User:Siriverbs1903/sandbox

RING HOMOMORPHISM:

Let R1 and R2 be rings. The function f:R1→R2 is said to be ring homomorphism if for any r,s belongs to R1, f(r+s)=f(r)+f(s) f(r.s)=f(r).f(s)

Examples :

1. The identity function defined on a ring      R is a ring homomorphism f:R→R       f(x)=x for all x belongs to R     f(r+s)=f(r)+f(s) f(r.s)=f(r).f(s). 2. Let S be a sub-ring of R the inclusion function j:S→R; j(x)=x for all x belongs to S is a ring homomorphism.

 RING ISOMORPHISM :

A ring homomorphism f:R1→R2 is said to be an Isomorphism if it is one-one and onto.RING HOMOMORPHISM: A ring R1 is said to be isomorphic to ring R2 if there is a ring isomorphism f:R1→R2. Examples: 1. Identity function on any ring is ring isomorphism. 2. Define f:k→k by, f(z)=z. That is f(a+ib)=a-ib, for all a+ib belongs to k then f is a ring isomorphism.

Remark:

therefore we have: 1. f(0)=0, the additive identity in S.  2. f(-r)=-f(r) for all r belongs to R.
 * There does not exist any ring isomorphism from Q→R.
 * Z and 2Z are isomorphic as groups where as they are not isomorphic as rings.
 * Suppose f:R→S is a ring homomorphism then we know that it is an additive group homomorphism.

THEOREM: Let f:R→S be a ring homomorphism then the f image of R is a sub-ring of S. PROOF: Let f:R→S be a ring homomorphism then f(R)={f(r)/r belongs to R}      since f is a function and R is a non-empty it is clear that f(R) is a non-empty subset of S.       Now for any two elements f(r1)+f(r2) belongs to f(R) we see that, f(r1)-f(r2)=f(r1)+f(-(r2)) =f(r1)+f(-r2) [since f is a ring homomorphism] =f(r1+(-r2)) =f(r1-r2) thus, f(r1)-f(r2)=f(r1-r2) belongs to f(R) and also f(r1).f(r2)=f(r1.r2)  [since f is a ring homomorphism] thus, f(r1).f(r2) belongs to f(R) therefore By sub-ring test f(R) is a sub-ring of S.     therefore f(R) is a sub-ring of S.