Suppose $V$ and $W$ are vector spaces and $\bigwedge V$ and $\bigwedge W$ their exterior algebras. Then it is known that $\bigwedge (V \oplus W) \simeq \bigwedge V \otimes \bigwedge W$. Now my question is:
Is $\bigwedge V \otimes \bigwedge V \simeq \bigwedge V$ true?
The reason why I think that: The elements of $\bigwedge (V \oplus W)$ are linear combinations of wedges like $v_1 \wedge \ldots \wedge v_n \wedge w_1 \wedge \ldots \wedge w_m$ but on $\bigwedge (V \oplus V)$ the wedges $v_1 \wedge \ldots \wedge v_n \wedge v'_1 \wedge \ldots \wedge v'_{n'}$ can be reduced (using $v_i \wedge v_i =0 $) to something like $v''_1 \wedge \ldots \wedge v''_{n''} \in \bigwedge V$
No, that is not true.
We can see this using two pieces of information:
If $V$ is of finite dimension $n$, then $\Lambda V$ has dimension $2^n$.
On the other hand, if $U$ and $W$ are vector spaces of finite dimension, we have $\dim U\otimes W=\dim U\cdot\dim W$.
If your isomorphism existed, we would then have that $2^n2^n=2^n$, which we don't for most values of $n$.