Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am thinking about the proof of the second isomorphism theorem, and something isn't very clear to me.

Let $R$ be a ring ,$S\subset R$ a subring and $I\subset R$ an ideal. We have the natural homomorphism $f:R\rightarrow R/I$. The theorem states that $Im(S)=S+I/I$. My question is why not simply $Im(S)=S/I$?

I understand that it is not true (for starters, $S/I$ need not be a subring), but I cannot explain that to my self in a convincing way.

share|cite|improve this question
up vote 10 down vote accepted

So you define $f : R \to R/I$ by $f(r) = r+I$. Then $$f(S) = \{ s+I\, :\, s \in S \}$$ It's tempting to say that this is just $S/I$, but that implicitly assumes that $I \subseteq S$. However, this may not be true: an ideal of a ring is not necessarily contained in all subrings of the ring.

But calling it $(S+I)/I$ takes care of this, because we certainly have $I \subseteq S+I$ (because $0_R \in S$), and $S+I$ is a subring of $R$ (as [usually] shown in the proof).

That is, the problem isn't something like '$S/I$ not being a subring' as you suggest, but rather that the notation $S/I$ needn't make any sense!

share|cite|improve this answer
Of course, we could also say $S / (I \cap S)$ instead, and the two are isomorphic. – Zhen Lin Sep 10 '12 at 13:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.