$$ 2 x^2 + 2 xy + 3 y^2 $$
Apparantly these are the norms of the non-principal ideals of $\Bbb Z[\sqrt{-5}]$.
Why is that ? How is that computed ?
Apparantly the ideals of prime norm in this ring have norms $2$, $5$ and $p$ with $p\equiv1,3,7,9\pmod{20}$.
I assume this follows from quadratic reciprocity and the norm of the ring being $x^2 + 5 y^2$. Is that correct ?
Apparantly the principal ideals cannot have norms $\equiv3,7$ and the non-principal ones cannot have norms $\equiv1,9$.
Again , why is that ?
I think I need to understand ideals better. These questions will help me.
0. Number field properties and ideal class
Let $w = \sqrt{-5}$, $K = \mathbb Q(w)$ and $R = \mathbb Z[w]$. We can compute the Minkowski's bound as $M_K <3$, then checking the ideal factorization of $(2)$ we find that $(2) = (2, w+1)^2 = J^2$ and that $J$ is non-principal. We remark that its norm is $N(J)=2$.
Hence the number field $K$ has class number $2$ and we can think of the ideal class as having two representatives: $P(K)$ the principal ideal group and $P(K)J$. In particular, multiplication by $J$ switches the principality of a given ideal $I$ since $$P(K)J\cdot J=P(K)J^2=P(K)(2)=P(K)$$
1. Norms of ideals in $R$
For a principal ideal $(a+bw)\subseteq R$, the norm is straightforward:
$$ N(a+bw) = a^2+5b^2$$ So we are mostly concerned about $N(I)$ for a non-principal ideal $I\subseteq R$. From previous paragraph, $IJ$ is principal so we have $IJ=(a+bw)$ for some $a,b\in\mathbb Z$. Therefore using the multiplicative property of norm: $$ 2N(I)=N(J)N(I)=N(IJ)=N(a+bw)=a^2+5b^2 $$ Taking modulo 2, we see that $$ a\equiv b \pmod 2 \implies a = b+2u $$ for some integer $u$. Now $$ \begin{align*} 2N(I) &= a^2+5b^2 = (b+2u)^2+5b^2 = 4u^2+4ub+6b^2\\ N(I) &= 2u^2 + 2ub + 3b^2 \end{align*} $$ giving us the required form.
2. Prime norms
We have see that $N(J) = N(2,w+1)=2$. Norm $5$ comes from $N(w)=-w^2 = 5$. Ideals are either principal or not, so it comes down to solving for prime $p$: $$ \begin{align*} p &= x^2+5y^2\\ p &= 2x^2+2xy+3y^2 \end{align*} $$ For the first, $p\equiv x^2\pmod 5$ so since $p$ is prime we must have $p\equiv 1$ or $4$ mod $5$. Using quadratic reciprocity: $$ \left(\frac{5}{p}\right) = \left(\frac{p}{5}\right) = 1 $$ so $5$ is a square mod $p$. But then $$ -1 \equiv (\sqrt 5 y x^{-1})^2 \pmod p $$ $-1$ is a square iff $p\equiv 1 \pmod 4$, so combining both we get $$ p \equiv 1, 9\pmod {20} $$
For the other case, we can try to transform into the first case and use what we found. The obvious way is to multiply by 2: $$ \begin{align*} 2p &= 4x^2 + 4xy + 6 y^2 = (2x+y)^2 + 5y^2 \\ 2p &\equiv (2x+y)^2 \pmod 5 \end{align*} $$ Now since $2$ is a quadratic non-residue, this time round we require $p\equiv 2$ or $3$ mod $5$. Using quadratic reciprocity again, we have $$ \left(\frac{5}{p}\right) = \left(\frac{p}{5}\right) = -1 $$ Now taking modulo $p$, we have $$ \begin{align*} (2x+y)^2 + 5y^2 &\equiv 0 \pmod p \\ ((2x+y)y^{-1})^2 &\equiv -5 \pmod p \end{align*} $$ Since $-5$ is a square and we have seen earlier that $5$ is not, hence $-1$ is a quadratic non-residue. This is possible iff $p\equiv 3 \pmod 4$. Now combining both we have $$ p \equiv 3, 7\pmod{20} $$ Therefore combining both sets of solutions the primes must be $$ p\equiv 1,3,7,9 \pmod {20} $$
Notice that $1,9\pmod{20}$ came from $p=x^2+5y^2$, which is from principal ideals. On the other hand $3,7\pmod{20}$ came from $p=2x^2+2xy+3y^2$, which is for non-principal ideals. This explains your last question: principal ideals cannot go to the $\{3,7\}$ class and vice-versa.