# Thread: Prime ideal but not maximal ideal

1. ## Prime ideal but not maximal ideal

In $Z \oplus Z$, let I = $\{ (a,0):a \in Z \}$. 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

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

I'm having problem understanding what the I actaully equal to here.
$\mathbb{Z}\times \mathbb{Z}$ is a set of ordered pairs $(x,y)$ which are integers. The set $I = \{..., (-1,0),(0,0),(1,0),...\}$. We need to show it is a prime ideal. Say that $\bold{a},\bold{b}$ are elements in ring $\mathbb{Z}\times \mathbb{Z}$ say $\bold{a}=(a_1,a_2),\bold{b}=(b_1,b_2)$. Then if $\bold{a}\bold{b} \in I$ it means $(a_1b_1,a_2b_2)\in I$ so $a_2b_2 = 0 \implies a_2 \mbox{ or }b_2=0$. Thus, $(a_1,b_1)\in I \mbox{ or }(a_2,b_2)\in I$. However, it is not a maximal ideal because consider $N = \{(x,y)|x\in \mathbb{Z} , y \in \{-1,0,1\} \}$.

3. Originally Posted by ThePerfectHacker
However, it is not a maximal ideal because consider $N = \{(x,y)|x\in \mathbb{Z} , y \in \{-1,0,1\} \}$.
That N is not an ideal. Replace it by something like $N = \{(x,2y):x,y\in \mathbb{Z}\}$.

4. Originally Posted by Opalg
That N is not an ideal. Replace it by something like $N = \{(x,2y):x,y\in \mathbb{Z}\}$.
Thank you for the correction.