I want to prove that the graded homology ring with $\mathbb{Z}$ coefficients of $S^1 \times S^1$ is $\mathbb{Z}[\alpha, \beta]/(\alpha^2, \beta^2)$ with $\alpha$ and $\beta$ in degree 1.
My attempt:
Let's consinder the projections $\pi_1, \pi_2: S^1 \times S^1 \to S^1$. They induce ring homomorphisms $$\pi_1^*,\pi_2^*:H^*(S^1;\mathbb{Z}) \cong \mathbb{Z}[\alpha]/(\alpha^2) \to H^*(S^1 \times S^1; \mathbb{Z}).$$ Analogously, the inclusions $\iota_1, \iota_2: S^1 \to S^1 \times S^1$ induce ring homomorphisms $$\iota_1^*,\iota_2^*: H^*(S^1 \times S^1; \mathbb{Z})\to H^*(S^1;\mathbb{Z}) \cong \mathbb{Z}[\alpha]/(\alpha^2).$$
As $\pi_i \circ \iota_i = Id$, we have that $\iota_i^* \circ \pi_i^* = Id$. So $$\langle \iota_1^*, \iota_2^* \rangle: H^*(S^1 \times S^1; \mathbb{Z}) \to H^*(S^1; \mathbb{Z}) \otimes H^*(S^1; \mathbb{Z}) \cong \mathbb{Z}[\alpha]/(\alpha^2) \otimes \mathbb{Z}[\beta]/(\beta^2)$$ has a section, which means that it is surjective. Maybe with Kunneth we can show that it must be injective as well. Then, I need to prove that
$$\mathbb{Z}[\alpha]/(\alpha^2) \otimes \mathbb{Z}[\beta]/(\beta^2) \cong \mathbb{Z}[\alpha, \beta](\alpha^2, \beta^2)$$
Any hint?
Best Answer
By Kunneth theorem $H^*(S^1 \times S^1) \cong H^*(S^1) \otimes H^*(S^1) \cong \mathbb{Z}[\alpha]/\alpha^2 \otimes \mathbb{Z}[\beta]/\beta^2$.
The tensor product is to be intended as tensor product of graded $\mathbb{Z}$-algebras.
Note that $\mathbb{Z}[\alpha]/\alpha^2 \cong \mathbb{Z} \cdot \{1, \alpha\}$ and $\mathbb{Z}[\beta]/\beta^2 \cong \mathbb{Z} \cdot \{1, \beta\}$ as graded $\mathbb{Z}$-modules (with gradings 0, 1), hence $\mathbb{Z}[\alpha]/\alpha^2 \otimes \mathbb{Z}[\beta]/\beta^2 \cong \mathbb{Z} \cdot \{1, \alpha \otimes 1, 1 \otimes \beta, \alpha \otimes \beta\}$ as graded $\mathbb{Z}$-modules (with gradings 0, 1, 1, 2).
At the same time it holds that $\mathbb{Z}[\alpha\, \beta]/(\alpha^2, \beta^2) \cong \mathbb{Z} \cdot \{1, \alpha, \beta, \alpha\beta\}$ as graded $\mathbb{Z}$-modules (with gradings 0, 1, 1, 2).
Hence they are isomorphic as graded $\mathbb{Z}$-modules, but the isomorphism is not a isomorphism of graded $\mathbb{Z}$-algebras.
In fact $(1 \otimes \beta)\cdot(\alpha\otimes 1) = (-1)^{|\beta||\alpha|} \alpha \otimes \beta = - \alpha \otimes \beta \neq \alpha \otimes \beta$ while $\beta \cdot \alpha = \beta\alpha = \alpha\beta$.
In general, there could not be an isomorphism of graded $\mathbb{Z}$-algebras between $\mathbb{Z}[\alpha]/\alpha^2 \otimes \mathbb{Z}[\beta]/\beta^2$ and $\mathbb{Z}[\alpha\, \beta]/(\alpha^2, \beta^2)$ because the first one is graded-commutative with terms of odd degrees, while the second one is commutative.