If $A \subset B$, both commutative rings, then we define $\bar{A}_B = \{x \in B| b \text{ is integral over A}\}$ where $x$ being integral over $A$ means there is a monic relation $a_0 + a_1x + \ldots + a_{n-1}x^n = 0$ with all $a_i$ in $A$. There is a profound analogy between the concept of an element being integral over a ring and an element being algebraic over a field. For instance if $A$ and $B$ are fields then an element of B is integral over A iff it is algebraic over A. However the similarities between the two concepts seem to break down when it comes to defining the "closure" operation. The way Atiyah-MacDonald presents things he only talks about integral closure of $A$ over $B$, there is no presentation of an intrinsic algebraic closure (so not canonically embedded into a larger ring) the same way there is for fields. Why is that so?
1 Answers
If you restrict to integral domains $A$, you could consider the integral closure in an algebraic closure of the fraction field of $A$ to be the "absolute integral closure". At least among integral domains, you won't find a larger integral extension. But this absolute integral closure does not play the same fundamental role the algebraic closure of a field plays for Galois theory (except when you go deeper, see the comments), hence Atiyah-MacDonald don't mention it.
Without this restriction, there is no maximal integral extension. You can, for example, always add more nilpotents or idempotents, both of which are always integral over the base ring. To add nilpotents, go from $A$ to $A[x]/(x^n)$, to add idempotents go from $A$ diagonally to $A \times A$.
However, one "intrinsic integral closure" that is often used is the normalization, which in the case on an integral domain is the integral closure in its field of fractions. It's the maximal integral extension with the same fraction field as the original domain. This is the minimal integrally closed extension, i.e. it embeds into any integrally closed extension. Here integrally closed means normal, i.e. integrally closed in its field of fractions.
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Just to complement Lukas's nice answer, let me mention that absolute integral closures do play an important/fundamental role, just not one covered in a first course in algebraic geometry/commutative algebra. For example, if $R$ is an integral domain (or more generally a ring) it has an etale fundamental group $\pi_1(R)$ that plays the analogue of the classical fundamental group in algebraic topology. For theoretical reasons, it is very useful to find a surjective map $X\to\mathrm{Spec}(R)$ where $X$ is 'simply connected' (e.g., this provides a sort 'path-connectedness' statement – Alex Youcis Oct 11 '25 at 15:11
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allowing one to prove that the fundamental group is independent of base point). Perhaps the most natural example of this is $X=\mathrm{Spec}(\overline{R})$, where $\overline{R}$ is the absolute algebraic closure of $X$. For more discussion on this, you can see this answer: https://math.stackexchange.com/questions/4492569/independence-of-etale-fundamental-group-on-base-point/4499399#4499399 – Alex Youcis Oct 11 '25 at 15:14
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1@AlexYoucis that's a good addition! I've learned about étale cohomology, but not about concrete étale paths before. – Lukas Heger Oct 11 '25 at 15:16
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1Your answer is nice but I do want to nitpick that the normalization is maximal in a wider sense, if $K$ is the field of fractions and $R\subset \overline{R}\subset S\subset K$ are subrings, then $\overline{R}_S = \overline{R}$. However, outside of subrings of the field of fraction, it is not maximal -- perhaps this is what you meant. But, thinking of the integral closure as the largest integral extension in the fraction field is an extremely important perspective, just as important imo as thinking of it as the minimal integrally closed extension. – walkar Oct 12 '25 at 00:57
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1@walkar you're right of course. This maximality follows directly from the definition of a relative integral closure. – Lukas Heger Oct 12 '25 at 01:21