Mod p cohomology ring of real projective space

Question

I have looked at five algebraic topology books, but the only computation of cohomology ring of $\mathbb{RP}^\infty$ which I could found is with $\mathbb{Z}_2$-coefficient : $$ H^*(\mathbb{RP}^\infty, \mathbb{Z}_2) \cong \mathbb{Z}_2[\alpha] $$ where $\alpha$ generator of degree 1.

From this computation, using the homomorphismes $\mathbb{Z} \to \mathbb{Z}_2$ or $\mathbb{Z}_{2k} \to \mathbb{Z}_2$, looking at induced map of chain complexes, I believe we can compute cohomology with coefficient in $\mathbb{Z}$ or $\mathbb{Z}_{2k}$.

However, the question : how to compute cohomology ring of real projective space with $\mathbb{Z}_p$ coefficient? We have no such morphism of groups.

Answer 1

For odd $p$, one has $H^*(\Bbb RP^\infty; \Bbb Z_p)=0$.

Consider the cellular cochain complex of $\Bbb RP^\infty$ with $\Bbb Z_p$-coefficients:

$$\cdots \stackrel{\cdot 2}{\leftarrow} \Bbb Z_p \stackrel{0}{\leftarrow} \Bbb Z_p \stackrel{\cdot 2}{\leftarrow} \Bbb Z_p\stackrel{0}{\leftarrow} \Bbb Z_p\leftarrow 0.$$

Note that $2\Bbb Z_p = \Bbb Z_p$ and $\ker(\Bbb Z_p \stackrel{\cdot 2}{\leftarrow}\Bbb Z_p)=0$, hence all cohomology groups vanish.

Answer 2

I see that an answer has already been posted, but I am posting my answer because I wrote it up and used a different technique.

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By universal coefficient theorem, we have the following short exact sequence

$$0 \to \text{Ext}^1_{\Bbb Z}(H_{n-1}(\Bbb{RP}^\infty), \Bbb Z/p) \to H^n(\Bbb{RP}^\infty; \Bbb Z/p) \to \hom(H_n(\Bbb{RP}^\infty), \Bbb Z/p) \to 0$$

Whatever the parity of $n$ is, either the Ext term or the hom term is zero because $\Bbb {RP}^\infty$ do not have nontrivial cohomology in adjacent dimensions, and $\text{Ext}^1(\Bbb Z/2, \Bbb Z/p)$ and $\hom(\Bbb Z/2, \Bbb Z/p)$ are both trivial if $p$ is odd because the former is just $\Bbb Z_p/\text{ker}(\Bbb Z_p \stackrel{\times 2}{\to} \Bbb Z_p)$ and the latter is $0$ too since $\Bbb Z/p$ has no element of order $2$, so the generator must go to $0$.

Thus, all the cohomology groups of $\Bbb{RP}^\infty$ with mod $p$ coefficients are $0$. Hence $H^*(\Bbb{RP}^\infty; \Bbb Z/p) \cong 0$.