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The proof that the preimage of a prime ideal $P \subset S$ under say the ring homomorphism $\phi: R \rightarrow S$ is prime is straightforward. However, it only seems to show that if $xy \in \phi^{-1}(P)$ then $x$ or $y \in \phi^{-1}(P).$ Shouldn't it also show that $\phi^{-1}(P) \neq R$ as part of the definition of prime ideal is that the ideal is not the entire ring? I'm assuming that the conventional proof disregards this part of the definition...

For example, $\phi^{-1}(5Z)$ under the map $\phi: 5Z \rightarrow Z$ is the entire ring $5Z.$

green frog
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    Can the preimage of a prime ideal contain $1$? – Bernard Aug 23 '18 at 12:19
  • Well in general the proof doesn't assume unital rings. But if the map is not surjective I don't see a problem as 1 might not map to a unit. – green frog Aug 23 '18 at 12:20
  • Many authors use ring to mean unital ring. In this case, a ring homomorphism must map $1$ to $1$, and as Bernard suggests, $1$ cannot be contained in the preimage of a prime ideal. I would look carefully at the context for this proof, I've been caught out by confusions like this many times! – Daniel Mroz Aug 23 '18 at 12:22
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    In commutative algebra, unless explicitly otherwise stated, all rings have unit elements, and ring homomorphisms map a unit onto a unit. Now if that's a question in non-commutative algebra, it's a different problem. – Bernard Aug 23 '18 at 12:23
  • Please state the source of your proof. In particular, find out if you are working in a unital ring setting or in a more general setting. – Sarvesh Ravichandran Iyer Aug 23 '18 at 12:26

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