Let $M$ be a closed, orientable, connected manifold of dimension $3$, such that $\pi_1(M)=\mathbb Z$. Find its cohomology ring $H^*(M;\mathbb Z)$.
Clearly $H^0=H^3=\mathbb Z$. Now since connected implies path connected on a manifold and we know $H_1=\pi_1/[\pi_1,\pi_1]=\mathbb Z$, and by Poincaré duality we have $H^2=\mathbb Z$. Since there is no torsion in $H_0$, then $H^1(M)=\hom(H_1,\mathbb Z)=\mathbb Z$. So the cohomology ring has one generator in each positive degree, let's call them $\alpha\in H^1,\beta\in H^2,\gamma\in H^3$. The only nontrivial relations to find are $\alpha\cup\beta$ and $\alpha\cup\alpha$.
By singularity of the cup product pairing $H^1\times H^2\to \mathbb Z$ with $(\phi,\psi)\mapsto(\phi\cup\psi)[M]$, and $[M]$ the fundamental class, we have that $\alpha$ is a generator if and only if $\alpha\cup n\beta=\gamma$ for some integer $n$. If it were true that $\alpha\cup \alpha=\beta$, then we could conclude that $n=\pm1$, but I don't know how to show (or not show) the latter.