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I'm curious if there is some sort of distributive law for ideals.

If $I,J,K$ are ideals in an arbitrary ring, does $I(J+K)=IJ+IK$?

The containment "$\subset$" is pretty clear I think. But the opposite ontainment doesn't feel like it should work. I couldn't work out a counterexample with ideals in $\mathbb{Z}$ however. So does such an equality always hold or not?

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up vote 7 down vote accepted

Note that if $A$, $B$, and $C$ are ideals, and $B\subseteq C$, then $AB\subseteq AC$; and if $A$ and $B$ are both contained in $C$, then $A+B\subseteq C$.

Since $J\subseteq J+K$, then $IJ\subseteq I(J+K)$. Since $K\subseteq J+K$, then $IK\subseteq I(J+K)$. Therefore, $IJ$ and $IK$ are both contained in $I(J+K)$, so $IJ+IK\subseteq I(J+K)$.

For the converse inclusion, a general element of $I(J+K)$ is of the form $$\sum a_i(j_i+k_i)$$ with $a_i\in I$, $j_i\in J$, and $k_i\in K$. And we have $$\sum a_i(j_i+k_i) = \sum\Bigl( a_ij_i + a_ik_i\Bigr) = \left( \sum a_ij_i\right) + \left(\sum a_ik_i\right) \in IJ + IK.$$

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Makes perfect sense, thanks. –  chelsea cain Jun 15 '12 at 6:21

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