# Poincare polynomial of product is product of Poincare polynomials

## Statement

### 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 Poincare polynomial (?)s $PX$ 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.