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Let $A\subset B$ be rings, and let $\mathfrak{a}$ be an ideal of $A$. Under what circumstances does $\mathfrak{a}B\cap A = \mathfrak{a}$? More precisely, are there conditions on $A,B$ that guarantee this for all ideals $\mathfrak{a}\triangleleft A$? If so, what are the most general such conditions?

(I had a hunch that if $A$ is integrally closed and $B$ is integral over $A$, this would be guaranteed, but have thus far been unable either to prove it or form a counterexample.)

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up vote 3 down vote accepted

Your hunch is in the right direction, but further conditions are required. Below is an excerpt from the introduction of R. Gilmer's Contracted ideals with respect to integral extensions.

Theorem $\ $Let $D$ be an integrally closed domain with quotient field $K$; let $L$ be an extension field of $K$ such hat $L/K$ is separable algebraic, and let $D’$ be the integral closure of $D$ in $L.$ Then each ideal of $D$ is the contraction of an ideal of $D’$ -- that is, if $A$ is an ideal of $D,$ there is an ideal $B$ of $D’$ such that $A = B \cap D.$

We later show (Example 2) that this conjecture is false in general. But we show that in several important cases, each ideal of $ D$ is the contraction of an ideal of $D’$. Notably, this is true in case $D$ is a Prüfer domain (Corollary 2) or if $D’$ has an integral basis over $D$ (Theorem 6).

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I know of one fairly general criterion. This is taken from Atiyah-MacDonald, Chapter 3, Exercise 16:

If $B$ is a flat $A$-algebra then the following conditions are equivalent:

i) $\mathfrak{a}^{ec}=\mathfrak{a}$ for all ideals $\mathfrak{a}$ of $A$;

ii) Spec($B)\rightarrow$Spec($A$) is surjective;

iii) For every maximal ideal $\mathfrak{m}$ of $A$ we have $\mathfrak{m}^e\neq 1$;

iv) if $M$ is an $A$-module, then $M_B\neq 0$

v) for every $A$-module $M$, the mapping $x\mapsto1\otimes x$ of $M$ into $M_B$ is injective.

If $A$ and $B$ satisfy any of these equivalent conditions then $B$ is said to be a faithfully flat $A$-algebra.

Solutions to this exercise can be found here:

As a useful condition for flatness, if $A$ is a Noetherian ring, and $B$ is finitely generated as an $A$-module then $B$ is a flat $A$-module $\Leftrightarrow$ $B_\mathfrak{m}$ is a free $A_\mathfrak{m}$-module of each maximal ideal $\mathfrak{m}$ of $A$. (This is also found in A-M, Chapter 7, exercise 16).

For example, let $A$ be a Dedekind domain, $K$ its field of fractions, $L$ a field of extension of $K$. Then if the integral closure of $A$ in $L$ is finitely generated as an $A$-module (e.g. if $L/K$ is finite, separable and algebraic) it is a Dedekind domain and a faithfully flat $A$-algebra.

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Conditions (iii) and (iv) look like the definition of a faithfully flat algebra. I knew that these implied (i), but had never thought about whether the converse was true. Interesting! [In your condition after the link, by the way, do you want $B$ to be finitely generated as a module?] – Dylan Moreland Sep 11 '12 at 19:18
yeah, will change it now. thanks – M Davolo Sep 11 '12 at 19:23

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