# What is the difference between total integral closure and integral closure?

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...

## 2 Answers

It seems that there is no reference where the notion of total integral closure is discussed in detail. But a good place to look at is Bhatt's notes on perfectoid spaces, especially at Proposition 5.2.5. It explains the main usage of total closures in theory of perfectoid spaces.

Definition: Let $$A \subset B$$ be an extension of rings then the total integral closure $$A_{tic}$$ of $$A$$ in $$B$$ is the set of all $$f\in B$$ such that $$f^{\mathbf N}$$ is contained in a finitely generated $$A$$-submodule of $$B$$.

Remark: It can be easily checked that $$A_{tic}$$ is an $$A$$-subalgebra of $$B$$.

Remark: The total integral closure coincides with the usual integral closure of $$A$$ in $$B$$ when $$A$$ is noetherian. So this definition is interesting only for non-noetherian rings.

Warning: The name may be a bit misleading as it need not be the case that $$A_{tic}$$ is itself totally integrally closed in $$B$$.

Even though this definition makes perfect sense for any arbitrary extension of abstract rings, we will use it only for a Tate ring (and a ring of definitions therein). So suppose that we have a Tate ring $$A$$ and a couple of definition $$(A_0, \varpi)$$ ($$\varpi$$ is assumed to be a pseudo-uniformizer). Then one of the main applications of our definition is that the subring $$A^{\circ}$$ can be recovered from this data as the total integral closure of $$A_0$$ in $$A$$.

Proposition 1: Let $$A, A_0, \varpi$$ be as above, then $$A^{\circ}$$ is equal to the total integral closure of $$A_0$$ in $$A$$.

Let us postpone the proof of this proposition and discuss its applications. We start with the following corollary:

Corollary 2: A ring of definition $$A_0$$ in a Tate ring $$A$$ is equal to $$A^{\circ}$$ if and only if it is totally integrally closed in $$A$$.

Note that Corollary $$2$$ is false if we assume that $$A_0$$ is only integrally closed in $$A$$ as there are examples of perfectoid pairs $$(R, R^+)$$ with $$R^+ \neq R^{\circ}$$.

Corollary 3: Let $$A$$ be a uniform Tate ring then $$A^{\circ}$$ is totally integrally closed in $$A$$.

Proof: We note that $$A^{\circ}$$ is a ring of definition in $$A$$ since $$A$$ is uniform. Then we can just apply Corollary $$2$$ to finish the proof.

Corollary $$2$$ is a pretty useful criterion to check that various rings of definitions are actually equal to the subrings of power-bounded elements. These two corollaries lie at the core of the proof of Proposition 5.2.5 in Bhatt's notes that describes the category of uniform Banach $$K$$-algebras in more algebraic terms. This, in turn, is used (together with a similar Proposition 5.2.6) in the proof of tilting equivalence for perfectoid $$K$$-algebras. Look at the proof of Theorem 6.2.5 for more details.

Proof of Proposition 1: The first thing to note is that $$f\in A$$ is power-bounded if and only if there is an integer $$M$$ such that $$f^{\mathbf N} \subset \varpi^{-M}A_0.$$ This follows from the definition of a power-bounded element taking into account that $$A$$ is Tate. We claim that this implies that
$$A^{\circ} \subset A_{0, tic}.$$ Indeed, for any element $$f\in A^{\circ}$$ we have an inclusion $$f^{\mathbf N} \subset \varpi^{-M}A_0$$ for some $$M$$. It is a finitely generated $$A_0$$-module (really it is just isomorphic to $$A_0$$) as there is a surjection $$A_0 \xrightarrow{\varpi^{-M}} \varpi^{-M}A_0,$$ so $$f\in A_{0, tic}$$.

Now we want to show that $$A_{0, tic} \subset A^{\circ}.$$ Pick any element $$f\in A_{0, tic}$$ then the set $$f^{\mathbf N}$$ is contained in some finitely generated $$A_0$$-submodule of $$A$$. Denote this submodule by $$M$$ and choose some set of generators $$x_1, \dots, x_n$$. Recall that $$A\cong A_0[\frac{1}{\varpi}]$$ as $$A$$ is Tate, so we conclude that each $$x_i$$ can be written as $$x_i=a_i/\varpi^{M_i}$$ for some $$a_i\in A_0$$ and $$n_i\in \mathbf N$$. Set $$M:=\max_{i=1}^n{M_i}$$ then it is easy to see that $$M\subset \varpi^{-M}A_0$$. Thus we see that $$f^{\mathbf N} \subset \varpi^{-M}A_0.$$ So $$f$$ is power-bounded by the observation at the beginning of the proof. We conclude that $$A^0\subset A_{0, tic}$$.

Finally, I need to say that in the context of rigid geometry many of those subtleties with total integral closure disappear.

Theorem: Let $$K$$ be a complete non-archimedean field with a pseudo-uniformizer $$\varpi$$, and let $$A$$ be a topologically finitely generated $$K$$-algebra. Suppose that $$A_0$$ is an open topologically finitely generated $$K^{\circ}$$-subalgebra of $$A$$ such that it is integrally closed in $$A$$ and $$A_0[\frac{1}{\varpi}]=A$$ (i.e. $$A_0$$ is a ring of definition in $$A$$). Then $$A_0$$ is equal to $$A^{\circ}$$.

Proof: Pick any surjection $$K^{\circ}\langle X_1, \dots, X_n \rangle \to A_0$$ that exists by the assumption on $$A_0$$. We use Theorem 6.3.5/1 from the book "Non-Archimedean analysis" by Bosch, Guntzer, and Remmert to conclude that $$A^{\circ}$$ is integral over $$K^{\circ}\langle X_1, \dots, X_n \rangle$$, in particular, it is integral over $$A_0$$. Finally, we recall that $$A_0$$ is integrally closed in $$A$$, so $$A^{\circ}$$ must be equal to $$A_0$$.

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 $$x^{-1}$$.