Let $A$ be a commutative ring with unit and $I \subseteq A$ an ideal of $A$. Prove that with $a,b,c \in I$ we have $$a(b\wedge c) = a \wedge (bc) = (bc) \wedge a $$
Here's what I thought: I know that $b\wedge c$ is the image of the pair $(b,c)$ under the morphism $fg$: $$ I^2 \xrightarrow{g} I^{\otimes2} \xrightarrow{f} \Lambda^2I $$ So if I take $a(b\wedge c) = a\cdot fg(b,c) = a\cdot f(b \otimes c) = f(ab \otimes c) = f(a \otimes bc ) = a \wedge bc$. This could be a possible solution? Any other method for proving that? Thank you very much.