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

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SUMMARY

This discussion focuses on the properties of homomorphisms and ideals in ring theory, specifically addressing the homomorphism $\phi:R\to S$. It establishes that the preimage $\phi^{-1}(J)$ is an ideal of $R$ and that the kernel $\ker(\phi)$ is a subset of $\phi^{-1}(J)$. Additionally, it clarifies that the image $\phi(I)$ of an ideal $I$ in $R$ is not necessarily an ideal in $S$. The examples provided, including the homomorphism $\phi(x)=diag(x,x)$, illustrate these concepts effectively.

PREREQUISITES
  • Understanding of ring theory and the definitions of homomorphisms
  • Familiarity with ideals in ring structures
  • Knowledge of kernel and image concepts in algebra
  • Basic proficiency in working with diagonal matrices
NEXT STEPS
  • Study the properties of ring homomorphisms in greater detail
  • Explore the relationship between ideals and their images under homomorphisms
  • Investigate examples of non-ideal images in various ring structures
  • Learn about the implications of the kernel in homomorphic mappings
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Mathematicians, algebra students, and educators focusing on abstract algebra, particularly those interested in ring theory and its applications.

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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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