12
$\begingroup$

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?

$\endgroup$
19
$\begingroup$

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 $$\sum a_i(j_i+k_i) = \sum\Bigl( a_ij_i + a_ik_i\Bigr) = \left( \sum a_ij_i\right) + \left(\sum a_ik_i\right) \in IJ + IK.$$

$\endgroup$
  • $\begingroup$ Makes perfect sense, thanks. $\endgroup$ – chelsea cain Jun 15 '12 at 6:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.