Restricting the quotient map of rings to a subring

Question

When $q$ maps $R$ to $R/I$ and $p$ is the restriction of $q$ to a subring $A$ of $R$, why is the image of $p$ $(A+I)/I$?

$q$ maps $r$ to $r+I$, so shouldn't $p$ map $a \in A$ to $a+I$, so image of $p$ is the set $\{a+I : a\in A\}?$

Answer 1

Consider $p : A \rightarrow R/I$ given by $p = q\vert_A$. The first isomorphism theorem provides $A/\ker(p) \simeq p(A)$. It should be evident that $\ker(p) = I \cap A$, so $A/(I \cap A) \simeq p(A)$. The second isomorphism theorem provides $(A + I)/I \simeq A/(A \cap I)$, so $(A + I)/I \simeq p(A)$