$ I(J+L)=IJ+IL$ if $I,J,L$ are ideals of $K$ Given that $I,J,L$ are ideals of $K$, do we have $I(J+L)=IJ+IL$? 
I am confused how to do it.
 A: $J \subseteq J+L$ implies $IJ \subseteq I(J+L)$. Similarily, we get $IL \subseteq I(J+L)$. Thus, $IJ+IL \subseteq I(J+L)$. Conversely, $I(J+L)$ is generated by elements of the form $i \cdot (j+l)$ with $i \in I, j \in J, l \in L$, which we can write as $i \cdot j + i \cdot l$, and therefore belong to $IJ + IL$. This proves $I(J+L) \subseteq IJ+IL$.
Optional remark (for the more advanced readers): We can prove it even more abstractly, getting rid of elements completely, for a monoid object $K$ in an abelian $\otimes$-category: If $I,J,L$ are ideals, then $I(J+L)$ is by definition the image of the natural map $I \otimes (J+L) \to K$, which in turn is the image of the natural map $I \otimes (J \oplus L) \to K$. But $I \otimes (J \oplus L) \cong (I \otimes J) \oplus (I \otimes L)$, so that the image coincides with $IJ+IL$.
A: $\rm\, H \supseteq I(J\!+\!L)\!\!\iff\!\! H\!:\!I \supseteq J\!+\!L \!\!\iff\!\! H\!:\!I \supseteq J,L \!\!\iff\!\! H\supseteq IJ,IL \!\!\iff\!\! H\supseteq  IJ\!+\!IL.\  $ QED
Remark $\ $ We employed the universal property of ideal sums, and ideal quotients. Recall that the ideal quotient $\rm\,H\!:\!I \,=\, \{ r\in R\ |\ rI\supseteq H\},\:$ hence $\rm\:H\!:\!I \supseteq J\!\iff\! H\supseteq IJ.$
