Hi I am needing some help and guidance on determining how to know when something is an ideal. We are working with commutative rings.
Our definition of an Ideal I, of a commutative ring R , is a subset of R such that $0 \in I$,
$s,t \in I \to s+t \in I$
and for all $s \in I$, any $r \in R, sr=rs \in I$
I am trying to see if I understand the problem; 
My attempt:
For the first one, at first glance both option b and d seem to be candidates. If I had to pick however I would go with b, because it seems to fit all the criteria and doesnt just have a 2 infront of the a. What do you guys think?
Also, I do know that $\mathbb{Z}$ is a principal ideal rings, so maybe that can help to ?
0 is in all of the choices. All are closed under addition.
(a) wont work because if we multiply by something in our ideal by Z we may not be in ideal.
(b) Seems to work
(c) Doesn't seem to work because $0+a\sqrt{2}$ is in our ideal but this times (c) will give us ab but no root.
(d) I am not as sure. This would be my other choice. Unless im way off, I am between b and d and looking for help
Please everyone my test is tomrrow. I have been trying to know if I am right for over a day . Can anyone please just tell me if I am right by thinking it is b?
Caution:Some authors do not include the ring $R$ among the ideals of $R$..... If $R$ is commutative and has a multiplicative identity $1$, then $1\in I\implies I=R $ because $1\in I\implies I\supset I R=$ $\{i r :i\in I\land r\in R\}\supset \{1 r:r\in R\}=R. $ Also if $R$ is commutative and $x\in I$ where $xy=1$ for some $y, $ then $I=R$ because $1=xy\in \{x r:r\in R\}\subset I.$ For (a) if $Z$ were an ideal of $R=Z[\sqrt 2] $ then $Z=R=Z[\sqrt 2]$ because $1\in Z. $ For (c), if it were an ideal $I, $ then $2=(\sqrt 2)^2\in (\sqrt 2)R=\{\sqrt 2\cdot r :r\in R\}\subset I R\subset I$ which implies $2=b\sqrt 2$ for some integer $b.$ For (b) (hint) For $u,v \in Z, $we have $u+v\sqrt 2\in I\iff u\equiv 3v \pmod 7.$ For (d) (hint) If $p,q\in Z$ and $(2+3\sqrt 2)(p+q\sqrt 2)=r+s\sqrt 2$ with $r.s\in Z$, then is $s$ necessarily divisible by $3$?