Geometric characterization of integral ring extensions In commutative algebra, one usually learns about Lying Over, Incomparability, Going Up and Going Down for integral extensions. These are, in their spirit, quite geometrical statements (as is e.g. demonstrated in Gathmann's lecture notes).
I'm wondering if there is any kind of converse statement: Something in the sort of "For (sufficiently nice) rings $R \hookrightarrow R'$ that satisfy (some of) Lying Over, Incomparability, Going Up, Going Down and some property, the extension is integral."
 A: Unfortunately, as demonstrated in the comments by uno with the extension $k\subset k(t)$, the statement "if $R\to S$ is a ring map satisfying Going Up, Going Down, and Lying Over, then $R\to S$ is integral" is false. There is a geometric characterization of integral ring maps, though: $R\to S$ is integral iff $\operatorname{Spec} S \to \operatorname{Spec} R$ is universally closed. Full details are available at Stacks 01WM. Here's an outline of the proof:
To show integral implies universally closed, it's enough to show that integral implies closed, as a base change of an integral ring extension is again integral. Since going up is equivalent to the map on spectra being closed, we're done.
To show that $\operatorname{Spec} A\to\operatorname{Spec} R$ universally closed implies $R\to A$ integral, we need to show that an arbitrary element $a\in A$ is integral over $R$, or that the kernel of $R[x]\to A$ by $x\mapsto a$ contains a monic polynomial. Let $B=A[x]/(ax-1)$, and let $J$ be the kernel of $R[x]\to A[x]\to B$. If $f\in J$, there exists some $q\in A[x]$ such that $f=(ax-1)q$ in $A[x]$, so if $f=\sum f_ix^i$ and $q=\sum q_ix^i$, for all $i\geq 0$ we have $f_i=aq_{i-1}-q_i$. For $n\geq \deg q+1$, the polynomial $$\sum f_ix^{n-i} = \sum (aq_{i-1}-q_i)x^{n-i} = (a-x) \sum q_ix^{n-i-1}$$ is in $I$, and if $f_0=1$ this polynomial is also monic. So we are reduced to showing that $J$ contains a polynomial with constant term $1$. We do this by proving $\operatorname{Spec} R[x]/(J+(x))$ is empty.
Since $f$ is universally closed, the base change $\operatorname{Spec} A[x]\to\operatorname{Spec} R[x]$ is closed, and the image of the closed subset $\operatorname{Spec} B \subset \operatorname{Spec} A[x]$ is the closed subset $\operatorname{Spec} R[x]/J \subset \operatorname{Spec} R[x]$, and in particular, $\operatorname{Spec} B$ surjects on to $\operatorname{Spec} R[x]/J$. Consider the following diagram, where every square is a fiber product square:
$$\require{AMScd}
\begin{CD}
\emptyset @>{}>> \operatorname{Spec} R[x]/(J+(x)) @>>> \operatorname{Spec} R \\
@VVV @VVV @VVV \\
\operatorname{Spec} B @>{}>> \operatorname{Spec} R[x]/J @>>> \operatorname{Spec} R[x]
\end{CD}$$
The top left corner is empty, since it's the spectrum of $R\otimes_{R[x]} B$ where $R[x]\to B$ sends $x$ to an invertible element and $R[x]\to R$ sends $x$ to $0$. Since the lower left horizontal map is surjective, it's base change is too, and thus we have the desired conclusion.
