$\wedge(R^{\oplus d})=0$?

Question

Let $R$ be a commutative ring. We have for all $M$, $N$ $R$-modules $\wedge_R (M\oplus N)=\wedge_R M\otimes\wedge_R N$. So what wrong with that calculus: $$ \wedge_R R^{\oplus d} =(\wedge_R R)^{\otimes d}= 0 ^{\otimes d}=0 $$ I write $\wedge_R R= 0$ because for all $a_1,\ldots,a_k\in R$ $$ a_1\wedge a_2\wedge \ldots \wedge a_k=(a_1 a_2\ldots a_k) 1\wedge 1\wedge\ldots \wedge 1=0$$ What's going wrong?

Answer 1

$\bigwedge R \neq 0$ : $R$ is a submodule of it (twice, actually, if $\bigwedge$ is the full exterior algebra, there is $\bigwedge^0 R \oplus \bigwedge^1 R$, and both of these are isomorphic to $R$)

Answer 2

The formula $$ \Lambda_R^\bullet(M\oplus N)\cong\Lambda_R^\bullet M\otimes\Lambda_R^\bullet N $$ of graded $R$-modules means $$ \Lambda_R^p(M\oplus N)\cong\bigoplus_{j=0}^p\Lambda_R^j M\otimes\Lambda_R^{p-j} N $$

Going back to your example. The graded module $\Lambda^\bullet_R R$ is $$ \Lambda_R^d R=\begin{cases} R & d=0,1\\ 0 & \text{otherwise} \end{cases} $$ and so $$ \Lambda_R^\bullet(R^{\oplus d})=\bigotimes^{d}\Lambda_R^\bullet R $$ is $$ =\begin{cases} R^{\oplus\binom{d}{k}} & \text{degree }k\in\{0,1,\dots,d\}\\ 0 & \text{otherwise} \end{cases}. $$