Homomorphisms and ideals

Krizalid

Active member
Let $\phi:R\to S$ be a homomorphism of rings. Let $I$ be an ideal of $R$ and $J$ be an ideal of $S.$ Prove that $\phi^{-1}(J)$ is an ideal of $R$ and $\ker(\phi)\subset\phi^{-1}(J).$ Also prove that $\phi(I)$ is not necessarily an ideal of $S.$

ModusPonens

Well-known member
Let $\phi:R\to S$ be a homomorphism of rings. Let $I$ be an ideal of $R$ and $J$ be an ideal of $S.$ Prove that $\phi^{-1}(J)$ is an ideal of $R$ and $\ker(\phi)\subset\phi^{-1}(J).$ Also prove that $\phi(I)$ is not necessarily an ideal of $S.$
The second is the easiest: since $0\in J$ then $\phi (ker (\phi))=0\in J$. The third is not complicated: think of the homomorphism $\phi (x)=diag(x,x)$, where $diag(x,x)$ is the 2x2 diagonal matrix with x's on the diagonal. It can't be an ideal.

The first: Let $J^{-1}:=\phi^{-1}(J)$. If $a\in J^{-1}$ and $r\in R$. Then $\phi(ra)=\phi(r) \phi(a) \in J$, since J is an ideal. So $ra\in J^{-1}$.