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Let $I,\, J$ and $K$ be ideals in a commutative ring $R$. Could you please give an example such that $(I\cap J)K = IK\cap JK$ is not true?

Daniel Fischer
  • 206,697
PK Sharma
  • 39

2 Answers2

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Let $R$ be $\Bbb Z[x,y]$, $I = (x), J = (y), K = (x,y)$.

Then:

$$I \cap J = (xy), \quad (I \cap J)K = (x^2y,xy^2)$$

while:

$$IK = (x^2,xy), \quad JK = (xy,y^2),\quad IK \cap JK = (xy)$$

NB. Since $I \cap J\subseteq I,J$, one always has $(I\cap J)K \subseteq IK, JK$, hence:

$$(I \cap J)K \subseteq IK\cap JK$$

Lord_Farin
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Take $R = k[x,y,z,w]$ and $I = (x,y)$, $J = (z,w)$ and $K = (x,z)$. Then $I \cap J = (xw,xz,yz,yw)$ and so $(I \cap J)K = (x^2w,x^2z,xyz,xyw,zxw,xz^2,yz^2,yzw)$. But now $IK = (x^2,xy,xz,yz)$ and $JK = (zx,xw,z^2,wz)$ and so

$$zx \in IK \cap JK.$$

However $(I \cap J)K$ only can contain degree three terms and above and so it cannot be that $(I \cap J)K = IK \cap JK$.