TITLE: What is the difference between total integral closure and integral closure?
QUESTION [6 upvotes]: I was advised here to make this a new question:
What is the difference between total integral closure and integral closure (geometrically, in the context of rigid analytic geometry)? I have read in multiple places that the difference is basically rank one valuations vs higher rank valuations but how does one make this precise?
It is surprisingly hard to even find a definition of total integral closure...

REPLY [4 votes]: A basic example: Let $R$ be the ring of Laurent series $k[x^{\pm}, y^{\pm}]$ with coefficients in some field $k$. Let $A$ be the subring generated by $x$ and by all monomials of the form $x^{-j} y$. The element $x^{-1}$ is not integral over $A$; equivalently, the $A$-submodule $M$ of $R$ generated by all powers of $x^{-1}$ is not finitely generated as an $A$-module.
However, $M \subset y^{-1} A$, so $M$ is contained in a finitely generated $A$ module, and $x^{-1}$ is in the total integral closure of $A$. 
One can see the connection to valuations easily in this example: Let $\Lambda$ be $\mathbb{Z}^2$ ordered lexicographically (with the second coordinate more important than the first) and define $v : R_{\neq 0} \to \Lambda$ by 
$$v\left( \sum_{(i,j)} c_{ij} x^i y^j \right) = \min\left\{ \vphantom{\sum_{(i,j)} c_{ij} x^i y^j} (i,j) : c_{ij} \neq 0 \right\}.$$
Then $v(x^{-1}) = (-1, 0)$, detecting that $v$ is not in the integral closure. However, no rank $1$ valuation takes negative value on $x^{-1}$.