Prime ideal but not maximal ideal

**tttcomrader** · November 13th 2007, 05:48 PM

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

**ThePerfectHacker** · November 13th 2007, 06:46 PM

Originally Posted by tttcomrader

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 $N = \{(x,y)|x\in \mathbb{Z} , y \in \{-1,0,1\} \}$ .

**Opalg** · November 14th 2007, 12:58 AM

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 $N = \{(x,2y):x,y\in \mathbb{Z}\}$ .

**ThePerfectHacker** · November 14th 2007, 09:50 AM

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.

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