2
$\begingroup$

I have quite some problems understanding the definition/derivation of the gaussian integers.

We defined the notation $\mathbb{Z}[i]\subset \mathbb{C} $ the following way: $\mathbb{Z}[i]:=<\mathbb{Z} \cup i>$ and $<\mathbb{Z} \cup i>:=\bigcap_{R\subset \mathbb{C} , E \subset R}R$.

Now as far as I understand it here $\mathbb{Z}[i]$ is the smallest subring of $\mathbb{Z} \cup i$. But how do you get from this to $<\mathbb{Z} \cup i>= \{a+b*i | a,b\in \mathbb{Z}\}$? Why can't there be a subring like {...,-2,-1,0,1,2,...,i} or would this just not contain everything needed? Or is it just not a ring? And if so how do I get to the mentioned as the smallest subring?

  • 1
    For example, $i+i$ must be in the ring again, so the subring you suggest is not sufficient2017-01-16

2 Answers 2

0

Passing to the quotient rings it is enough to show that the smallest field containing $\mathbb{Q}$ and $i$ is given by $$ \mathbb{Q}(i)=\{a+bi\mid a,b\in \mathbb{Q}\}. $$ This is clear, because $i$ has minimal polynomial $x^2+1$ of degree $2$, so that $(1,i)$ is a basis of the field extension $\mathbb{Q}(i)$ over $\mathbb{Q}$. Hence the ring of integers in $\mathbb{Q}(i)$, given by $\{a+bi\mid a,b\in \mathbb{Z}\}$ is the smallest ring containing $\mathbb{Z}$ and $i$.

  • 0
    Thanks, with the comment of Peter this solved my question!2017-01-16
1

In general, if you have a ring $R$ and you would like to add an element $\alpha$ to $R$, you also have to add the elements

$$r_0 + r_1\alpha + r_2\alpha^2 + \ldots + r_n\alpha^n,\ n\in\mathbb{N},\ r_i\in R$$

to $R$ to get another ring. (A ring has to be closed under addition and multiplication.) In the case $R=\mathbb{Z}$ and $\alpha = \mathrm{i}$ we have $\alpha^2=-1$, so every element above has the form $r_0 + r_1\alpha$ with $r_i\in\mathbb{Z}$.