The rational cohomology ring of complex projective plane $\mathbb{CP}^{2}$ is truncated polynomial ring $\frac{\mathbb{Q}[X]}{(X)^{3}},\,\,deg(X)=2$. In this case, the degree of a generator is 2. Is there any closed oriented 2m-manifold with the truncated polynomial ring over a single odd degree generator and has exactly three Betti numbers. For example 4m+2-dimensional closed oriented manifold with non-zero Betti numbers are $\beta_{0}=1,\,\beta_{2m+1}=1,\,\beta_{4m+2}=1.$
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Such a manifold does not exist. Suppose that $M$ is $4n+2$ dimensional closed orientable manifold. Then the intersection form on $H^{2n+1}(M,\mathbb{R})$ is symplectic*, hence this vector space has even dimension implying that the Betti number $b_{2n+1}(M) = \dim(H^{2n+1}(M,\mathbb{R}))$ is even.
Maybe such examples can exist if we drop the assumption of orientability.
*The relevant facts to prove this are in Hatcher's book, or if you prefer here the fact that it is skew-symmetric is stated in the first paragraph, that it is non-degenerate is proven in Proposition 1.2, hence it is symplectic.
