17

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?

chelsea cain
  • 173

2 Answers2

30

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

Arturo Magidin
  • 420,020
0

An arbitrary element of $IJ +IK$ is of the form $$\sum_{i} x_i y_i + \sum_j w_j y_j',$$ with $x_i, w_j \in I$, $y_i \in J$ and $y_j' \in K$. Simply write $y_i = y_i + 0_R$ and $y_j' = 0_R + y_j'.$ Since $0_R \in K$, we have $y_i \in J + K$; since $0_R \in J$, we also have $y_j' \in J + K$.

hdecristo
  • 1,277