# In $\mathbb{Z}[\sqrt{d}]$, $d$ is not divisible by the square of any prime. Why?

In Gallian's Abstract Algebra, $$\mathbb{Z}[\sqrt{d}]$$ is defined as the ring

$$\mathbb{Z}[\sqrt{d}] = \{a+b\sqrt{d}, a,b\in \mathbb{Z}\},$$

where $$d$$ is required to not be divisible by the square of any prime. Why? What would happen to this ring is if $$d$$ was divisible by the square of some prime?

• In your case $√d$ is irrational. Oct 24, 2018 at 19:26
• What would happen to this ring? Say, we have $d=p^2$. Then $d=1$ gives the same ring. In general, it is convenient to assume that $d$ is squarefree. Oct 24, 2018 at 19:42

We enlarge $$R = \Bbb Z[\sqrt{a^2d}]$$ to $$\,\Bbb Z[\sqrt d]\,$$ because the larger ring might enjoy unique factorizarion, whereas $$R$$ never does when $$\,a> 1.\,$$ Indeed $$\,w = \sqrt{a^2d}/a\not\in R$$ is a $$\rm\color{#c00}{proper\ fraction}$$ over $$R$$ but $$\,w^2 = (a^2d)/a^2 = d\in R,\,$$ so $$\,w\,$$ is a root of the $$\rm\color{#c00}{monic}$$ polynomial $$\,x^2-d\in R[x].\,$$ This implies that unique factorization fails in $$R$$ because RRT = Rational Root Test fails, but RRT is true in any unique factorization domain (with same proof as in $$\Bbb Z)$$, i.e. UFDs, being gcd domains, are integrally closed.