Cup length over a coefficient ring

Let $X$ be nonempty and $R$ a commutative unital ring. In positive degrees reduced and ordinary cohomology coincide; use the multiplication of Singular cohomology ring. Define a set of nonnegative integers by $L_R(X)=\{0\}\cup \{k\ge 1:\text{there exist }{a}_j\in H^{d_j}(X;R),d_j>0(1\le j\le k),a_1⌣\cdots ⌣a_k\mathrm{\ne }0\}\mathrm{.}$ The cup length over $R$ is ${\mathrm{cl}}_R(X)=\sup L_R(X)$ in $\mathbb{N}\cup \{\mathrm{\infty }\}$. More explicitly it is $\mathrm{\infty }$ if the set is unbounded, and otherwise its largest member, which exists for a bounded nonempty subset of $\mathbb{N}$. If there are no nonzero products of positive length, it is $0$. The inserted $0$ is a convention and does not assert that an empty product is nonzero over the zero ring.

Associativity makes the unparenthesized finite product well-defined. By Singular cohomology is graded commutative, permuting homogeneous factors changes it by a unit sign, hence does not change whether it is zero. Positive degrees exclude padding a product with degree-zero units. If a product of length $k$ is nonzero, every initial subproduct is nonzero, since multiplying a zero subproduct by the remaining factors would give zero. Thus $L_R(X)$ is downward closed.

Cup length is a homotopy invariant: Cup product is natural, unital and associative supplies graded ring pullbacks and their homotopy invariance. A homotopy equivalence and its inverse induce inverse graded ring maps, which carry nonzero products of positive-degree classes to nonzero products of the same length in both directions. Hence their sets $L_R$ agree.

A zero cohomology ring, in particular $R=0$, has cup length zero. Length one asks only for a nonzero positive-degree class. The value $\mathrm{\infty }$ means that each finite bound is exceeded; it does not assert the existence of one infinite sequence with all products nonzero. Degrees and factors are chosen only for a particular finite witness, so no AC is used. No cup length for empty $X$ is assigned by this definition.