In numerous places it is claimed that the cohomology ring with coefficients in $R$ of an even dimensional sphere is equal to $R[\alpha]/(\alpha^2)$ where the degree of $\alpha$ is the dimension of the sphere, e.g. here.
However, in the case of $S^0$ this is not my conclusion. Let us consider $S^0=\{-1,1\}\subset \mathbb{R}$. We can generate $H_0(S^0)$ by the unique zero-simplices $a:\Delta^0\rightarrow \{1\}$ and $b:\Delta^0\rightarrow \{-1\}$. The other homology rings are zero.
By the universal coefficient theorem, we have that $H^0(S^0;R)\cong\operatorname{Hom}(H_0(S^0),R)$, so we can generate the zeroth cohomology group by the dual basis given by maps $\alpha,\beta$ defined by $$\alpha(a)=1,\alpha(b)=0$$ and $$\beta(a)=0,\beta(b)=1$$ If we compute their cup products, we get $$\alpha\smallsmile \beta=\beta\smallsmile \alpha =0, \qquad \alpha\smallsmile\alpha =\alpha, \qquad \beta\smallsmile\beta = \beta$$
Hence, this does not have a non-trivial element for which the product with itself is zero.
Is my reasoning correct and is it possible to write this as a polynomial ring?
Best Answer
Good catch!
Yes, the case of the zero dimensional sphere is a bit different. You're computation is exactly right, and it shows that $\mathrm{H}^*(\mathrm{S}^0;R) \cong R[\alpha]/(\alpha^2 - \alpha)$, and $\beta = 1-\alpha$.
By the Chinese Remainder Theorem, we can rephrase this as $\mathrm{H}^*(\mathrm{S}^0;R) \cong R \times R$, which we might have expected since $\mathrm{H}^*(X \sqcup Y; R) \cong \mathrm{H}^*(X;R) \times \mathrm{H}^*(Y;R)$.