Suppose we have ideals $I_1, I_2$. Is there any way to simplify $(I_1+I_2)/I_1$?
I'm thinking since all elements are of the form $i_1+i_2+I_1$ it should be simplified to $I_2/I_1$ but since $I_1$ is not necessarily in $I_2$ I am confused.
2026-04-07 11:08:46.1775560126
Quotient of sum of ideals
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As abelian group, this is isomorphic to $I_2/(I_1 \cap I_2)$, using the second isomorphism theorem : look at the group morphism $$f : I_2 \to (I_1+I_2)/I_1$$ defined by $f(x)=x+I_1$. Is it surjective? What is its kernel? Using the first isomorphism theorem $$f/\ker(f) \cong \mathrm{im}(f)$$ you should get $I_2/(I_1 \cap I_2) \cong (I_1+I_2)/I_1$.
Notice that ideals are not subrings (in general), so there is no ring structure on your quotient. Then, it makes no sense to ask "What about as rings?".
You can however ask: is it an isomorphism of $R$-modules, where $I_1,I_2$ are ideals of a commutative ring $R$?