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.
2026-04-02 10:40:11.1775126411
If rings R and S are isomorphic, prove that if R has 3 prime ideals, then S does too?
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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 $n = m$.