Special Case of Lying Over Theorem

In Pete Clark's commutative algebra lecture notes which can be found here. He proves the following lemma (14.12)

Let $R$ be a local ring with maximal ideal $\mathfrak{p}$ and $S/R$ an integral ring extension. Then the pushed forward ideal $\mathfrak{p}S$ is proper.

The majority of the proof is clear to me except the induction argument. The proof argues by contradiction that if not we may find $p_i \in \mathfrak p$ and $s_i \in S$ such that $$1=\sum_i p_i s_i.$$ So the pushforward is already not proper in the ring $R[s_1,\dots,s_d]$. Clark says that by induction we need only consider the case $S=R[s]$. I understand the proof in this case, but it's not clear to me how to extend it to say $R[s_1,s_2]$. The argument uses very directly the fact that $R$ is local and if I wanted to extend it in a natural way I would need the stronger statement that $R[s]$ is a local ring. Is there some easy way to extend this argument in general that I'm missing?

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"pushed forward ideal" is called "the extension of ideal" in Commutative Algebra. (See Atiyah & MacDonald.) – user26857 Jan 10 '13 at 11:13
Furthermore, the proof of LO in the book of Clark is unnecessary complicated. Look at the proof of the same result in Atiyah & MacDonald, Theorem 5.10. It's almost an one-liner proof (forget about the diagram)! – user26857 Jan 10 '13 at 11:21
@YACP, I agree that the proof is much simpler. Although could you clarify why it's safe to assume the localization maps are injective? – JSchlather Jan 10 '13 at 11:39
The localization preserves injectivity. – user26857 Jan 10 '13 at 11:42

Let's try to extend to $R[s_1,s_2]$. What we know is that $pR[s_1]$ is proper in $R[s_1]$. Let $m$ be a maximal ideal in $R[s_1]$ that contains $pR[s_1]$, and consider $R[s_1]_m \to R_[s_1,s_2]_m$. Using $d=1$ case again we see that $mR[s_1,s_2]_m$ is proper in $R[s_1,s_2]_m$. This implies that $mR[s_1,s_2]$ is proper in $R[s_1,s_2]$, so $pR[s_1,s_2]$ is also proper.
The same argument will work for any $d$.