For an ideal $I \subset R$ with relative K-groups $K\_i(R,I)$ we have an exact sequence


$K\_2(R) \to K\_2(R,I) \to K\_2(R/I) \to K\_1(R) \to K\_1(R,I) \to K\_1(R/I)$


$\to K\_0(R) \to K\_0(R,I) \to K\_0(R/I)$


This is reminiscent of the relative homotopy exact sequence and suggests - perhaps not alone - that we are dealing with homotopy groups. Let's say $K\_i(R) = \pi\_i(X\_R)$ for some space $X\_R$. 


There also exist 'multiplication' maps $K\_0 \times K\_0 \to K\_0$, $K\_1 \times K\_0 \to K\_1$, $K\_0 \times K\_1 \to K\_1$, $K\_1 \times K\_1 \to K\_2$, $K\_2 \times K\_0 \to K\_2$, $K\_0 \times K\_2 \to K\_2$ (as given in Milnor's book), which suggests that $\pi\_{\bullet}(X\_R)$ should have a graded ring structure. 


Now, if $X\_R$ is an H-space, then its multiplication induces a ring structure on its homology, but does this relate to the homotopy as well?



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> How might one reason that $X\_R$ ought to be an H-space, given that $\pi\_{\bullet}(X\_R)$ gives a graded ring?
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> What sort of classes of 'nice' spaces naturally have a graded ring structure on their homotopy?
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Intuition as to why the K-theory of a ring should be the homotopy theory of an H-space
For based spaces $X$ and $Y$ there is a map $\pi\_i(X)\times \pi\_j(Y)\to \pi\_{i+j}(X\wedge Y)$ given by 
$$
(a:S^i\to X,b:S^j\to Y)\mapsto (a\wedge b:S^i\wedge S^j\to X\wedge Y).$$ 
This function of $a$ (resp. $b$) is a homomorphism if $i$ (resp. $j$) is positive.
Thus a space-level multiplication of the form $X\wedge X\to X$ gives a graded ring structure to $\pi\_\ast(X)$ (except maybe on $\pi\_0$ and $\pi\_1$). 


In particular this is so if $X=\Omega^\infty E$ is the infinite loopspace of a spectrum and you have a spectrum map $m: E\wedge E\to E$. In this case you have abelian groups $\pi\_i(E)$ for all $i\in \mathbb Z$, not just $i\ge 2$, and $m$ makes this a graded ring.


Note: smash product, not cartesian product.


Edit: Here is the promised elaboration. This is extremely sketchy. It would probably be pointless to try to avoid mentioning spectra here.


When you have a spectrum you have a very special sort of $H$-space, an infinite loopspace. If $X$ is a loopspace, say $X=\Omega Y$, then this gives a group structure to $\pi\_0X$ and makes the group $\pi\_1X$ abelian (and has lots of other consequences, too, for the homotopy type of $X$). If $X$ is a double loopspace, the loopspace of a loopspace, then it is an even more special kind of loopspace. $\pi\_0X$ is now abelian. In fact, the multiplication $X\times X\to X$ is now commutative up to homotopy. When you have a spectrum $E$, you have a sequence of spaces $E\_n$ in which $E\_n$ is (at least weakly homotopy equivalent to) $\Omega E\_{n+1}$. The homotopy groups of the spectrum are defined by $\pi\_kE=\pi\_{k+n}E\_n$. The space $E\_0$ is then homotopy-commutative in a very strong sense. 


You can think of infinite loopspaces as a generalization of topological abelian groups. Let's think of the 'multiplication' in such an $H$-space as addition, because we now want to think of the possibility of another binary operation called multiplication that makes this generalized abelian group into a generalized ring. A ring spectrum is a spectrum $E$ together with a map of spectra $E\wedge E\to E$. The smash product $E\wedge F$ of two spectra is somewhat awkward to define. Its homotopy groups can be described by saying that $\pi\_k(E\wedge F)$ is the direct limit of $\pi\_{k+m+n}(E\_m\wedge F\_n)$. A map $E\wedge F\to G$ gives you a map $\pi\_iE\otimes \pi\_jF\to \pi\_{i+j}G$. The homotopy groups of a ring spectrum form a graded ring. As a very special case, suppose that $A$ is a simplicial abelian group, i.e. a simplicial set with a suitable addition law. It is an infinite loopspace: you can deloop it by explicitly writing down another such object; if you replace simplicial abelian groups by nonnegatively graded chain complexes using the Dold-Kan equivalence, then it's just a matter of regrading, shifting the chain groups by one dimension. (The homology groups of the chain complex are the homotopy groups of the realization of the simplicial set.) 


As a very special case: If $A$ is not just a simplicial abelian group but a simplicial ring, then the spectrum becomes a ring spectrum. The multiplication in homotopy groups can be described either by writing down a multiplication on the chain complex level or by noting that the multiplication $A\times A\to A$ takes $A\times 0\cup 0\times A$ to $0$ and so gives a map $|A|\wedge |A|=|A\wedge A|\to |A|$. (Now look at the beginning of my answer above.) 


More generally, if $X$ is the infinite loopspace $E\_0$ of a ring spectrum $E$, then you always get a map $E\_0\wedge E\_0\to E\_0$.


The algebraic $K$-groups of a ring are in fact the homotopy groups of an infinite loopspace, or of a spectrum. You make the spectrum by using a category of $R$-modules. If $R$ is commutative then the spectrum is a ring spectrum. The multiplication is related to tensor product of $R$-modules.


