# When does $\mathfrak{a}B\cap A = \mathfrak{a}$?

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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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:

http://www-users.math.umd.edu/~karpuk/chap3solns.pdf

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