Let phi: R --> R' be a homomorphism where both R and R' are commutative rings and I is a subring of R. Show that if J is a prime ideal of R' and I = phi^-1(J), then I is a prime ideal of R.

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- Apr 14th 2009, 07:12 PMCoda202Prime Ideals of Commutative Rings
Let phi: R --> R' be a homomorphism where both R and R' are commutative rings and I is a subring of R. Show that if J is a prime ideal of R' and I = phi^-1(J), then I is a prime ideal of R.

- Apr 14th 2009, 07:27 PMNonCommAlg
$\displaystyle I$ is an ideal of $\displaystyle R$ because $\displaystyle J$ is an ideal of $\displaystyle R'$ and $\displaystyle I=\phi^{-1}(J)=\{a \in R: \ \phi(a) \in J \}.$ we don't need any assumptions on $\displaystyle I$ (including assuming that $\displaystyle I$ is a subring!!).

to show that $\displaystyle I$ is prime, just follow the definition again: if $\displaystyle ab \in I,$ then $\displaystyle \phi(a) \phi(b)=\phi(ab) \in J.$ but $\displaystyle J$ is a prime ideal of $\displaystyle R'.$ so either $\displaystyle \phi(a) \in J$ or $\displaystyle \phi(b) \in J.$ hence either $\displaystyle a \in I$ or $\displaystyle b \in I.$