The Künneth formula states that $$\ \Phi: H^*(M) \otimes H^*(F) \to H^*(M×F)$$ is an isomorphism, where $\ \Phi $ is the induced map of $$\ \omega \otimes \tau \to \pi ^* \omega \wedge\rho^*\tau, $$ where $\ \pi$ and $\ \rho$ are projections from $\ M× F $ to M and F respectively.
My book (differential form in algebraic topology, page 49) says that if $\ M = \mathbb{R}^n$, then the Künneth formula is just the Poincaré lemma, which says $\ H^*(M) \cong H^*(M×\mathbb{R}^n)$
I don't see how this is the case. As far as I know, $\ \dim(V \otimes W) = \dim(V) × \dim(W)$, and $\ H^*(\mathbb{R}^n)$ is zero in dimensions greater than zero. That means $\ \dim(H^*(M) \otimes H^*(F)) $ is zero when the dimension of $\ H^*(M = \mathbb{R}^n)$ is nonzero.
Does tensor products work differently on de Rham cohomologies?
Note that $H^*(M)\otimes H^*(\mathbb{R}^n) \cong H^*(M)\otimes\mathbb{R} \cong H^*(M)$. In particular, $\dim (H^*(M)\otimes H^*(\mathbb{R}^n))^k = \dim H^k(M)$ where the exponent denotes the degree $k$ part of the graded ring $H^*(M)\otimes H^*(\mathbb{R}^n)$. For $k > \dim M$ we get zero, but we don't necessarily get zero for $k \leq \dim M$.