Show that $(p,\sqrt{d})$ is a prime ideal in $Z[\sqrt{d}]$

We will use the fact that $R/I$ is a domain if and only if $I$ is a prime ideal of the ring $R$.

Let $f:\mathbb Z\to \mathbb Z[\sqrt d]/(p,\sqrt d)$ be a mapping given by $f(x)=x+(p,\sqrt d)$. It is routine to check that this is an onto ring homomorphism. Moreover, $\ker(f)=(p,d)\subset \mathbb Z$. By the First Isomorphism Theorem,

$$\mathbb Z/(p,d)\cong\mathbb Z[\sqrt d]/(p,\sqrt d).$$

In $\mathbb Z$, $(p,d)=(\gcd(p,d))$. Since $(d/p)=0$, $\gcd(p,d)\neq 1$, so we must have $\gcd(p,d)=p$. Since $p$ is prime, $\mathbb Z/(p)$ is a domain, so $\mathbb Z[\sqrt d]/(p,\sqrt d)$ is a domain. Hence, $(p,\sqrt d)$ is a prime ideal of $\mathbb Z[\sqrt d]$.

To find $\ker(f)=\{x\in\mathbb Z\mid f(x)=\overline 0=(p,\sqrt d)\}$:

Suppose $x\in(p,d)\subset\mathbb Z$. Then $x=ap+bd$, where $a,b\in\mathbb Z$. Therefore $$f(x)=x+(p,\sqrt d)=ap+bd+(p,\sqrt d)=(p,\sqrt d),$$

since $ap\in (p)$ and $bd\in(\sqrt d)$. Therefore, $(p,d)\subseteq\ker(f)$.

Now, suppose $x\in\ker(f)$. Then

\begin{align*} f(x)=(p,\sqrt d)\implies x&\in(p,\sqrt d)\\ \implies x&=ap+b\sqrt d&a,b\in\mathbb Z[\sqrt d]\\ &=a_1p+a_2p\sqrt d+b_1\sqrt d+b_2\sqrt d\sqrt d&a_i,b_i\in\mathbb Z\\ &=(a_1p+b_2d)+(a_2p+b_1)\sqrt d \end{align*}

Since $x\in\mathbb Z$, we have $x=a_1p+b_2d$ for some $a_1, b_2\in\mathbb Z$. Hence, $x\in(p,d)\subset\mathbb Z$ and $\ker(f)\subseteq (p,d)$.

Thus, $\ker(f)=(p,d)$.