In $\displaystyle Z \oplus Z$, let I = $\displaystyle \{ (a,0):a \in Z \} $. Show that I is a prime ideal but not a maximal ideal.

I'm having problem understanding what the I actaully equal to here.

Thanks

Printable View

- Nov 13th 2007, 05:48 PMtttcomraderPrime ideal but not maximal ideal
In $\displaystyle Z \oplus Z$, let I = $\displaystyle \{ (a,0):a \in Z \} $. Show that I is a prime ideal but not a maximal ideal.

I'm having problem understanding what the I actaully equal to here.

Thanks - Nov 13th 2007, 06:46 PMThePerfectHacker
$\displaystyle \mathbb{Z}\times \mathbb{Z}$ is a set of ordered pairs $\displaystyle (x,y)$ which are integers. The set $\displaystyle I = \{..., (-1,0),(0,0),(1,0),...\}$. We need to show it is a prime ideal. Say that $\displaystyle \bold{a},\bold{b}$ are elements in ring $\displaystyle \mathbb{Z}\times \mathbb{Z}$ say $\displaystyle \bold{a}=(a_1,a_2),\bold{b}=(b_1,b_2)$. Then if $\displaystyle \bold{a}\bold{b} \in I$ it means $\displaystyle (a_1b_1,a_2b_2)\in I$ so $\displaystyle a_2b_2 = 0 \implies a_2 \mbox{ or }b_2=0$. Thus, $\displaystyle (a_1,b_1)\in I \mbox{ or }(a_2,b_2)\in I$. However, it is not a maximal ideal because consider $\displaystyle N = \{(x,y)|x\in \mathbb{Z} , y \in \{-1,0,1\} \}$.

- Nov 14th 2007, 12:58 AMOpalg
- Nov 14th 2007, 09:50 AMThePerfectHacker