I am attempting to verify that the product variety $G \times G'$ of algebraic groups with the direct product group structure is an algebraic group, though I'm running into trouble. In particular, I'm having trouble showing that $$\mu: (G\times G') \times (G\times G') \rightarrow G\times G'$$ is a morphism of ringed spaces.
The definition of morphism that I am using is that $\mu$ is a morphism if for every open $V \subseteq G \times G'$, $$\mu\big\vert_{\mu^{-1}V}: f \mapsto f\mu$$ maps $\mathcal{O}_{G\times G'}(V)$ into $\mathcal{O}_{(G \times G') \times (G \times G')}(\mu^{-1}V)$ (where $\mathcal{O}$ is the sheaf of regular functions).
EDIT: I'm sure there is an easier way to do this using the universal property of $G \times G'$ as a direct product of varieties, but I'm still trying to do it without category theory to force myself to develop a better understanding of the concepts.
So far what I've done is show that if you have a regular function $f$ on $G \times G'$, then you can get a regular function on $G \times G$. Specifically, I've shown that if $f$ is regular at $(x_{1}x_{2}, x_{1}'x_{2}')$, there is an open neighborhood $B_{G}$ of $(x_{1}, x_{2})$ in $G \times G$ so that $$f\Big(\mu_{G}(b_{1}, b_{2}), x_{1}'x_{2}'\Big)$$ is a rational function for all $(b_{1}, b_{2}) \in B_{G}$.
The problem is that I don't see any way to extend this open neighborhood to one of $(x_{1}, x_{2}, x_{1}', x_{2}') \in (G \times G) \times (G' \times G')$ on which $f$ will be a rational function.
I assume I should use that $$\mu_{G}: G \times G \rightarrow G,\qquad \mu_{G'}: G' \times G' \rightarrow G'$$ are morphisms of ringed spaces, but I can't see how to separate the two. It seems like it might be alright in the case that $G, G'$ are affine and $V = G \times G'$, since in this case $$\mathcal{O}(G \times G') \approx k[G \times G'] \approx k[G] \otimes k[G']$$ and I can break a regular function $f$ into $\sum f_{i} \otimes g_{i}$. But otherwise, I don't see how to get something to apply $\mu_{G}$ to and something to apply $\mu_{G'}$ to.
I would like to mention that I am reading Springer's book on algebraic groups and he does not use schemes when he develops algebraic geometry.
Note that $$(G\times G')\times (G\times G')\cong (G\times G)\times (G'\times G')$$ as algebraic varieties.