Prime Ideals of Commutative Rings

**Coda202** · April 14th 2009, 07:12 PM

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.

**NonCommAlg** · April 14th 2009, 07:27 PM

Originally Posted by Coda202

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.

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

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

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