Poincare polynomial of product is product of Poincare polynomials

For two spaces with finitely generated homology, over the integers
Suppose $$X$$ and $$Y$$ are (possibly homeomorphic/equal) topological spaces and both have finitely generated homology over the integers, i.e., at most finitely many of the homology groups of $$X$$ are nonzero, and all these are finitely generated, and the same holds for $$Y$$. In particular, this means that the fact about::Poincare polynomials $$PX$$ and $$PY$$ are defined.

Then, the Cartesian product $$X \times Y$$, equipped with the product topology, is also a space with finitely generated homology over the integers, and its Poincare polynomial is given by:

$$P(X \times Y) = (PX)(PY)$$

where the multiplication on the right is as multiplication of polynomials in $$\mathbb{Z}[x]$$.

For finitely many spaces with finitely generated homology, over the integers
Suppose $$X_1,X_2,\dots,X_n$$ are all (possibly homeomorphic/equal) topological spaces, each of which has finitely generated homology over the integers. Then, the Cartesian product $$X_1 \times X_2 \times \dots \times X_n$$ is also a topological space with finitely generated homology, and its Poincare polynomial is the product of the Poincare polynomials of each of the $$X_i$$s, i.e.:

$$P(X_1 \times X_2 \times \dots X_n) = (PX_1)(PX_2)\dots (PX_n)$$

where the multiplication on the right is carried out as polynomials in $$\mathbb{Z}[x]$$.

Over an arbitrary commutative unital ring
Analogous statements to the above hold if we replace the ring of integers by an arbitrary commutative unital ring.

Related facts

 * Poincare series of product is product of Poincare series