Localization and extension of rings Is $\mathbb{Z}_{(3)}[i,\sqrt{2}]=(\mathbb{Z}[i,\sqrt{2}])_{(3)}$ (where by subscript $(3)$ we mean localization at the ideal generated by $3$)? 
Do both of these rings contain elements like
$$
\frac{4}{1-\sqrt{2}i}?
$$
 A: Suppose $S$ is a multiplicative subset of a commutative ring $R.$ Intuitively, $S^{-1}R$ is a construction similar to taking a field of fractions, but we only allow things in $S$ to be denominators. 
Exercise: An ideal $I$ of $R$ is prime if and only if $R\setminus I$ is a multiplicative subset of $R.$ 
From that fact, we define the notion of localizing at a prime ideal. We define $R_P=(R\setminus P)^{-1}R.$
So $\mathbb{Z}_{(3)}$ is the subfield of the rationals that contains the elements which, when written in lowest terms, are of the form $a/b$ where $a$ is any integer and $b$ is not a multiple of $3.$ 
Let $B$ be a subring of $A$ and $a\in A.$ Then the ring $B[a]$ is defined to be the intersection of all subrings of $A$ containing $B$ and $a.$ Constructively this is the subset of $A$ with elements of the form $b_0 +b_1 a + b_2 a^2 + b_3 a^3\cdots $ where $b_i\in B$ (only finitely many non-zero). One can check these two definitions are the same by verifying this last set is indeed a subring containing $B$ and $a,$ and that any other such subring contains elements of that form. 
Since $\mathbb{Z}_{(3)}$ is a subring of the complex numbers, when we adjoin $i$ we see $\mathbb{Z}_{(3)}[i]$ has elements of the form $a_0 + a_1 i + a_2 i^2 \cdots $ but since $i^2=-1$ the elements are just of the form $\alpha + i\beta$ where $\alpha,\beta \in \mathbb{Z}_{(3)}.$ 
Now adjoining $\sqrt{2}$ to that, we see the elements of $\mathbb{Z}_{(3)}[i,\sqrt{2}]$ are of the form $(\alpha_0+i\beta_0)+(\alpha_1+i\beta_1)\sqrt{2} + (\alpha_2+i\beta_2) \sqrt{2}^2 + \cdots$ and since $\sqrt{2}^2 =2$ we can always write the elements in the form $\alpha+\beta i+\gamma\sqrt{2}+\delta i\sqrt{2}$ where $\alpha,\beta,\gamma,\delta \in \mathbb{Z}_{(3)}$ i.e. are rationals which can be written with a denominator not divisible by $3.$ So we have a complete description of what the elements of this ring look like. 
For your example, $$ \frac{4}{1-i\sqrt{2} i}=\frac{4}{1-i\sqrt{2} i} \cdot \frac{1+i\sqrt{2}}{1+i\sqrt{2}} = \frac{4+4i\sqrt{2} }{1+2} = \frac{4}{3} + \frac{4}{3} i \sqrt{2}.$$
So the ring on the left hand side does not contain that element, since we can't express $4/3$ as a fraction where the denominator is not a multiple of $3.$
Now consider the ring of the right hand side. $\mathbb{Z}[i,\sqrt{2}]$ consists of elements of the form $a+bi+c\sqrt{2}+di\sqrt{2}$ where $a,b,c,d\in \mathbb{Z}.$ The localization will allow denominators that are not divisible by $3$ in $\mathbb{Z}[i,\sqrt{2}].$ So for example, $\dfrac{1+i}{2+i}= 1+\frac{1}{3}i  $ is in the ring on the right, but it is not in the ring on the left. So the two rings are not equal. 
