2 Update to reflect Tyler's point.

I was discussing with a student today the nature of the non-commutativity of cup-product at the level of cochains. In trying to explain what happens, I came up with the following statement.

Theorem. Fix a commutative ring $R$. Suppose $F:\mathrm{Top}^{\mathrm{op}}\to \mathrm{cDGA}_R$ is a contravariant functor from spaces to commutative DGAs over $R$, such that $X\mapsto H^*(F(X))$ is ordinary cohomology with coefficents in $R$. Then $R$ contains $\mathbb{Q}$ as a subring.

I'm pretty sure this "Theorem" is true. But I don't have a proof at hand.

The question is: does anyone have a proof, or know of one in the literature? I'm particularly interested in seeing a proof which is relatively "elementary", in the sense of not requiring much more heavy machinery than is needed in order to make the statement.

Added. Tyler points out in his answer that this can't be true as stated. We should add a hypothesis, such as: F takes homotopy pushouts of spaces to homotopy pullbacks of chain complexes.

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# On-the-nose commutative cup product $\Longrightarrow$ characteristic $0$?

I was discussing with a student today the nature of the non-commutativity of cup-product at the level of cochains. In trying to explain what happens, I came up with the following statement.

Theorem. Fix a commutative ring $R$. Suppose $F:\mathrm{Top}^{\mathrm{op}}\to \mathrm{cDGA}_R$ is a contravariant functor from spaces to commutative DGAs over $R$, such that $X\mapsto H^*(F(X))$ is ordinary cohomology with coefficents in $R$. Then $R$ contains $\mathbb{Q}$ as a subring.

I'm pretty sure this "Theorem" is true. But I don't have a proof at hand.

The question is: does anyone have a proof, or know of one in the literature? I'm particularly interested in seeing a proof which is relatively "elementary", in the sense of not requiring much more heavy machinery than is needed in order to make the statement.