Lying over and going up for integral extensions

Let \(R\subseteq S\) be an integral extension of commutative rings.

1. Lying over: for every prime ideal \(\mfp\subseteq R\) there is a prime ideal \(\mathfrak q\subseteq S\) with \(\mathfrak q\cap R=\mfp\). Equivalently, the map \(\Spec S\to\Spec R\), \(\mathfrak q\mapsto\mathfrak q\cap R\), is surjective.

2. Going up: if \(\mfp_1\subseteq\mfp_2\) are prime ideals of \(R\) and \(\mathfrak q_1\) is a prime ideal of \(S\) with \(\mathfrak q_1\cap R=\mfp_1\), then there is a prime ideal \(\mathfrak q_2\supseteq\mathfrak q_1\) of \(S\) with \(\mathfrak q_2\cap R=\mfp_2\).