[proofplan] We verify the three ring axioms — associativity, existence of a unit, and the bilinear (distributive) property — for the cup product $\smile: H^k(X; R) \times H^\ell(X; R) \to H^{k+\ell}(X; R)$. Associativity and distributivity are verified at the cochain level and then inherited by cohomology via the [Cup Product Descends to Cohomology](/theorems/2270) theorem. The unit is the class $[1] \in H^0(X; R)$ of the cocycle $1 \in C^0(X; R)$ that sends every $0$-simplex to $1_R$. [/proofplan] [step:Verify associativity at the cochain level] Let $\phi \in C^k(X; R)$, $\psi \in C^\ell(X; R)$, and $\chi \in C^m(X; R)$. For a singular $(k+\ell+m)$-simplex $\sigma: \Delta^{k+\ell+m} \to X$ with vertices $v_0, \ldots, v_{k+\ell+m}$, we compute both bracketings. **Left bracketing:** $(\phi \smile \psi) \smile \chi$ first evaluates $\phi \smile \psi$ on the front $(k+\ell)$-face of $\sigma$ and $\chi$ on the back $m$-face: \begin{align*} ((\phi \smile \psi) \smile \chi)(\sigma) &= (\phi \smile \psi)(\sigma|_{[v_0, \ldots, v_{k+\ell}]}) \cdot \chi(\sigma|_{[v_{k+\ell}, \ldots, v_{k+\ell+m}]}) \\ &= \phi(\sigma|_{[v_0, \ldots, v_k]}) \cdot \psi(\sigma|_{[v_k, \ldots, v_{k+\ell}]}) \cdot \chi(\sigma|_{[v_{k+\ell}, \ldots, v_{k+\ell+m}]}). \end{align*} **Right bracketing:** $\phi \smile (\psi \smile \chi)$ evaluates $\phi$ on the front $k$-face and $\psi \smile \chi$ on the back $(\ell+m)$-face: \begin{align*} (\phi \smile (\psi \smile \chi))(\sigma) &= \phi(\sigma|_{[v_0, \ldots, v_k]}) \cdot (\psi \smile \chi)(\sigma|_{[v_k, \ldots, v_{k+\ell+m}]}) \\ &= \phi(\sigma|_{[v_0, \ldots, v_k]}) \cdot \psi(\sigma|_{[v_k, \ldots, v_{k+\ell}]}) \cdot \chi(\sigma|_{[v_{k+\ell}, \ldots, v_{k+\ell+m}]}). \end{align*} Both expressions are the product of the same three elements of $R$. Since $R$ is associative (and indeed commutative), the two products are equal. Therefore $(\phi \smile \psi) \smile \chi = \phi \smile (\psi \smile \chi)$ as cochains. [guided] The key observation is that both bracketings of the triple cup product decompose a $(k+\ell+m)$-simplex into the same three pieces: the front $k$-face $[v_0, \ldots, v_k]$ probed by $\phi$, the middle $\ell$-face $[v_k, \ldots, v_{k+\ell}]$ probed by $\psi$, and the back $m$-face $[v_{k+\ell}, \ldots, v_{k+\ell+m}]$ probed by $\chi$. The two bracketings differ only in the order of evaluation (first combine $\phi$ with $\psi$ then multiply by $\chi$, versus first combine $\psi$ with $\chi$ then multiply by $\phi$), which yields the same result by associativity of multiplication in $R$. This is entirely analogous to how the product of three numbers $(ab)c = a(bc)$ is independent of the grouping, and the proof is just as straightforward once one writes out the definition. [/guided] [/step] [step:Verify the unit element] Define $1 \in C^0(X; R)$ by $1(\sigma) = 1_R$ for every singular $0$-simplex $\sigma: \Delta^0 \to X$. We verify that $1$ is a cocycle and that it acts as a two-sided identity for $\smile$. **$1$ is a cocycle:** For any singular $1$-simplex $\tau: \Delta^1 \to X$ with vertices $w_0, w_1$, \begin{align*} (d \cdot 1)(\tau) = 1(d\tau) = 1(\tau|_{[w_1]}) - 1(\tau|_{[w_0]}) = 1_R - 1_R = 0. \end{align*} So $d(1) = 0$ and $[1] \in H^0(X; R)$. **Left identity:** For $\phi \in C^k(X; R)$ and $\sigma: \Delta^k \to X$ with vertices $v_0, \ldots, v_k$, \begin{align*} (1 \smile \phi)(\sigma) = 1(\sigma|_{[v_0]}) \cdot \phi(\sigma|_{[v_0, \ldots, v_k]}) = 1_R \cdot \phi(\sigma) = \phi(\sigma). \end{align*} **Right identity:** For $\phi \in C^k(X; R)$ and $\sigma: \Delta^k \to X$, \begin{align*} (\phi \smile 1)(\sigma) = \phi(\sigma|_{[v_0, \ldots, v_k]}) \cdot 1(\sigma|_{[v_k]}) = \phi(\sigma) \cdot 1_R = \phi(\sigma). \end{align*} Therefore $1 \smile \phi = \phi = \phi \smile 1$ for all $\phi$, and $[1]$ is the multiplicative identity in $H^*(X; R)$. [/step] [step:Verify distributivity (bilinearity)] The cup product is $R$-bilinear at the cochain level. For $\phi_1, \phi_2 \in C^k(X; R)$, $\psi \in C^\ell(X; R)$, and $r \in R$: \begin{align*} ((\phi_1 + \phi_2) \smile \psi)(\sigma) &= (\phi_1 + \phi_2)(\sigma|_{[v_0, \ldots, v_k]}) \cdot \psi(\sigma|_{[v_k, \ldots, v_{k+\ell}]}) \\ &= \phi_1(\sigma|_{[v_0, \ldots, v_k]}) \cdot \psi(\sigma|_{[v_k, \ldots, v_{k+\ell}]}) + \phi_2(\sigma|_{[v_0, \ldots, v_k]}) \cdot \psi(\sigma|_{[v_k, \ldots, v_{k+\ell}]}) \\ &= (\phi_1 \smile \psi + \phi_2 \smile \psi)(\sigma), \end{align*} using the distributive law in $R$. Similarly, $(r\phi_1) \smile \psi = r(\phi_1 \smile \psi)$ and $\phi_1 \smile (r\psi) = r(\phi_1 \smile \psi)$ by the commutativity of $R$. The same identities hold in the second variable. Since these identities hold at the cochain level, they descend to cohomology: the quotient map $Z^n(X; R) \to H^n(X; R)$ is an $R$-module homomorphism, so the induced cup product on cohomology inherits bilinearity. [/step] [step:Conclude that $H^*(X; R)$ is a graded ring] Associativity, the unit, and bilinearity together make $H^*(X; R) = \bigoplus_{n \geq 0} H^n(X; R)$ into a graded $R$-algebra. The grading is respected: the cup product sends $H^k(X; R) \times H^\ell(X; R)$ into $H^{k+\ell}(X; R)$, which is immediate from the definition (the cup product of a $k$-cochain and an $\ell$-cochain is a $(k+\ell)$-cochain). The unit $[1] \in H^0(X; R)$ sits in degree $0$. All three ring axioms were verified at the cochain level and inherited by cohomology via the [Cup Product Descends to Cohomology](/theorems/2270) theorem. [/step]