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

Question

I'm working on some exercises in the book "Problems in algebraic number theory" by Murthy and Esmonde. In exercise 7.4.3 (Show that $(d/p)=0$ iff $pO_K=\wp^2$, where $(d/p)$ is the Kronecker symbol, $p$ is a prime number, $O_K$ is the number ring of the number field $K=Q(\sqrt{d})$, $d$ square free, and $\wp$ a prime ideal) they use the fact that $(p,\sqrt{d})$ is a prime ideal of $Z[\sqrt{d}]$, where $p$ is a prime number and $d$ is square free. But how does one show this fact? As $Z[\sqrt{d}]$ is not necessarily commutative, I can't think of an easy way to show this. (Maybe I've just forgot to prove simple things about rings, as it's long ago since I've studied rings in depth, so please bear with me.)

Answer 1

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 $\mathbb Z[\sqrt d]$.

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