State and prove fundamental theorem of homomorphism of rings ?
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Answer:
The Fundamental Theorem of Ring Homomorphisms
On the Quotient Rings page said that if (R,+,∗) is a ring and (I,+,∗) is an ideal, then the quotient of R by I is the set R/I of equivalence classes a+I={a+i:i∈I} with the operations of addition and multiplication defined for all (a+I),(b+I)∈R/I by:
(1)
(a+I)+(b+I)=(a+b)+I
(2)
(a+I)∗(b+I)=(a∗b)+I
So in order to define a quotient ring we need a ring (R,+,∗) and an ideal (I,+,∗). Now suppose that R and S are homomorphic rings with homomorphism ϕ:R→S. Then ker(ϕ) is a subring of R (this is relatively easy to show). We would ultimately like ker(ϕ) to be an ideal of R to define the quotient R/ker(ϕ).
This is not true in general though. Recall that when we defined a ring, we required that there exists an element 1∈R such that a∗1=a and 1∗a=a for all a∈R, that is, we required the existence of a multiplicative identity. However, by definition, ker(ϕ) in general cannot be a subring of R because if 1 is the identity of R, then by definition of a ring homomorphism, ϕ(1) is mapped to the multiplicative identity of S and not to the additive identity of S.
To establish a fundamental theorem of ring homomorphisms, we make a small exception in not requiring that ker(ϕ) is an ideal for the quotient R/ker(ϕ) to be defined.
Theorem 1 (The Fundamental Theorem of Ring Homomorphisms): Let (R,+1,∗1) and (S,+2,∗2) be homomorphic rings with ring homomorphism ϕ:R→S. Then R/ker(ϕ)≅ϕ(R).
Proof: Let K=ker(ϕ) and let ψ:R/K→ϕ(R) be defined for all (a+K)∈R by:
(3)
ψ(a+K)=ϕ(a)
Then ψ is well-defined, for if a+K=b+K then a=b+k for some k∈K and so:
(4)
ψ(a+K)=ψ((b+k)+K)=ψ((b+K)+K)=ψ(b+K)
Now let (a+K),(b+K)∈R/K. Then:
(5)
ψ((a+K)+(b+K))=ψ((a+b)+K)=ϕ(a+b)=ϕ(a)+ϕ(b)=ψ(a+K)+ψ(b+K)
(6)
ψ((a+K)∗(b+K))=ψ((a∗b)+K)=ϕ(a∗b)=ϕ(a)∗ϕ(b)=ψ(a+K)∗ψ(b+K)
We lastly show that ψ is bijective.
Let (a+K),(b+K)∈R/K and suppose that ψ(a+K)=ψ(b+K). Then ϕ(a)=ϕ(b). So ϕ(a−b)=0. So a−b∈K. So a+K=b+K. Hence ψ is injective.
Furthermore, for all a∈ϕ(R) we have that (a+I)∈R/K is such that ψ(a+K)=a. So ψ is surjective.
Hence ψ is bijective and so ψ is an isomorphism from R/K to ϕ(R), that is:
(7)
R/ker(ϕ)≅ϕ(R)■
For example, consider the rings Z and Zn. We can define a homomorphism ϕ:Z→Zn for all z∈Z by:
(8)
ϕ(z)=zmodn
Indeed ϕ is a homomorphism, as you should verify. Note that:
(9)
kerϕ(z)={0,±n,±2n,..}=nZ
So by the fundamental theorem of ring homomorphisms we have that:
(10)
Z/nZ≅Zn
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Answer:
The Fundamental Theorem of Ring Homomorphisms. ... However, by definition, in general cannot be a subring of because if is the identity of , then by definition of a ring homomorphism, is mapped to the multiplicative identity of and not to the additive identity of .
Step-by-step explanation:
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