I have a question about an argument using the Künneth Theorem for cohomologies: Let $X, Y$ topological spaces and $X$ has finite homology groups. Then Künneth provides for every degree $n \ge 0$ we have exact sequence
(*) $$0 \to \bigoplus_{p,q: p+q=n} H^p(X;R) \otimes H^q(Y;R) \to H^n(X \times Y; R) \to \bigoplus_{p,q: p+q=n+1}Tor^1 (H^p(X;R), H^q(Y;R)) \to 0 $$
For our purposes we can assume $R = \mathbb{Z}$.
Futhermore, the cohomology ring $\bigoplus_n H^n(X; R)$ is endowed with graded multiplication, the cup-product $\cup$ induced in every grade by
(**) $$H^k(X;R) \otimes H^m(X;R) \to H^{k+m}(X \times X; R) \to H^{k+m}(X; R)$$
where the right map is induced on cohomology by diagonal map $X \to X \times X$.
Now my question: For arbitrary $n,m \ge0$ it is stated by Künneth that for complex projective spaces $\mathbb{CP}^n$ their graded rings holds
$$H^*(\mathbb{CP}^n \times \mathbb{CP}^m, \mathbb{Z}) = H^*(\mathbb{CP}^n , \mathbb{Z}) \otimes H^*(\mathbb{CP}^m , \mathbb{Z})$$
We know (using the fact that $\mathbb{CP}^n$ is compact, orientable manifold) that the graded ring structure with respect the $\cup$-product is:
$$H^*(\mathbb{CP}^n , \mathbb{Z}) = \mathbb{Z}[x]/(x^{n+1})$$
with $x \in H^2(\mathbb{CP}^n, \mathbb{Z})$. The question is: Why Künneth states that
$$H^*(\mathbb{CP}^n \times \mathbb{CP}^m, \mathbb{Z}) = \mathbb{Z}[x]/(x^{n+1}) \otimes \mathbb{Z}[y]/(y^{m+1}) = \mathbb{Z}[x,y]/(x^{n+1},y^{m+1})$$
is graded ring with respect to the $\cup$-product ?
Indeed, the formula (*) just states that (since here $Tor$-terms vanish) in each grade $k$ the $k$-th kohomology can be identified as $\mathbb{Z}$-modules via
$$H^k(\mathbb{CP}^n \times \mathbb{CP}^m, \mathbb{Z})= \bigoplus_{p,q: 2p+2q=k} \mathbb{Z} \cdot x^p y^q$$
But from here I think that Künneth tells nothing about their $\cup$-multiplication structure. But the "identification"
$$H^*(\mathbb{CP}^n \times \mathbb{CP}^m, \mathbb{Z}) = \mathbb{Z}[x,y]/(x^{n+1},y^{m+1})$$
would state that here $ax^k \cup by^l= ab x^k y^l$. Finally, that's true, but I don't see why it should follow from Künneth's (*).
I think here we need an extra argument. Could anybody explain it with Künneth provides also that for this example the $\cup$-product coinsides indeed with naivve multiplication $ax^k \cup by^l= ab x^k y^l$ in ring $\mathbb{Z}[x,y]/(x^{n+1},y^{m+1})$?
Equivalently: Why $H^*(\mathbb{CP}^n \times \mathbb{CP}^m, \mathbb{Z}) = H^*(\mathbb{CP}^n , \mathbb{Z}) \otimes H^*(\mathbb{CP}^m , \mathbb{Z})$ holds as algebras and not only $\mathbb{Z}$-modules?
There is a ring map from $H^*(X;\Bbb Z) \otimes H^*(Y;\Bbb Z)$ to $H^*(X \times Y;\Bbb Z)$, the cross product map. This is an algebra homomorphism. The Kunneth theorem (in your case) guarantees that it is an isomorphism.
This is Theorem 3.15 of Hatcher, stated for coeffients in a ring $R$ and spaces $X, Y$ with $H^*(X;R)$ and $H^*(Y;R)$ both free and finitely generated over $R$.