Does the integral cohomology ring determine the ring structure with other coefficients?

Let $A$ be an abelian group. By the universal coefficient theorem, the cohomology group of a manifold with coefficient $A$ is determined by the integral cohomology group. How about the ring structure?

Let $M_1$ and $M_2$ be manifolds with $H^*(M_1; \mathbb{Z}) \cong H^*(M_2; \mathbb{Z})$ as a graded ring. Is it true that $H^*(M_1; R) \cong H^*(M_2; R)$ for any commutative ring $R$?

I guess the answer is no. What would be an example?

Solution 1:

Lemma. Every finite CW complex is homotopy-equivalent to a smooth manifold .

Proof. (Rourke and Sanderson's book "Introduction to Piecewise-Linear Topology" will contain all the necessary background information.) Every finite CW complex is homotopy-equivalent to the underlying space of a finite simplicial complex. Every finite simplicial complex is a subcomplex of the face-complex of a simplex. Thus, every finite simplicial complex embeds in $R^n$. (One can actually do a bit better: Each finite $k$-dimensional simplicial complex embeds in $R^{2k+1}$, this is a version of Whitney's embedding theorem in the PL category.) Now, let $X\subset R^n$ be a finite simplicial complex. Let $N(X)$ denote the regular neighborhood of $X$ in $R^n$. (The regular neighborhood of a subcomplex $X$ in a simplicial complex $Y$ is defined as follows. Take the 2nd barycentric subdivision $Y''$ of $Y$ and let $N(X)$ denote the subcomplex of $Y''$ consisting of simplices having nonempty intersection with $X''$.) Then $N(R)$ is homotopy-equivalent to $|X|$. (This is a general fact about regular neighborhoods of subcomplexes in simplicial complexes.) The regular neighborhood is a codimension zero PL manifold with boundary in $R^n$. Hence, the interior of $N(X)$ is an open subset in $R^n$ homotopy-equivalent to $|X|$. qed

Remark. One can do a bit better: If $X$ is a $k$-dimensional countable CW complex, then $X$ is homotopy-equivalent to a smooth $2k$-dimensional manifold. But you do not need this.

In view of this lemma, since you already have finite CW complexes answering your question ($RP^3$ and $RP^2\vee S^3$), you also have manifolds answering your question.