Let $I$ be an ideal of ring $R$ contained in another ring $S$: what is $IS$?

Question

In fact, I know the definition of $IS=\{$linear combinations of ab with a in I and b in S$\}$ and that it is an ideal in $S$. I was wondering what is the most general framework this fitted into? e.g. I initially thought it was a product of $R$-algebras but $I$ isn't a ring. Nor can it be thought of as a product of $R$-modules as multiplication of $R$-modules doesn't make sense in the general case even when one is a subset of the other.

This type of thing appears all the time when studying ramification in number fields.

Answer 1

If $f:R\to S$ is a ring homomorphism and $I$ an ideal of $R$, then we denote by $IS$ the ideal of $S$ consisting from all linear combinations of the form $\sum_{i=1}^nf(a_i)s_i$ with $a_i\in I$ and $s_i\in S$.
The ideal $IS$ is called the extension of $I$ to $S$.

Answer 2

Let $R \subset S$ be rings and $I$ and ideal in $R$.

Define $J$ to be the smallest ideal of $S$ which contains $I$ (equivalently you can define it as the intersection of all ideals of $S$ which contain $I$).

Then, it is an easy exercise to show that $I \cdot S =J$.