Map from exterior power to the tensor product

Question

Let $R$ be a commutative ring such that $2\in R^\times$ and let $M$ be a free $R$-module of finite rank. Let $e_i\wedge e_j$ for $i

I have a few questions. To show that it's a homomorphism, does it suffice to show linearity just on the basis elements?

My other confusion is about what the things in the source look like. If the rank of $M$ is $n$, then there are ${n\choose 2}$ pairs of indices such that $1\leq i_1

Answer 1

Your expression for an arbitrary element in $\wedge^2 M$ looks fine to me.

To show linearity, you should note that $$ \varphi \left( \sum_{1 \leq i_1 < i_2 \leq n} a_{i_1 i_2} e_{i_1} \wedge e_{i_2} \right) = \sum_{1 \leq i_1 < i_2 \leq n} a_{i_1 i_2} \varphi\left ( e_{i_1} \wedge e_{i_2} \right)$$ for every possible choice of $a_{i_1 i_2} \in R$. This isn't something you want to show; this should be a part of your definition of $\varphi$.

Now check that this implies linearity, i.e check that this implies that $$ \varphi( \omega_1 + \omega_2 ) = \varphi(\omega_1) + \varphi(\omega_2)$$ for $\omega_1, \omega_2 \in \wedge^2 M$ and $$ \varphi( \alpha \omega ) = \alpha \varphi(\omega) $$ for $\alpha \in R, \omega \in \wedge^2 M$