a. Find a non-trivial element in the class group of $\mathbb{Q}(\sqrt{−5})$.
b. Show that the class group of $\mathbb{Q}( \sqrt{−5})$ has order two.
For part a: I know that the class group is the factor group $I_K/P_K$, where $I_K$ is the group of fractional ideals of $K$ and $P_K$ is the subgroup of principal fractional ideals. So a non-trivial element is represented by a fractional ideal (e.g. an ideal of $\mathcal{O}_K$) that is not principal (= not generatable by a single element).
So is $(2, 1 +\sqrt{−5})$ non-trivial element in the class group of $\mathbb{Q}( \sqrt{−5})$? and if so, it's because $(2) = (2, 1 + \sqrt{−5})^2$?
For part b: I know that if $D < 0$ is square-free and $d = D$ if $D \equiv 1\bmod 4$ then the class group of a quadratic number field is equal to the class number of an integer. But I don't know how to show that the class group of $\mathbb{Q}(\sqrt{−5})$ has order two.
Just the first part.
In general, if $R$ is a commutative ring, and $\alpha,\beta\in R,$ then the product of principal ideals is the principal ideal of the product: $\langle \alpha\rangle\langle \beta\rangle = \langle \alpha\beta\rangle.$
So if $\langle \alpha\rangle=\langle 2,1+\sqrt{-5}\rangle,$ then we get $\langle \alpha^2\rangle = \langle\alpha\rangle^2=\langle 2\rangle.$ So $\alpha^2$ must be a factor of two, and hence $\alpha^2\overline{\alpha}^2=(\alpha\overline\alpha)^2\mid 4.$
Since $N(\alpha)=\alpha\overline\alpha$ is an integer, it must be a factor of $2,$ and it can't be $\pm 1,$ because we can easily show our ideal is not the whole ring, so $\alpha$ cannot be a unit.
But is also not hard to show that for $\alpha\in\mathcal O_K,$ it is not possible for $N(\alpha)=\pm 2.$
Another approach is to show $2$ is irreducible in $\mathbb O_K.$ It hinges on the same fact: That $N(\alpha)=\pm 2$ is not possible.
If $2=\alpha\beta,$ with $\alpha,\beta\in\mathcal O_K$ not units, then $4=N(2)=N(\alpha)N(\beta).$ If $N(\alpha)=\pm 1,$ then $\alpha$ is a unit. Likewise, for $N(\beta).$ So we must have $N(\alpha)=\pm 2.$
But there is no such $\alpha.$
Since $2$ is irreducible, it cannot be a member of any principal ideals other than $\langle 1\rangle$ and $\langle 2\rangle.$
So $\langle 2,1+\sqrt{-5}\rangle$ cannot be a principal ideal.