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Math Help - Prime ideal but not maximal ideal

  1. #1
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    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.

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    Quote Originally Posted by tttcomrader View Post
    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\} \}.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    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}\}.
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    Quote Originally Posted by Opalg View Post
    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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