How do I go about finding these? I know that the prime and maximal ideals in this case are the same, and that an ideal $M$ is only a maximal ideal of $R$ iff $R/M$ is a field, but I don't really know where to start from there, apart from trying each ideal individually.
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The key is the Lattice Isomorphism Theorem.
In fact the correspondence between the ideals given in the theorem extends to prime and also to maximal ideals. The latter should be obvious, you may wish to think about why the former is also true.
Now you are reduced to finding the prime/maximal ideals of $\mathbf{Z}$ that contain the ideal $(12)$.
Future
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There is an order preserving bijection between prime ideals of $\mathbb{Z}$ which contain $12\mathbb{Z}$, and prime ideals of $\mathbb{Z}/12\mathbb{Z}$, given by $$\mathfrak p \mapsto \{x + 12\mathbb{Z} : x \in \mathfrak p \}$$ If $\mathfrak p$ is a prime ideal of $\mathbb{Z}$, then $\mathfrak p = p\mathbb{Z}$ for some prime number $p$. The ideal $p\mathbb{Z}$ contains $12\mathbb{Z}$ if and only if $p$ divides $12$, if and only if $p$ is equal to $2$ or $3$. Thus the prime ideals of $\mathbb{Z}/12\mathbb{Z}$ are $$2\mathbb{Z}/12\mathbb{Z} = \{\bar{0}, \bar{2}, \bar{4}, \bar{6}, \bar{8}, \bar{10} \}$$ and $$3\mathbb{Z}/12\mathbb{Z} = \{ \bar{0}, \bar{3}, \bar{6}, \bar{9} \}$$
D_S
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sorry I don't get why you get to $2\Bbb{Z}/12\Bbb{Z}$. I thought it should be ${2\Bbb{Z}+12\Bbb{Z}}$ and ${3\Bbb{Z}+12\Bbb{Z}}$. Could you maybe explain this quickly – user1294729 Jul 04 '22 at 06:46
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They are the same. $2\mathbb Z / 12 \mathbb Z$ is by definition the set of cosets in $\mathbb Z/12\mathbb Z$ of the form $x + 12\mathbb Z$, where $x \in 2\mathbb Z$. This is the same as what you call ${2\mathbb Z + 12 \mathbb Z}$. I am using more traditional notation than you, but you are right. – D_S Jul 04 '22 at 10:09
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Ah perfect thanks you! – user1294729 Jul 04 '22 at 11:26