Question

(од +5) Х (е“-4 R+4+)​

Answer 1

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

)