Why is $R/ (I \cap R)$ a subring of $S/I$?

Question

Let $R \subset S$ be a ring extension. Let $I \subset S$ be an ideal. Then why do we have $$R/(R \cap I) \subset S/I ~~.$$

Surely, $R \cap I$ is an ideal in $R$. However, the elements of $R/(R \cap I)$ are of the form $\overline{r} := r + (R \cap I) = \{r + a \vert a \in R \cap I \}$ with $r \in R$, while the elements of $S/I$ are of the form $\overline{s} := s + I = \{s + b \vert b \in I \}$.

Let $\overline{r} \in R/(R \cap I)$. Given that $R/(R \cap I)$ is (in particular) a subset of $S/I$, there has to be a $\overline{s} \in S/I$ s.t. $\overline{r} = \overline{s}$.

My question is, what is the element $\overline{s}$?

It isn't $r + I$ since $r+i \not\in \overline{r}$ whenever $i \in I \setminus R$.

Answer 1

That inclusion is just a matter of notation. Actually, what happens is that there is a natural injective ring homomorphism $R/(R\cap I)\to S/I$ and this is precisely the unique map $f$ such that: $$(R\hookrightarrow S\xrightarrow{\pi_I}S/I)=(R\xrightarrow{\pi_{R\cap I}}R/(R\cap I)\xrightarrow{f}S/I). $$ This map $f$ exists and is unique because of the universal property of the quotient map $R\xrightarrow{\pi_{R\cap I}}R/(R\cap I)$, as $R\cap I \subset I$. In fact $f$ is just the map sending a class $x + R\cap I$ to the corresponding $x + I$. It's easy to prove that it is indeed well-defined and injective.

It is well defined because $R\cap I\subset I$ and it is injective because, if $x + R\cap I$ and $y+ R\cap I$ are elements of $R/(R\cap I)$ such that $x+ I=y+ I$, then $x-y \in I$. Since $x-y \in R$ as well, it holds that $x-y \in R \cap I$, that is $x+R\cap I=y+R\cap I$.

Answer 2

Two elements $x, y\in R$ are in the same coset of $R\cap I$ if $x-y\in R\cap I$. Since $x-y\in I$ in particular, $x$ and $y$ are in the same unique coset of $I$ in $S$. Thus it makes sense to associate $x+R\cap I$ with the coset $x+I$ containing it.

If $x-y\notin R\cap I$, then $x-y\notin I$ since $x-y\in R$, hence $x+I$ and $y+I$ are different cosets.

Answer 3

As abelian groups, you should have no qualms with the idea that $\frac{S}{I}\supseteq\frac{R+I}{I}\cong \frac{R}{R\cap I}$ by the second isomorphism theorem. The thing on the right hand side can be identified with the thing in the middle via the isomorphism.

Now the fact that $I\lhd S$ and $I\lhd R+I$ imbues the things on the left and center with ring structure, in fact a subring structure. The thing on the right is ring isomorphic with the thing in the middle.