Question:
Show that $3$ is not a prime in $\mathbb Q [\sqrt 7] $.
To show this, should I start by assuming that $3 = ab$ where $a$ and $b$ are integers in $\mathbb Q[\sqrt{7}]$ and then try to show they are not units?
What should I do to show that? Or is there another better way to do this problem?
On
I believe 3 is small enough that you can just use the continued fraction digits of $\sqrt{7} = [2;\overline{1,1,1,4}]$.
The approximations are $2, 3, \frac{5}{2}, \frac{8}{3},\frac{37}{14},\dots$ Indeed $37^2 - 7\cdot 14^2 = -3$ and even $2^2 - 7*1^2 = -3$.
These lead to factorizations $3 = (\sqrt{7}-2)(\sqrt{7}+2) =(14\sqrt{7}-37)(14\sqrt{7}+37) $
On
Recall that in a commutative ring $R$, an ideal $I$ is prime iff $R/I$ is a domain. Also note that $\mathbb{Z}[\sqrt{7}] = \frac{\mathbb{Z}[x]}{(x^2-7)}$. Then \begin{align*} \frac{\mathbb{Z}[\sqrt{7}]}{(3)} &\cong \frac{\mathbb{Z}[x]/(x^2-7)}{(3,x^2-7)/(x^2-7)} \cong \frac{\mathbb{Z}[x]}{(3,x^2-7)} \cong \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x^2-7)} = \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x^2-1)}\\ &=\frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x-1)(x+1)} \cong \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x-1)} \times \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x+1)} \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \end{align*} where the second isomorphism holds by the Third Isomorphism Theorem, and the second-to-last by the Chinese Remainder Theorem. Since $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ is not a domain, then $(3)$ is not prime in $\mathbb{Z}[\sqrt{7}]$.
Moreover, we can recover a factorization of $3$ from this isomorphism. Since $$ x^2 - 7 \equiv x^2 - 1 = (x+1)(x-1) \equiv (x-2)(x+2) \pmod{3} $$ replacing $x$ by $\sqrt{7}$ yields the factorization $3 = (\sqrt{7}-2)(\sqrt{7}+2)$.
$$3 = (\sqrt 7 - 2)(\sqrt 7 + 2)$$
And $2^2 - 1^2 \dot \, 7 \neq \pm4 $ or $2^2 - 1^2 \dot\, 7 \neq 1$.
Edit: A fundamental unit in $\mathbb Q[\sqrt 7]$ is $8 + 3\sqrt 7$. Therefore the units are of the form $(8 + 3 \sqrt 7)^n , n\in \mathbb Z$.