Some practical questions on cohomology and the ring $\mathbf{Z}[x]/(x^2)$

Question

So I know that the cohomology ring of $S^n$ is $\mathbf{Z}[x]/(x^2)$ with "$x$ in degree $n$"; if we ignore the grading then this ring fails to distinguish the spheres. What is actually meant by "in degree $n$"? Isn't it the case that $\mathbf{Z}[x]/(x^2)$ somehow has a standard grading, what is going on here, do we shift the grading or what? Perhaps $\mathbf{Z}[x]/(x^2)[+n]$? (twist)

Does anyone have an example of spaces $X$ and $Y$ whose cohomology rings are isomorphic (in the graded sense) but which are not homotopy equivalent?

Now forget about the grading and consider the ring $\mathbf{Z}[x]/(x^2)$ purely algebraically. I would like to fit its spectrum into $\mathrm{Spec} \ \mathbf{Z}[x]$, in which it is homeomorphic to $V(x^2)$ (should be a curve innit?). Is there a way to tell which prime ideals (using the standard classification) of $\mathbf{Z}[x]$ contain $x^2$?

Answer 1

As an abelian group, $\mathbf{Z}[x] / x^2 \cong \mathbf{Z} \oplus x \mathbf{Z}$. The first summand has to be in the $0$-graded part, but there is no restriction on the possibilities for the $x\mathbf{Z}$ summand, except that it too has to be contained in some graded part.

For the third question, $\mathrm{Spec} \mathbf{Z}[x]/(x^2)$ is homeomorphic to $\mathrm{Spec} \mathbf{Z}$: as $x^2 = x \cdot x$, any prime ideal of $\mathbf{Z}[x]$ containing $x^2$ must contain $x$ or $x$. In fact, the natural map $\mathrm{Spec} \mathbf{Z}[x]/(x^2) \to \mathrm{Spec} \mathbf{Z}[x]$ factors through $\mathrm{Spec} \mathbf{Z}[x]/(x)$

Answer 2

I'll only bite on the first part. Given a graded ring $$R = ... \oplus R_{-1} \oplus R_0 \oplus R_1 \oplus ... $$ We can fix any non-zero integer $n$ and create an isomorphic graded ring $S$ with $S_{nk} \cong R_{k}$ and all other graded components zero. Now, of course the polynomial ring has an "obvious" grading. But there is a much more canonical grading induced by the dimension of the cohomology structure.

Answer 3

For your second question: there exist connected acyclic spaces (i.e. with vanishing cohomology in dimension $>0$) which are not contractible (i.e. not homotopy equivalent to a point). For instance, the Poincaré sphere with a point deleted from it is acyclic, but it is not contractible because it has nonvanishing $\pi_1$. (The $\pi_1$ of the non-punctured Poincaré sphere is the binary isocahedral group of order 120, whose abelianization is trivial. Therefore, $H^1$ vanishes, and since the Poincaré sphere is a $3$-fold, Poincaré duality implies that $H^2$ also vanishes, and thus the cohomology is concentrated in degree 0 and in the top degree $3$. Then, deleting a point kills the top cohomology, while not affecting $\pi_1$.)