Seu SlideShare está sendo baixado. ×

Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio
Anúncio

1 de 1 Anúncio

# Let R and S be rings- with phi- R -- S a ring homomorphism- Suppose th.docx

Let R and S be rings, with phi: R -> S a ring homomorphism. Suppose that T is a subring of S. Let phi^-1(T) = {r E R: phi(r) E T} . Prove that phi^-1(T) is a subring of R. When is phi^-1(T) the whole ring R? page 222 of A first course in abstract algebra number 28.
Solution
Let X = phi^-1(T) be set. We need to prove it is a subring. Let x E X, then r E R such that phi(r) = x. Now phi(a*r) = a*phi(r), as phi is a ring homomorphism. Hence phi(a*r) = a*x E X, since phi(a*r) E X. Similarly we can prove that if x and y E X, then x + y E X.
Hence X = phi^-1(T) is a subring.

phi^-1(T) is the whole ring when T = R(the complete set), and phi is an isomorphism
.

Let R and S be rings, with phi: R -> S a ring homomorphism. Suppose that T is a subring of S. Let phi^-1(T) = {r E R: phi(r) E T} . Prove that phi^-1(T) is a subring of R. When is phi^-1(T) the whole ring R? page 222 of A first course in abstract algebra number 28.
Solution
Let X = phi^-1(T) be set. We need to prove it is a subring. Let x E X, then r E R such that phi(r) = x. Now phi(a*r) = a*phi(r), as phi is a ring homomorphism. Hence phi(a*r) = a*x E X, since phi(a*r) E X. Similarly we can prove that if x and y E X, then x + y E X.
Hence X = phi^-1(T) is a subring.

phi^-1(T) is the whole ring when T = R(the complete set), and phi is an isomorphism
.

Anúncio
Anúncio