Maximal ideals in the ring of Gaussian integers

Question

Let $R= \{ a+bi : a,b \in \mathbb{Z} \}$ be a subring of $\mathbb{C}$. Consider two principal ideals $I=(7)$ and $J=(13)$ in $R$. Is the ideal $I$ maximal? How about $J$?

I don't understand what $I=(7)$ or $J=(11)$ means. Aren't there both $a$ and $b$ to be considered? Why is there only one number?

I know that an ideal $M$ in a ring $R$ is maximal if $M \ne R$ and whenever if $N$ is also an ideal such that $M \subseteq N \subseteq R$, then $M=N$ or $N=R$. So if I can understand what these ideals are I can probably figure out which one is maximal.

Any help will be greatly appreciated.

Thanks in advance!!

$(7)$ is the ideal generated by $7$. In this case $$(7)={ 7a+7bi|a,b \in \mathbb{Z} } ,.$$ — N. S., Dec 03 '13 at 21:45

N. S. · Accepted Answer · 2013-12-05T01:18:45.993

5

Hint

$\mathbb Z[i]$ is a PID. If $(7)$ or $(13)$ is not maximal, it can be included in some larger ideal $(m)$.

This means that $m|7$ or $m|11$ in $\mathbb Z[i]$. Now all you need is to figure out if you can find $a+bi$ and $c+di$ which are not units and

$$(a+bi)(c+di)=7$$ respectively $$(a+bi)(c+di)=13$$

edited Dec 05 '13 at 01:18

answered Dec 03 '13 at 21:48

N. S.

132,525

thank you! So I was able to find 3 and 2 for 11 therefore 11 is not maximal. 7 is indeed a maximal. – PandaMan Dec 05 '13 at 01:04

LASV · Answer 2 · 2013-12-03T21:55:37.200

3

hint: $13=(2+3i)(2-3i)$. Does this help you? Is the ideal (13) maximal from this? It might help to note that the Gaussian integers are an Euclidean domain and hence prime ideals are maximal.

edited Dec 03 '13 at 21:55

answered Dec 03 '13 at 21:49

LASV

3,108

Maximal ideals in the ring of Gaussian integers

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