Affine scheme as a cogroup Let $k$ be a field and $A$ be a $k$-algebra. If $A$ has a Hopf algebra structure, then the affine scheme $X = \mathrm{Spec}(A)$ became an affine group scheme and the group operation $X\times X\to X$ corresponds to the comultiplication $\Delta:A\to A\otimes_{k} A$. There are a lot of examples of affine group scheme.
However, any (associative) $k$-algebra has a multiplcation $A\otimes_{k}A\to A$ which induces a map $X\to X\times X$. This may induces a cogroup structure on $X$ and we may call $X$ as affine cogroup scheme. (In this sense, any affine group scheme is an affine cogroup scheme.) Is there any interesting theory in this direction? Actually, I didn't see cogroup much while I'm studying math, so I hope that there might be some interesting theory for affine cogroup schemes. 
 A: Every object in a category with finite products is canonically a comonoid with comultiplication induced by the diagonal map $\Delta : X \to X \times X$. This applies in particular to the category of affine schemes, where (check this!) the diagonal map is dual to multiplication, hence is precisely the map you describe. 
This comonoid structure is not necessarily a cogroup structure. There's not a whole lot to say about it. 
A: The multiplication $A\otimes_k A\to A$ always gives a comonoid structure on $X=\operatorname{Spec}(A)$.  As Qiaochu Yuan said, this is nothing special, though: in any category with finite products, every object has a unique comonoid structure (with respect to the categorical product) given by the diagonal map $X\to X\times X$ (and the unique map from $X$ to the terminal object as the counit).
Getting a cogroup structure, though, requires a lot more.  Notice in particular that to say that a map $i:X\to X$ is the "coinverse map", you need to say that a certain composition $$X \to X\times X\stackrel{1\times i}\to X\times X\to X$$ is the "trivial" map.  (This is dual to saying that the map $G\to G$ which takes an element and multiplies it by its inverse always outputs the identity element, so by "trivial map" I mean the map that plays the dual role to the map $G\to G$ sending every element to the identity.) What are the first and last maps in this diagram?  Well, the first one is just our comultiplication $X\to X\times X$.  But we don't have any canonical choice for the last map!  So actually, to define a cogroup structure, you need not just a comultiplication but also a multiplication!  And moreover, to define what the "trivial map" means, it turns out that you also need a unit in addition to your counit.  (You might wonder why we don't need a comultiplication and counit to define a group structure.  The answer is that we actually do, but we can just use the diagonal map $X\to X\times X$, at least as long as our group structure is with respect to the categorical product.  Similarly, the counit is just the unique map $X\to 1$ to the terminal object.)
As a result, the notion of a "cogroup object" (in a monoidal category) is actually exactly the same as the notion of a "group object".  So, "cogroup schemes" in your sense are exactly the same thing as group schemes.
Now, you do sometimes see people talking about cogroup objects.  Why would they do that, if cogroup objects are the same as group objects?  The reason is that to talk about group objects, you need to pick a monoidal structure on your category.  Normally when we talk about group objects, we by default just use the categorical product (and as mentioned before, this is why we don't have to worry about having a comultiplication).  So dually, when we talk about cogroup objects, that by default means we're talking about the coproduct.  So a cogroup object normally has a comultiplication of the form $X\to X\coprod X$ rather than $X\to X\times X$.  (As mentioned above, it also needs a multiplication $X\coprod X\to X$, but that is canonically provided by the codiagonal map.)
