MHB How to Prove Certain Properties of Homomorphisms and Ideals in Ring Theory?

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To prove that $\phi^{-1}(J)$ is an ideal of $R$, it is shown that for any $a \in \phi^{-1}(J)$ and $r \in R$, the product $ra$ remains in $\phi^{-1}(J)$ due to the properties of ideals. Additionally, it is established that the kernel of the homomorphism $\phi$, $\ker(\phi)$, is a subset of $\phi^{-1}(J)$ since the image of the kernel under $\phi$ is zero, which is contained in any ideal $J$. However, $\phi(I)$ is not necessarily an ideal of $S$, illustrated by the example of the homomorphism $\phi(x) = diag(x,x)$, which does not satisfy the ideal properties in the codomain. The discussion emphasizes the importance of understanding the relationships between homomorphisms, ideals, and their inverses in ring theory. Overall, these properties are crucial for analyzing the structure of rings and their homomorphic images.
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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 $S.$
 
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Krizalid said:
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 $ra\in J^{-1}$.
 
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