Cohomology ring of product of even dimensional sphere with odd sphere The cohomology ring of an even dimensional sphere is $\mathbb{Z}[\alpha]/(\alpha^2)$ where the degree of $\alpha$ is the dimension of the sphere. From the Künneth formula, the cohomology ring of a product of even dimensional spheres is $\mathbb{Z}[\alpha]/(\alpha^2) \otimes \mathbb{Z}[\beta]/(\beta^2) \simeq \mathbb{Z}[\alpha,\beta]/(\alpha^2,\beta^2)$.
I want to focus on the isomorphism between the above tensor product of polynomial rings with the two variable polynomial ring.
In particular the case of an odd dimensional sphere, according to Hatcher, the cohomology ring of an odd dimensional sphere is the exterior algebra $\Lambda[\alpha]$ where the degree of $\alpha$ is odd. For an exterior algebra of more than one variable, the generators anti-commute which, from my understanding, is the real distinction between an exterior algebra and the polynomial ring $\mathbb{Z}[\alpha]/(\alpha^2)$. In the case of a one variable exterior algebra, it seems there is no difference between the polynomial ring and the exterior algebra.
If, for example, one wishes to compute the polynomial ring of, say, $S^3 \times S^2$, we have by Künneth that the polynomial ring is $\Lambda[\alpha] \otimes \mathbb{Z}[\beta]/(\beta^2)$ where the degrees of $\alpha$ and $\beta$ are 3 and 2, respectively.
My question is, can we say that the above graded tensor product is isomorphic to $\mathbb{Z}[\alpha,\beta]/(\alpha^2,\beta^2)$?
I believe we can, because the cup product multiplication satisfies $\alpha\beta = (-1)^{|\alpha||\beta|}\beta\alpha$ and, because the degree of $\beta$ is even, this means that $\alpha$ and $\beta$ commute.

Is this logic correct?

 A: Let's consider $S^{2n}\times S^{2m+1}$, where $m,n\in\Bbb Z^+$, then $H^*(S^{2n}\times S^{2m+1})\cong \Lambda_{\Bbb Z}[\alpha]\otimes \Bbb Z[\beta]/(\beta^2)$, where $|\alpha|=2m+1,|\beta|=2n$.

*

*There is a difference between $\Lambda[\alpha]$ and $\Bbb Z[\alpha]/(\alpha^2)$ (degree odd). The reason is that for the latter quotient ring to make sense, we may expect $\Bbb Z[\alpha]$ to exist, but $\alpha^2=(-1)^{(2m+1)^2}\alpha^2=-\alpha^2\implies 2\alpha^2=0$, which is impossible. It doesn't make much sense to take the quotient of something that doesn't really exist, so I think Hatcher avoids writing exterior algebra generated by one odd degree element as a truncated polynomial ring of $R[\alpha]$ unless $2=0$ is true in $R$, in which case either way is fine. But we still have $\Lambda[\alpha]\cong\Bbb Z[\alpha]/(\alpha^2)$ in any case as $\alpha^2$ gets quotient out, so I think it's more of a matter of 'notation'. When it comes to the structure (operations, etc.), the addition is the same but the multiplication in $\Lambda[\alpha]$ is just the usual multiplication of polynomial with an additional relation $\alpha^2=0$.

*It's true that $\Lambda_{\Bbb Z}[\alpha]\otimes \Bbb Z[\beta]/(\beta^2)\cong \Bbb Z[\alpha,\beta]/(x^2,y^2)$, and I think that it can be justified by your observation. Alternatively, we can define the isomorphism, say $\phi$. Let $\psi:\Lambda[\alpha]\to\Bbb Z[\alpha]/(\alpha^2)$ with $\alpha\mapsto\alpha$ be an isomorphism, then $\phi:=\psi\otimes\operatorname{id_{\Bbb Z[\beta](\beta^2)}}$, then it becomes the tensor product of two truncated polynomial rings.

*For the same reason as in 1, it makes more sense to write $H^*(S^{2n}\times S^{2m+1})\cong \Lambda_{\Bbb Z}[\alpha]\otimes \Bbb Z[\beta]/(\beta^2)$ than the other way (in Hatcher's book).

