A question about integral ring extension

Question

Lemma: Let $R\subset S$ be an integral ring extension. Let $I$ be a proper ideal of $R$. Then $IS$ is a proper ideal of $S$.

I'm reading the proof from following link

http://www.math.rwth-aachen.de/~zerz/ast10/dim1.pdf

I have some questions about it. First we suppose that $IS=S$. It is stated that "we can write $1=\sum_{i=1}^{n}r_is_i$ where $r_i\in I$ and $s_i\in S$".

1. My first question is that why we need to choose $n$ elements? Since $IS=S$, we can choose $r\in I$ and $s\in S$ such that $1=rs$.

2. Then one defines $S'=R[s_1,...,s_n]$ and says that $S'=IS'$. Why?

Answer 1

1. Here's the crux of your confusion - this is not how this product is defined, in fact $IS$ is the set of such finite sums. To see that this is necessary, consider $R=\Bbb Q[x,y]$, $S=\Bbb Q[x,y,z,w]$ and $I=(x,y)$. Then, $xz \in IS$ and $yw \in IS$, so clearly $xz + yw \in IS$ as well. However, you can not express $xz + yw$ as a single product of one element in $I$ and one element in $S$.
2. Since $s_1,\ldots,s_n\in S'$, you have $1=\sum_{i=1}^n r_i s_i \in IS'$, hence $IS'=S'$. In other words, since $S'$ contains all the necessary elements, it is a ring where you can use the expression from the first part to generate the element $1$.