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I understand how to prove that S has at least 3 prime ideals. How to I prove that there are no more? I'm thinking of doing this by contradiction but am not entirely sure how.

rschwieb
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weepyhollow
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    If the rings are isomorphic, they have the same algebraic structure. Just apply the isomorphism to the ideals to show that $S$ has three prime ideals, and then apply inverse isomorphism to $S$ to show that the inverse image of all of its ideals must also be ideals of $R$. – Ashwin Trisal Dec 14 '16 at 23:52
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    "isomorphic" is a symmetric relationship, so if you were able to get that far, you should be done. If $S$ had 4 prime ideals, then by the same reasoning $R$ would have at least 4 prime ideals, too. –  Dec 14 '16 at 23:53
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    Suppose that $S$ has at least 4 prime ideals. Then, by the same argument you use, you can conclude that $R$ has at least 4 prime ideals. But this is not the case, hence $S$ has exactly 3 prime ideals. – Crostul Dec 15 '16 at 00:07
  • Related, essentially an answer: http://math.stackexchange.com/questions/2039702/what-is-an-homomorphism-isomorphism-saying/2039715#2039715 – Ethan Bolker Dec 15 '16 at 00:15
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    Ah, okay. I understand. That made intuitive sense, just wasn't sure how to put it into proof words. Thank you! – weepyhollow Dec 15 '16 at 00:21

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This fact is obvious, since isomorphisms preserve ring structure.

We note that the $n$ prime ideals in $R$ must map to at least $n$ prime ideals under an isomorphism; a fact you stated yourself. Let $m$ be the amount of prime ideals in $S$. Since the inverse of an isomorphism is an isomorphism, we have that the $m$ prime ideals must map to at least $m$ prime ideals in $R$. However, since there are $n$ prime ideals in $R$, we have that

$$n \leq m \leq n$$

and thus $n = m$.

Meow Mix
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