# Associated primes and integral closure

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?

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 $BI = \langle uv, v^2 \rangle = \langle v \rangle \cap \langle u, v^2 \rangle$.
• 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? – Bonanza Aug 31 '11 at 20:38