(од +5) Х (е“-4 R+4+)
Answers
Explanation:
Let (R, +_1, *_1)(R,+
1
,∗
1
) and (S, +_2, *_2)(S,+
2
,∗
2
) be homomorphic rings with ring homomorphism \phi: R \to S.ϕ:R→S. Then R/ker(\phi) \cong \phi (R)R/ker(ϕ)≅ϕ(R)
.
Let R_4 = ker(\phi)R
4
=ker(ϕ) and let \Phi :R_2/R_4 \to \phi(R_2)Φ:R
2
/R
4
→ϕ(R
2
) be defined for all (a + R_4) \in R_2(a+R
4
)∈R
2
by \phi (a+R_4)=\phi(a)ϕ(a+R
4
)=ϕ(a)
Then \psiψ is well defined, for if a+R_4=b+R_4 \implies a=b+ ra+R
4
=b+R
4
⟹a=b+r for some r \in R_4r∈R
4
and so:
\psi (a+R_4)=\psi ((b+r)+R_4)=\psi((b+R_4)+R_4)=\psi(b+R_4)ψ(a+R
4
)=ψ((b+r)+R
4
)=ψ((b+R
4
)+R
4
)=ψ(b+R
4
)
To show that R_2/R_4R
2
/R
4
is an isomorphism to \phi (R_2)ϕ(R
2
), we need to show that \psiψ is bijective.
Let (a+R_4), (b+R_4) \in R_2/R_4(a+R
4
),(b+R
4
)∈R
2
/R
4
and suppose that
\psi(a+R_4)=\psi(b+R_4).ψ(a+R
4
)=ψ(b+R
4
).
Then \phi(a)=\phi(b)ϕ(a)=ϕ(b)
So,
\phi(a-b)=0ϕ(a−b)=0
So,
a-b \in K,a−b∈K,
So, a+R_4=b+R_4a+R
4
=b+R
4
Hence \psiψ is injective
Furthermore, \forall a \in \phi(R_2)∀a∈ϕ(R
2
) we have that (a +I) \in R_2/R_4(a+I)∈R
2
/R
4
is such that \psi (a + R_4)=aψ(a+R
4
)=a.
So \psiψ is surjective.
Hence \psiψ is bijective and so R_2/R_4R
2
/R
4
is proven to be an isomorphism, that is
R_2/R_4\cong \phi(R_2)R
2
/R
4
≅ϕ(R
2
)