Rings' second isomorphism theorem

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.

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