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Let $A$ be an integral domain which is finitely generated as a $k$-algebra and let $I\subset A$ be an ideal. Let $B$ be its integral closure (in the fraction field $\mathrm{Frac}\ A$) - in this case $B$ is finite as an $A$-module and a finitely generated $k$-algebra.

Are there any relations between the associated primes of $I$ and the associated primes of $IB$ (as an ideal in $B$)?

For example, I'd expect that the number of minimal primes is the same in both cases, but is this true for embedded primes also?

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It is not true that there is a bijection between the minimal primes in the two cases. For example, let $A = k[x,y] / (y^2 = x^2 + x^3)$. The normalization is $k[t]$, with $t = y/x$; we have $x = t^2-1$, $y=t^3-t$. If you look at the ideal $\langle x,y \rangle$, its preimage is $\langle t^2-1, t^3-t \rangle = \langle t^2-1 \rangle = \langle t-1 \rangle \cap \langle t+1 \rangle$. The conceptual way to think about this is that the curve $y^2 = x^2+x^3$ has a crunode at $(0,0)$, and the normalization separates the two branches.

It is also certainly possible to get embedded components. I just chose the first example which came to mind and it worked: Let $A = k[x,y,z]/(x^2 = y^2 z)$. This is sometimes known as the Whitney umbrella. Its normalization is $k[u,v]$, with $u = y$ and $v = x/y$; we have $(x,y,z) = (uv, u, v^2)$. Let $I$ be the prime ideal $\langle x, z \rangle$. Then $BI = \langle uv, v^2 \rangle = \langle v \rangle \cap \langle u, v^2 \rangle$.

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    $\begingroup$ Very nice! I guess I have to improve my intution about normalization and finite morphisms. A related question: Is it even possible the normalization can destroy some of the emebedded components, i.e., to have an ideal $I$ with an embedded prime such that $IB$ has no embedded primes? $\endgroup$ – Bonanza Aug 31 '11 at 20:38

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