Contraction of (non-prime) ideals in integral extensions

Question

If $A \subset B$ is an integral extension, then any prime $p \subset A$ is the contraction of some prime of $B$ (by lying-over property). Does this hold for more general ideals? That is, given an ideal $I \subset A$, is there always an ideal $J \subset B$ such that $I = J^c$ ?

If not, do we at least have $I^{ec} = I$ for all ideals $I$ of $A$? Possibly if we assume that $A \subset B$ is of finite type (hence finite)?

Thank you!

Just a remark: if $I=J^c$ then $I=I^{ec}$. So the two claims are equivalent. — user26857, Nov 02 '18 at 06:10

Servaes · Accepted Answer · 2018-11-16T23:47:08.040

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The answer to all your questions is no; a simple counterexample is the integral extension of finite type $\Bbb{Z}[2i]\subset\Bbb{Z}[i]$ with the ideal $(2)\subset\Bbb{Z}[2i]$. Then $(2)^{ec}$ contains the element $$2\cdot i=2i,$$ but $2i$ is not contained in $(2)\subset\Bbb{Z}[2i]$. So we have $(2)^{ec}\neq(2)$. It easily follows that $(2)\subset\Bbb{Z}[2i]$ is not the contraction of any ideal of $\Bbb{Z}[i]$.

edited Nov 16 '18 at 23:47

answered Nov 01 '18 at 18:16

Servaes

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Thank you! Would you have an example where $I^e$ is prime in $B$? – Alphonse Nov 01 '18 at 19:08
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Sure; simply replace all $2$'s by $3$'s in the example; the ideal $(3)\subset\Bbb{Z}[3i]$ does the trick. – Servaes Nov 01 '18 at 21:54

Contraction of (non-prime) ideals in integral extensions

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