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Ideals and Rings : Homomorphisms

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Prove the Second Isomorphism Theorem: If A is an ideal of R and S is a subring of R, then S+A is a subring, A, and (S intersecting A) are ideals of S+A and S, respectively, and (S+A)/A isomorphic to A/(S intersecting A).

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Solution Summary

The Second Isomorphism Theorem is proven.

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Commutative Rings, Ideals, Kernels, Matrices and Injective and Surjective Ring Homomorphisms

If n Є R and R is a commutative ring we indicate by Mn(R) the ring of allnxn entries wrt the usual operations on matrices. If n>1 this ring is commutative even if R is.
Let S={(aij)ЄMn(R)|i≠j=>aij=0}
Let k be an integer 1≤k≤n. Show that
a) S is a commutative subring of Mn(R)
b) The function f: S-->R defined by f((aij))=akk is a surjective ring homomorphism
c) The set defined by IK={(aij) Є S | akk=0} is the kernel of f
d) IK is an ideal of S

What are necessary conditions for S to be an integral domain?

e) Show that R ={[a 0]|a,b,c Є R} is a subring of M2(R). Is it commutative? Find a non trivial ideal of R.
[b c]

f) Is S ={[a b]|a,b,c Є R} is a subring of M2(R)?
[c 0]

g) Show that the function C-->M2(R) a+bi -->[a b] is an interjecting homomorphism.
[-b a]

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