$\langle p\rangle$ is a prime ideal in $\Bbb{Z}[\sqrt{d}]$ iff $x^2-d$ is irreducible in $\Bbb{F}_p[x]$ There is a theorem (Theorem 13.6.1 in the book "Algebra" by Artin) that states:

Let $d\equiv 2\text{ or }3\pmod{4}$
  and $\Bbb{Z}[\sqrt{d}]=\{a+b\sqrt{d}\mid a, b\in \Bbb{Z}\}$
  and $p$ is a prime integer.
$$\langle p\rangle\text{ is a prime ideal in }\Bbb{Z}[\sqrt{d}]\Leftrightarrow x^2-d\text{ is irreducible in }\Bbb{F}_p[x].$$

Artin proved the theorem by the following commutative diagram
$$
\begin{array}{rcl}
\Bbb{Z}[x] & \stackrel{\langle p\rangle}{\longrightarrow} & \Bbb{F}_p[x] \\
/\langle x^2-d\rangle \downarrow& & \downarrow /\langle x^2-d\rangle \\
\Bbb{Z}[\sqrt{d}] & \stackrel{\langle p\rangle}{\longrightarrow} & \tilde{R}
\end{array}
$$
According to the diagram,
he proved $x^2-d$ is irreducible in $\Bbb{F}_p[x]$ by proving that $\langle p\rangle$ is a maximal ideal in $\Bbb{Z}[\sqrt{d}]$.
But the prime ideal $\langle p\rangle$ doesn't necessarily be maximal because $\Bbb{Z}[\sqrt{d}]$ is not necessarily a P.I.D..
So is the ''$\Rightarrow$'' statement true?
If it is true, how can I prove it?
 A: Seems wrong to state anything about maximal ideals.
The diagram shows:


*

*$x^2-d$ is irreducible in $F_p[x]$ iff $\tilde R$ is not an integral domain. 

*$\langle p\rangle$ is a prime ideal in $\mathbb Z[\sqrt{d}]$ if and only if $\tilde R$ is an integral domain.


If $x^2-d$ is irreducible, then since $F_p[x]$ is a PID, then $ \langle x^2-d\rangle$ is a prime ideal.
If $\langle p\rangle$ is a prime ideal, then $\tilde R$ is an integral domain, which means that $F_p[x]/\langle x^2-d\rangle$ is an integral domain. So $x^2-d$ is irreducible.
We never need to talk about maximal ideals, but once we have this diagram, we do have that if $\langle p\rangle$ is prime in $\mathbb Z[\sqrt{d}]$ then it is also maximal, since then the quotient ring will be a field.
A: 13.6.1 (c)

Basically, if (p) is not maximal, then we cannot use it to construct a quotient ring, which will be a field.
$$\frac{Z}{pZ} \cong F_p$$
And if we cannot construct this $F_p$, then we cannot go on and construct $\hat{R} \cong Z[x]/(p,x^2-d)$. Both branches of diagram 13.6.3 start from the same set $Z[x]$, then use two maximal ideals to construct the final field $\hat{R}$.
We may introduce relations in different order with the same result (Artin page 337).
We use maximal ideals to construct fields by 11.8.2 page 344. Artin's diagram might be properly extended because we use First isomorphism theorem 11.4.2(b) page 335.

Still, I think parts (a), (b) of this theorem, which were skipped by Artin, are more interesting.
(a)

Key thing here is that we try to factor integer n in the set of integers, not in R, ring of integers of $\mathcal{Q}[\delta]$. This is because we may factor n in R, but we do not look at this. We only check if n factors in $\mathcal{Z} \subset \mathcal{Q}[\delta]$.
If n is an integer prime, then it matches Main Lemma 13.4.8.
If we can factor n in a few integers, then the number of these integers can only be two. The reason is Theorem 13.5.5, unique factorization of ideal into prime ideals. We can have $n=p_1 p_2$, and by 13.4.3 we have
$$(n)=(p_1 p_2)=(p_1)(p_2)$$
So
$$P\hat{P}=(n)=(p_1)(p_2)$$
In other words, $P=(p_1)$ or $P=(p_2)$. This is an equality because for factorization of ideals into prime ideals. When we factor ideals into prime ideals we do not have associates, "associate" ideals, as we do when we factor ring elements.
A complex conjugate of an ordinary integer is this very same integer, so $p_1=p_2$.

(b)

Case when principal ideal (p) is a prime ideal is the case of theorem 13.5.2(c).
In the other case, ideal (p) factors uniquely into a product of at least two prime ideals by 13.5.5.
$$(p)=P_1 P_2 \cdots P_k$$
We multiply the right side of this equation by the conjugates of these same prime ideals.
The picture looks like this.

The red marking shows that by the Main lemma 13.4.8 and unique factorization of ordinary prime integers, both the left and the right side is a product of ordinary integers.
And, without loss of generality, the product of one of ideals $P_1$ and its conjugate $\hat{P}_1$ equals $p$ or $p^2$.
The violet marking shows our assumption that (p) factors in some number of prime ideals.
The number of prime ideal factors for ideal (p) is at most two by the red marking, and at least two by the violet marking. So it is two and the factors, i.e. prime ideals, are conjugates.

PS. For 13.6.1 parts (a) and (b) we do not need to look at how element p or n may factor in ring R, look at lattices of ideals in this ring or look at the proof tactics of the Main lemma 13.4.8. All we need is the number of prime ideals then we factor (n) or (p), because by 13.5.5 this number is unique.
