What is the cohomology ring $H^*(S^1\times \mathbb{RP}^3, \mathbb{Z}_n)$? If $n$ is even, what is the cohomology ring $H^*(S^1\times \mathbb{RP}^3, \mathbb{Z}_n)$? We already know the cohomology rings $H^*(S^1\times\mathbb{RP}^3, \mathbb{Z}_2)$ and $H^*(S^1\times \mathbb{RP}^3, \mathbb{Z})$. Can we get $H^*(S^1\times \mathbb{RP}^3, \mathbb{Z}_n)$ from these?
 A: step one: $H^*(X\times S^1,\mathbb Z_n)=H^*(X,\mathbb Z_n)\otimes\mathbb Z[t]/(t^2)$ for every $X$. this is clear by considering the spectral sequense for cohomology of trivial fibration $X\times S^1\to X$ with coefficients in $\mathbb Z_n$.
step two: the ring $H^*(\mathbb RP^3,\mathbb Z_n)$ is isomorphic to $\mathbb Z_n[x,y,z]/(2x,2y,x^2,y^2,z^2,xz,yz,xy-\frac n2z)$ for even $\frac n2$, and to $\mathbb Z_n[x,z]/(2x,x^4,z^2,xz,x^3-\frac n2z)$ for odd $\frac n2$.
here $\deg x=1$, $\deg y=2$ and $\deg z=3$.
define the $\mathbb RP^3$ by $X$ for short. to prove the second step consider the $\smile$-product on $C^*(X,\mathbb Z)$. denote by $b_k$ the homomorphism $C^*(X,\mathbb Z)\to C^*(X,\mathbb Z_k)$ that reduces cochains modulo $k$.
we know that $H^*(X,\mathbb Z_2)=\mathbb Z_2[x]/(x^4)$. so, there is an element $\alpha\in C^1(X,\mathbb Z)$ such that classes of cocycles $b_2(\alpha)$, $b_2(\alpha^2)$ and $b_2(\alpha^3)$ are generators in $H^1(X,\mathbb Z_2)$, $H^2(X,\mathbb Z_2)$ and $H^3(X,\mathbb Z_2)$.
the class $[\frac n2 b_n(\alpha)]=:x$ is a generator of $H^1(X,\mathbb Z_n)$, and the class $[b_n(\alpha^2)]=:y$ is a generator of $H^2(X,\mathbb Z_n)$. the class $b_n(\alpha^3)=:z$ generates $H^3(X,\mathbb Z_n)$, and we can see that $xy=\frac n2 z$. 
finally, the difference between even and odd $\frac n2$ is the following: the element $x^2\in H^2$ may be presented by $(\frac n2 \alpha)^2$. since $H^2$ is generated by $y:=[\alpha^2]$ and $2y=0$, $x^2=y$ when $\frac n2$ is odd and $0$ otherwise.
