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In Atiyah-Macdonald I came across the following theorem:
Let $A \subseteq B$ be rings, $B$ integral over $A$.

  • If $\frak{b}$ is an ideal of $B$ and $\frak{a} = \frak{b}^c= \frak{b} \cap$$A$, then $B / \frak{b}$ is integral over $A/ \frak{a}$.
  • If $S$ is a multiplicatively closed subset of $A$, then $S^{-1}B$ is integral over $S^{-1}A$.

I was able to understand the proof of this statement yet I need some more clarity. The definition of integral extension in the textbook is given as follows:
Let $B$ be a ring, $A$ a subring of $B$ (so that $1 \in A$). An element $x$ of $B$ is said to be integral over $A$ if $x$ is a root of a monic polynomial with coefficients in $A$, that is if $x$ satisfies an equation of the form, $x^n + a_1x^{n -1} +...+ a_n = 0$ where the $a_i$ are elements of $A$.


Now $A/ \frak{a}$ need not be a subset of $B / \frak{b}$. So what I think Atiyah-Macdonald is saying is that since $A\hookrightarrow B \twoheadrightarrow\ B/\mathfrak b$ and under this composition of morphisms, $\frak{a}$ goes to zero of $B/\mathfrak b$, we have a morphism $A/\mathfrak a \to B/\mathfrak b$. Its easy to check that this map is injective. So we consider the image of this map, call it $C$ is a subring of $B/ \mathfrak b$ which is isomorphic to $A/ \mathfrak a$. So I think the theorem is saying that $B/ \mathfrak b$ is integral over $C$. Am I on the right path?


Similarly for second part, we consider the morphism $B \to S^{-1}B$ given by $b \mapsto \frac{b}{1}$ and consider the image of $A$ and proceed as above. But one of my friends said that since $S$ is a multiplicatively closed subset of $A$, we need to first find the localization of $A$ and then find an inclusion into $S^{-1}B$. So, I'm not sure which is the correct interpretation of the statement.
I'm still not confident about my reasoning. I still need some clarification about this proposition.

Ajin Shaji Jose
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1 Answers1

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In the first example, if we let $q \colon B\to B/\mathfrak b$ be the quotient map from $B$ to $B/\mathfrak b$, then $q$ induces an isomorphism between $A/\mathfrak a$ and $q(A) \subseteq B/\mathfrak b$, but it is also perfectly reasonable to consider $B/\mathfrak b$ as an $A$-algebra via the restriction of $q$.

In the second example, $S$ is a multiplicatively closed subset of $A$, and hence one can form the localizations $S^{-1}A$ of $A$ with respect to $S$ and $S^{-1}M \cong S^{-1}A\otimes_A M$ for any $A$-module $M$. Now localization is exact, in that if hence as the map from $A$ to $B$ is just inclusion, the induced map from $S^{-1}A$ to $S^{-1}B$ is again injective.

Thus in both of these examples, one has an injective ring map $\varphi\colon A \to B$, rather than an inclusion of a subring into a ring. But since $A$ is isomorphic to its image $\varphi(A)$, the generalization from the subring case is trivial.

In fact, in order to define the notion of a ring $R$ being integral over a ring $S$, one only needs $R$ to be an "$S$-algebra", that is, to have a ring map $\varphi\colon S \to R$. Given such a map, an element $a \in R$ is integral over $S$ if there is a monic polynomial $f(t) = t^d +\sum_{k=0}^{d-1} \alpha_k t^k \in S[t]$ such that $a$ satisfies $\tilde{\varphi}(f)$ where $\tilde{\varphi}$ where $\tilde{\varphi}$ is the natural extension of $\phi$ to $\tilde{\varphi}\colon S[t]\to R[t]$ which maps $t$ to itself. A ring $R$ is then integral over $S$ if every element of $R$ is integral over $S$.

krm2233
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