- #26

member 587159

Part 1 of (9)

##\phi: R \to S## is a ring epimorphism. Define ##I:= \phi^{-1}(J)##. It is well known that the inverse image of an ideal is an ideal, thus ##I## is an ideal.

Define ##\psi: R/I \to S/J: [r] \mapsto [\phi(r)]##

This is well defined: If ##r \in I##, then ##\phi(r) \in J##.

Clearly, this is also a ring morphism.

For injectivity, assume ##[\phi(r)] = 0##, then ##\phi(r) \in J##, and ##r \in \phi^{-1}(J) = I##, thus ##[r] = 0##. The kernel is trivial and the map is injective.

Surjectivity follows immediately by surjectivity of ##\phi##.

It follows that ##\psi## is an isomorphism, and thus ##R/I \cong S/J##.

Define ##\psi: R/I \to S/J: [r] \mapsto [\phi(r)]##

This is well defined: If ##r \in I##, then ##\phi(r) \in J##.

Clearly, this is also a ring morphism.

For injectivity, assume ##[\phi(r)] = 0##, then ##\phi(r) \in J##, and ##r \in \phi^{-1}(J) = I##, thus ##[r] = 0##. The kernel is trivial and the map is injective.

Surjectivity follows immediately by surjectivity of ##\phi##.

It follows that ##\psi## is an isomorphism, and thus ##R/I \cong S/J##.

Last edited by a moderator: