For our homework sheet it says: calculate the de Rham cohomology of $M = S^1 \times S^2$ or check a book if you are unsure. And then it continues. Since I didn't find the calculation in a book I wanted to ask here.
It is clear that $H_{dR}^0(M) = \mathbb{R}$, as $M$ is connected. And since $M$ has dimension $3$, that $H_{dR}^k(M) = 0$ if $k >3$.
I will be answering this question using only the Mayer Vietoris sequence. We can cover $S^1\times S^2$ with two open sets $U$ and $V$ which are homeomorphic to $(0,1)\times S^2$ which are both homotopic to $S^2$ and thus will have the same cohomology. We also have that $U\cap V$ will be homotopic to the disjoint union of two spheres. We also use a property of disjoint unions which is that for any set of manifolds $M_k$ we have that the DeRham cohomology of the disjoint union is given by \begin{equation*} H^i(\;\bigsqcup_{k\in I} M_k)\cong\bigoplus_{k\in I}H^i(M_k) \end{equation*} In our case we simply have that for all $i\in \mathbb{N}$ $H^i(S^2\sqcup S^2)\cong H^i(S^2)\oplus H^i(S^2)$. The sequence is non trivial in two segments
\begin{equation*} 0\rightarrow H^0(S^1\times S^2)\rightarrow H^0(S^2)\oplus H^0(S^2)\rightarrow H^0(S^2\sqcup S^2)\rightarrow H^1(S^1\times S^2) \rightarrow 0 \end{equation*} and \begin{equation*} 0\rightarrow H^2(S^1\times S^2)\rightarrow H^2(S^2)\oplus H^2(S^2)\rightarrow H^2(S^2\sqcup S^2)\rightarrow H^3(S^1\times S^2) \rightarrow 0 \end{equation*} By linear algebra reasons gives us that $H^0(S^1\times S^2)\cong H^1(S^1\times S^2)$ and $H^2(S^1\times S^2)\cong H^3(S^1\times S^2)$. But we have that since $S^1\times S^2$ is connected $H^0(S^1\times S^2)\cong \mathbb{R}$ By the fact that $S^1\times S^2$ is compact and orientable you can use Poincare duality and I get that $H^3(S^1\times S^2)\cong H^0(S^1\times S^2)$. Which gives us that the first four groups are of dimension $1$ and the rest are trivial.
If you have not seen Poincare duality you can notice that since it is compact and orientable it has a volume form $H^3(S^1\times S^2)$ has dimension at least $1$ but it is isomorphic to $H^2(S^1\times S^2)$ which injects into $H^2(S^2)\oplus H^2(S^2)$ which has dimension $2$. So the for $2$ and $3$ the cohomology groups will have both dimension $1$ or both dimension $2$. If you study the inclusion maps you might be able to improve this result without using the poincare lemma.
Another way is to see that you can decompose $S^1\times S^2$ into two opens sets $U$ and $V$ which are homeomorphic to $D^2\times S^1$ and so they will be homotopic to $S^1$ and their intersection will be homotopic to $S^1\times S^1$. Using the fact that $H^1(S^1\times S^1)\cong \mathbb{R}^2$ we have the following sequence \begin{equation*} 0\rightarrow H^1(S^1\times S^2)\rightarrow H^1(S^1)\oplus H^1(S^1)\rightarrow H^1(S^1\times S^1)\rightarrow H^2(S^1\times S^2)\rightarrow 0 \end{equation*} This is because the boundary map $\delta:H^0(U\cap V)\rightarrow H^1(S^1\times S^2)$ will have to be the zero map. Since $H^1(S^1)\oplus H^1(S^1)\cong H^1(S^1\times S^1)$ and we get that $H^1(S^1\times S^3)$ will have to have the same dimension as $H^2(S^1\times S^2)$. By using this you avoid having to use Poincare duality
If needed I am willing to clarify some steps which are not obvious