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I know that if $R$ is a domain, the following conditions are equivalent:

  • $R$ is a Prüfer Domain,
  • If $I,J,K \subset R$ are nonzero ideals, $I + (J \cap K) = (I \cap J) + (I \cap K)$
  • If $I,J,K \subset R$ are nonzero ideals, $I \cap (J + K) = (I + J) \cap (I + K)$

It is stated that $k[x,y]$ is not a Prüfer Domain where $k$ is a field in When do (multivariate) polynomial rings fail to be Prüfer rings?. However, I can not find any three nonzero ideals $I,J,K \subset k[x,y]$ such that $I + (J \cap K) \neq (I \cap J) + (I \cap K)$ or $I \cap (J + K) \neq (I + J) \cap (I + K)$.

Any ideas would be appreciated.

user26857
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Philip Johnson
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2 Answers2

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Let $$I=\langle x^2,y^2, xy \rangle $$ $$J=\langle x^2 \rangle$$ $$K=\langle y^2 \rangle$$ Then $$I+(J \cap K) = \langle x^2,y^2, xy \rangle + \langle x^2y^2 \rangle=\langle x^2,y^2, xy \rangle=I$$ $$I \cap J = J \qquad \qquad \qquad I \cap K = K$$ $$(I \cap J) + (I \cap K)= J+K=\langle x^2, y^2 \rangle \neq I$$

Indeed, using this argument you can prove a more general result:

In any Prufer domain you have the equality of ideals:$$\langle a, b \rangle^2 = \langle a^2, b^2 \rangle$$

Since in $k[x,y]$ this does not hold ($\langle x, y \rangle^2 = \langle x^2, y^2, xy \rangle$ ) this is not a Prufer domain.

Crostul
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  • As I remark here, a well known characterization is: $ $ a domain $D$ is Prufer $\iff D$ is integrally closed and $,ab\in (a^2,b^2),$ for all $,a,b\in D.,$ The inclusion hypothesis can be replaced by $,a^{n-1}b\in (a^n,b^n),,$ or $,(a,b)^n = (a^n,b^n),$ [Freshman's Dream], see statements $(8),(9)$ there. – Bill Dubuque Aug 19 '21 at 11:02
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A simple counterexample is the following: $I=(x^2)$, $J=(x)$, and $K=(y)$. Then $$I+(J\cap K)=(x^2)+(x)\cap(y)=(x^2,xy)$$ while $$(I\cap J)+(I\cap K)=(x^2)\cap(x)+(x^2)\cap(y)=(x^2,x^2y)=(x^2).$$

user26857
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