Suppose I have $\mathbb{Q}[\sqrt{d}]$, where d is negative. If $d \equiv 1$ mod 4 then the ring of integers is $\mathbb{Z}[\frac{1 + \sqrt{d}}{2}] \cong \mathbb{Z[x]}/(x^2 – x – \frac{1 – d}{4})$. It follows that $\mathbb{Z[\sqrt{d}]} \subset \mathbb{Z}[\frac{1 + \sqrt{d}}{2}]$, while the other direction is not true. Suppose p $\in \mathbb{Z}$ is prime and remains prime in $\mathbb{Z}[\frac{1 + \sqrt{d}}{2}]$. It follows by the subset relation that p remains prime in $\mathbb{Z[\sqrt{d}]} \cong \mathbb{Z[x]}/(x^2 + d)$. Does this mean that $x^2 + d$ is also irreducible in $\mathbb{Z}_{p}$? I feel like it should be because of the subset relation but I am not sure. Please let me know if my conclusion is true/valid.
Prime Ideals of Imaginary Quadratic Fields
abstract-algebramaximal-and-prime-ideals
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I'll give an "elementary" answer to this question based on the comments in the question:
Let $\alpha\in \mathcal{O}$ with $N(\alpha)=p$ a prime in $\mathbb{Z}$. Write $\mathcal{O}=\mathbb{Z}\oplus \lambda_d\mathbb{Z}$ and consider the abelian group homomorphism $\varphi:\mathbb{Z}\oplus \lambda_d\mathbb{Z}\to \mathbb{Z}\oplus \lambda_d\mathbb{Z}$ given by $\varphi(x)=\alpha x$. In the base $\{1,\lambda_d\}$ we can represent $\phi$ by a matrix $[\varphi]$, and a simple computation (considering $d\equiv 1$ (mod $4$) and $d\equiv 2,3$ (mod $4$) separately) gives $$ \det([\varphi]) = N(\alpha). $$ By using the Smith normal form, there are matrices $P,Q\in GL(n,\mathbb{Z})$ such that $\det(P)=\det(Q)=1$ and $$ [\varphi] = P\begin{pmatrix} d_1 & 0 \\ 0 & d_2 \end{pmatrix}Q, $$ and then $$ \varphi(\mathbb{Z}\oplus \lambda_d\mathbb{Z}) \cong (\mathbb{Z}/d_1\mathbb{Z})\oplus (\mathbb{Z}/d_2\mathbb{Z}), $$ so we have that $|\varphi(\mathbb{Z}\oplus \lambda_d\mathbb{Z})|=d_1d_2=\det([\varphi])=N(\alpha)=p$. But $\varphi(\mathbb{Z}\oplus \lambda_d\mathbb{Z})=(\alpha)$ is the principal ideal generated by $\alpha$ in $\mathcal{O}$, so, by the first isomorphism theorem, $|\mathcal{O}/(\alpha)|=p$. Then we can conlude (see for example this answer) that $\mathcal{O}/(\alpha)$ is isomorphic to the field $\mathbb{F}_p$, so $(\alpha)$ is a prime ideal and $\alpha$ is a prime elment in $\mathcal{O}$.
We can split $(2)$ as $\left(2,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1-\sqrt{-47}}{2}\right)$. Those are both nonprincipal because $2$ cannot be the norm of any principal element. If the class group has order $5$ like you claim, then either prime above $2$ is a generator because every non-identity element in a group of order $5$ is a generator. And in fact, $\left(2, \frac{1+\sqrt{-47}}{2}\right)^5=\left(\frac{9+\sqrt{-47}}{2}\right)$.
Now assume $p$ isn't prime in that ring. Say $x^2-x+12=(x-u)(x-v)\mod p$. Then the ideals $\left(p, \frac{1+\sqrt{-47}}{2}-u\right)$ and $\left(p, \frac{1+\sqrt{-47}}{2}-v\right)$ have norm $p$. If we look at them in the class group, since the prime above $2$ is a generator, we can multiply by it enough times so that we end up at the identity. This means that we have a principal ideal whose norm is some power of $2$ times $p$. But then we can complete the square and get $a^2+47b^2=2^kp$.
Your example with $3$ ends up being $(3)=\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(3,\frac{1-\sqrt{-47}}{2}\right)$. The prime $\left(3,\frac{1+\sqrt{-47}}{2}\right)$ is not principal, and neither is $\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1+\sqrt{-47}}{2}\right)$, but $\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1+\sqrt{-47}}{2}\right)^2$ is, as you showed: $\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1+\sqrt{-47}}{2}\right)^2=\left(\frac{1+\sqrt{-47}}{2}\right)$. So this has norm $12$ so multiplying by $2$ gives $1+\sqrt{-47}$ which has norm $48$ as you need.
You can also find a number of norm $96$ by using the other prime above $2$. You get $\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1-\sqrt{-47}}{2}\right)^3$ being principal, because it equals $\left(\frac{7+\sqrt{-47}}{2}\right)$. This has norm $24$ which means $7+\sqrt{-47}$ has norm $96$ as you wanted.
We could have also done the previous paragraph abstractly: if $\mathfrak{p}_1$ and $\mathfrak{p}_2$ are the two primes over $2$, and $\mathfrak{q}$ is the prime $\left(3, \frac{1+\sqrt{-47}}{2}\right)$ we talked about, then we know that in the group, $\mathfrak{q}+2\mathfrak{p}_1=0$ (where the group operation is addition instead of the natural multiplication) because we wrote $\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1+\sqrt{-47}}{2}\right)^2$ as a principal ideal. But we also know that $\mathfrak{p}_1+\mathfrak{p}_2=0$ because the product of the primes $\mathfrak{p}_1$ and $\mathfrak{p}_2$ was exactly the principal ideal $(2)$. So then these two are negatives of each other and $\mathfrak{q}-2\mathfrak{p}_2=0$. But the group has order $5$ so we can add $5\mathfrak{p}_2=0$ to both sides to get $\mathfrak{q}+3\mathfrak{p}_2=0$. This is how we knew that we could get $\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1-\sqrt{-47}}{2}\right)^3$ to be principal.
Finally, to make sure you understand the ideas here, here's a question to leave you with: If $2^kp=m^2+47n^2$, for some pair of $k$ values $k_1$ and $k_2$, with $m$ and $n$ both odd, show that either $5|k_1-k_2$ or $5|k_1+k_2+1$. Also the converse is true: if $2^{k_1}p=m_1^2+47n_1^2$ for some odd $m$ and $n$, then for any $k_2\geq 2$ satisfying either of the divisibilities above, there are another pair $(m_2, n_2)$ that have $2^{k_2}p=m_2^2+47n_2^2$.
So for example, knowing what you wrote in your question, you should be able to find $m$ and $n$ with $2^6\cdot7=448=m^2+47n^2$ with $m$ and $n$ odd.
Best Answer
If $p$ is an odd prime then:
$p$ will remain prime in $R=\mathbb{Z}[\frac{1 + \sqrt{d}}{2}]$
Iff
$(x^2 - x - \frac{1 - d}{4})$ is irreducible over $\Bbb Z_p$
Iff
$(x^2 - x - \frac{1 - d}{4}) \equiv 0 \pmod p$ has no solution.
Iff
$(4x^2 -4 x - ({1 - d})) \equiv 0 \pmod p)$ has no solution.
(Since $p$ is an odd prime )
Iff
$(2x-1)^2+d\equiv 0 \pmod p$ has no solution.
Hence,
$(y^2+d)$ is irreducible over $\Bbb Z_p$, where $y=2x-1.$