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A problem from introduction to abstract algebra by Hungerford.

It asks: If $p$ is a prime integer, prove that $M$ is a maximal ideal in $\mathbb Z \times \mathbb Z$, where $M =\{(pa,b)\mid a,b\in \mathbb Z\}$

Though, we know that an ideal is maximal iff $\frac{\mathbb Z \times \mathbb Z}{M}$ is a field. Isn't $\frac{\mathbb Z \times \mathbb Z}{M} = (\mathbb Z_{p}, \mathbb Z_{b})$? And since $b$ can be any integer, say $6$, then this is not a field. And so $M$ must not be maximal?

Can someone please show what I am misunderstanding? Thanks.

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    $\begingroup$ $b$ is not fixed - it is a general integer. The ideal has the form $p\mathbb Z \times \mathbb Z$. Don't forget to prove it is an ideal. $\endgroup$ – Mark Bennet Apr 15 '14 at 17:18
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    $\begingroup$ $M$ is the kernel of $\mathbb Z\times\mathbb Z\to\mathbb Z/p\mathbb Z$, $(x,y)\mapsto x+p\mathbb Z$ $\endgroup$ – Hagen von Eitzen Apr 15 '14 at 17:19
  • $\begingroup$ Right, thanks. :) $\endgroup$ – zzz2991 Apr 15 '14 at 17:28
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You're doing well, but this isn't right: $\frac{\mathbb Z \times \mathbb Z}{M} = (\mathbb Z_{p}, \mathbb Z_{b})$

It's true that everything in the left side of the pairs is of the form $pa$ for $a$ ranging over $\Bbb Z$, and those are exactly the elements of $(p)\lhd \Bbb Z$, so the left part is indeed $\Bbb Z_p$. But in the right hand side, $b$ can be anything, including $1$!

So $\Bbb Z_b$ is not right but $\frac{\mathbb Z \times \mathbb Z}{M} = (\mathbb Z_{p}, ?)$ (you suddenly realize what it actually should be in the comments below...)

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    $\begingroup$ Oh. I see. Thanks. So, just to make sure $\frac{\mathbb Z \times \mathbb Z}{M} = (\mathbb Z_{P}, 0) \cong \mathbb Z_{P}$ which is a field, hence $M$ must be maximal? $\endgroup$ – zzz2991 Apr 15 '14 at 17:23
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    $\begingroup$ @DavidJhoo Right! $\endgroup$ – rschwieb Apr 15 '14 at 17:41
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Add an ordered pair $(c,d)$, where $c$ is not a multiple of $p$. Using the Bézout Identity we find that $(1,k)$ is in the resulting ideal for some $k$. But $(0,k)$ is in the ideal, so $(1,0)$ is. Since $(0,1)$ is in the ideal, we get all of $\mathbb{Z}\times \mathbb{Z}$.

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  • $\begingroup$ Great, thanks. So we are letting $I = (c,d)$ be some other ideal s.t $M \subset I$. Then $\exists i \in I: i \nmid p$. So then by Bezout's identity $(1,k) \in I$ for some $k\in \mathbb Z$. Then we see that $(1,0)$ and $(0,1)$ are in the ideal, hence $(1,1) \in I$ and $I = \mathbb Z \times \mathbb Z$ and $M$ must be maximal. $\endgroup$ – zzz2991 Apr 15 '14 at 17:50
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    $\begingroup$ Yes, that's right. The answer was intended to confront the problem in a concrete "hands on" way. $\endgroup$ – André Nicolas Apr 15 '14 at 17:53
  • $\begingroup$ Yes, thanks for that. This proof follows more closely to the chapter proofs in the readings. $\endgroup$ – zzz2991 Apr 15 '14 at 17:54
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    $\begingroup$ I read your comment too quickly. In your notation, we suppose that $I$ is an ideal which is a proper extension of $M$, and we let $i=(c,d)$ be in $I$ but not in $M$. Then continue as you wrote. (I am objecting to the $I=(c,d)$.) $\endgroup$ – André Nicolas Apr 15 '14 at 17:59
  • $\begingroup$ Ok I see the mistake. Looks like if I try to define $I = \{...\}$ s.t its a proper extension of $M$, the only choice is $\mathbb Z \times \mathbb Z$... just like we proved. I guess I should have seen it that way faster since $\mathbb Z_{P}$ is always maximal. $\endgroup$ – zzz2991 Apr 15 '14 at 18:04

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