Let S be another
commutative ring which might be considered as an R-algebra via
some ring homomorphism 
. In this situation there is
a functor
leading to a map 
of monoids,
given by 
iff 
. In the second
step this induces a ring homomorphism
carrying the ideal 
into the ideal 
, since tensoring
short split exact sequences yields again short split exact sequences.
Thus we end up with a ring homomorphism
In the case where S is a field, this is a ring homomorphism to the
integers given on a residue class [M] in 
for some 
by
For a principal ideal domain you therefore have 
and
Turning to the algebra A we write 
for the
base extended algebra. As above, there is a functor
leading to a map 
.
This induces a 
-module homomorphism
and since 
finally
a -module homomorphism
Concerning the action of 
on 
one obviously has
for all 
and 
which leads to
If we interpret 
as an -algebra via 
,
we can restate this as
Now, on the other hand there are functors
carrying an S-module resp. 
-module M to the same abelian
group, but inflating the action of S resp. to an action
of R resp. A on M. Going again through the above procedure,
one obtains maps of abelian groups
If the map 
is surjective, then you have
and
for all S-modules U and -modules
M. Therefore in this case, one obtains
In the case where is surjektiv, there corresponds an
element in 
to S, i.e. the isomorphism class of the R-module
S. This leads to a corresponding element [S] in .
Denote the left multiplication by [S] in 
by 
.
PROOF:The first two statements are clear from definitions and what has
been said before (note that [S] is an idempotent and therefore
a projection map of -modules).
For the third one, it is enough to show that
defines an inverse
as a map of -modules. Now 
is directly clear from the above righthand
formula, whereas 
holds
by surjectivity of following from the lefthand
formula.
The set 
contains the set of fields F which
are epimorphic images of R and which we denote by 
.
These are precisely the elements of
corresponding to simple R-modules F=R/I, one for each
maximal ideal I of R. For any 
the -module
is the Grothendieck group of 
as mentioned
before. Therefore the proposition shows, that all such Grothendieck
groups are embedded in as direct summands and that a projection
on them is given by multiplication with the idempotent 
.
We set
As an intersection of the kernels of the -module homomorphisms
this is obviously a submodule. To see that for an R-algebra
S the map 
takes 
to the corresponding submodule
of 
, one uses the fact that the R-algebra structure map
induces an injective
map 
on a representative F=S/I
given by 
.
It is easily seen that the equation 
holds for all 
and 
. Thus, setting
it follows that factors to a map
. In the case where S is
a field you have 
and therefore 
.
This means that 
still maps to the Grothendieck group
for any field S that is an R-algebra.
