Question about the graded structure of the cohomology ring. Hatcher defines the cohomology ring $H^*(X;R)$ for a space $X$ and commutative ring $R$ to be the direct sum of the cohomology groups $H^n(X;R)$, and by his definition of a graded ring, elements of $H^n(X;R)$ will have dimension $n$. 
In his discussion of the cross product $H^*(X;R)\otimes_R H^*(Y;R) \rightarrow H^*(X \times Y; R); a \otimes b \mapsto a \times b$ where $\otimes_R$ denotes the tensor product as $R$-modules. 
He says that this becomes a ring homomorphism if we define the multiplication in a tensor product of graded rings by $(a \otimes b)(c \otimes d)=(-1)^{|b||c|}ac \otimes bd$ where $|x|$ denotes the dimension of $x$.

My problem here is that $a$ and $b$ are general elements of their respective cohomology rings, so are finite sum $a=\sum{\alpha_{n_i}}$ where each $\alpha_{n_i}$ is in some cohomology ring $H^{n_i}(X;R)$ and similarly for $b=\sum{\beta_{n_j}}$ For example, if $a=\alpha_1 +\alpha_2$ where $\alpha_i \in H^i(X;R)$, then what is the dimension of $a$ supposed to be? 


You are right, a element $x \in H^*(X,R)$ does not have a fixed degree. But since multiplication should be bilinear, it's enough to define the multiplication for "pure" elements on the form $a \otimes b$ where $a,b$ are of fixed degree. 
