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Ok, I am new to the concepts of maximal ideals and prime ideals. I know the definitions for both, but I am kind of stuck with understanding the examples. So, any help would be much appreciated.

Consider the ring $\mathbb{Z}$, which is commutative with unity. Let $I_1 = (0), I_2 =(−9), I_3 = (23)$ and $I_4 = (28)$

  1. List all $I_i$ that are prime ideals.
  2. List all $I_i$ that are maximal ideals.
  3. For each $I_i$ that is not a prime ideal, find $a_i, b_i \in \mathbb{Z}\setminus I_i$ such that $a_ib_i \in I_i$.
  4. For each $I_i$ that is not a maximal ideal, find a maximal ideal $M_i$ such that $I_i \subset M_i$.

My attempt:

  1. $I_1 = (0)$ and $I_3 = (23)$ are the prime ideals.
  2. $I_3 = (23)$ is the maximal ideal. I know $0$ is not a maximal ideal, but I exactly don't understand why.
  3. Still confused here. Does it mean like this: Since, $I_2=(-9)$ is not a prime ideal, then we can find $a_2 = -3$ and $b_2 = 3$, so that $a_2b_2 = (-9)$ and for $I_4 = 28, a_4 = 4$ and $b_4 = 7$(Again I am not exactly sure here).
  4. Confused.
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Your responses for the first three are exactly correct. To see why $\langle 0 \rangle$ is not maximal in $\mathbb{Z}$, notice that $\langle 0 \rangle \subsetneq \langle x \rangle$ for any nonzero $x \in \mathbb{Z}$.

For the fourth, the definition of a maximal ideal $I$ is that there are no other ideals $J$ such that $I \subsetneq J \subsetneq R$. For example, $\langle 4 \rangle$ is not maximal since $\langle 4 \rangle \subsetneq \langle 2 \rangle \subsetneq R$. If you can convince yourself that this is true, then you should be able to proceed with your problem.

As an aside, there are some very useful facts you should be aware of (and prove, if possible):

  • An ideal $I$ of a commutative ring $R$ is maximal $\iff$ $R/I$ is a field.
  • An ideal $I$ of a commutative ring $R$ is prime $\iff$ $R/I$ is an integral domain.
  • All maximal ideals are prime.
  • If $R$ is a principal ideal domain ($\mathbb{Z}$, e.g.), then all nonzero prime ideals are maximal.
  • If $R$ is a field, then $\langle 0 \rangle$ is the only maximal ideal.
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  • $\begingroup$ In a PID, all nonzero prime ideals are maximal, i.e. (Krull) dimension $\le 1.\ \ $ $\endgroup$ – Bill Dubuque Apr 4 '15 at 19:07
  • $\begingroup$ Thanks! That's what I intended...not what I wrote. $\endgroup$ – Kaj Hansen Apr 4 '15 at 19:07
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For 1. and 2. $I_1$ cannot be a maximal ideal since any maximal ideal is a proper and non-zero ideal. On the other hand, $(0)$ is prime since $ab\in(0)\implies ab=0\implies a$ or $b$ equal $0$ (Integral Domain).

For 3. your choice of $a_2,b_2$ is fine, but the right thing is $a_2b_2\in (-9)$, not $a_2b_2= (-9)$, the same goes for $a_4,b_4$.

For 4. I claim that $I_2$ is not a maximal ideal. Indeed $M_2:=(3)$ is an ideal such that $I_2\subsetneq M_2 \subsetneq \mathbb{Z}$. The same goes for $I_4$.

In fact, you are in a Principal Ideal Domain, i.e. an Integral Domain where each ideal is generated by only one element, in such rings, non-zero prime ideals are maximal (the converse holds in general).

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