proof of second isomorphism theorem for rings

In the context of rings, the Second Isomorphism Theorem can be phrased as follows:

If A is an ideal in a ring R and S is a subring of R, then

  • S+A is a subring of R,

  • A is an ideal in S+A,

  • SA is an ideal in S,

  • There is an isomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath S/(SA)(S+A)/A.

It should be emphasized that the proof of this is exactly the same as the proof of the corresponding statement for groups. Again, the main idea is to use the First Isomorphism TheoremPlanetmathPlanetmath. (It is quite routine to verify the above statements concerning what’s a subring of what and what’s an ideal in what. The heart of the matter is the isomorphism, and that is what we are up to here.)

Consider the mapping f:S(S+A)/A where f(s)=s+A. Note that this is a ring homomorphism. Furthermore, it is surjectivePlanetmathPlanetmath (or “onto”): If (s+a)+A is an arbitrary element of (S+A)/A with sS and aA, then f(s)=s+A=s+(a+A)=(s+a)+A.

Consequently, the First Isomorphism Theorem tells us that

S/ker(f)(S+A)/A, where ker(f) denotes the kernel of f.

So it all comes down to showing that ker(f)=SA.

Let xker(f). Then A=f(x)=x+A. So xA. Moreover, since ker(f)S, we have xS. Thus xSA and so ker(f)SA.

Conversely, suppose xSA. Then xA. So f(x)=x+A=A. That is, xker(f) and so SAker(f).

Therefore, ker(f)=SA.

Title proof of second isomorphism theorem for rings
Canonical name ProofOfSecondIsomorphismTheoremForRings
Date of creation 2013-03-22 15:22:39
Last modified on 2013-03-22 15:22:39
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 14
Author CWoo (3771)
Entry type Proof
Classification msc 20-00
Classification msc 16-00
Classification msc 13-00
Related topic ProofOfSecondIsomorphismTheorem