$, and $(3)=Q^2$ for some prime ideal $Q$. We also know that $\langle P \rangle$ and $\langle Q \rangle$ have order $2$ (otherwise $P$ and $Q$ are principal ideals dividing $(2)$ and $(3)$ respectively, but $2$ and $3$ do not admit any proper divisors in $R$).

All that is left therefore is to show that $\langle P \rangle \neq \langle Q \rangle$. Suppose that they are equal. Then $P = \lambda Q$ for some $\lambda \in \mathbb{C}^*$, which implies that $(2)=(3 \lambda^2)$ $-$ a contradiction.

Hence the class group is generated by two distinct elements of order $2$, and so it is $\mathbb{Z}_2 \times \mathbb{Z}_2$.