Talk:K-group, first (Ex)

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First we show that the determinant induces a well-defined morphism

\displaystyle \xymatrix{\det:K_1( R) \ar[r] & R^\times.}

For this we recall that K_1( R) = \mathrm{GL}( R)_{ab} is the abelianization of the infinite general linear group over R. Surely, since the ring is commutative, for every n we have the determinat as a map \det_n: \mathrm{GL}_n( R) \to R^\times and the diagram

\displaystyle \xymatrix{ \mathrm{GL}_n( R) \ar[r]^-{\det_n} \ar[d]_\iota & R^\times \\ \mathrm{GL}_{n+1}( R) \ar@/_1.3pc/[ur]_-{\det_{n+1}} & }

commutes, where \iota is the canonical inclusion as block matrix in the upper left corner.

Hence we get a well-defined map on the colimit \det: \mathrm{GL}( R) \to R^\times by standard properties of the determinant we see that this map factors over the abelianization to obtain the desired map

\displaystyle \xymatrix{\det: K_1( R) \ar[r] & R^\times.}

The fact that this map is surjective follows from the easy observation that we have the canonical map

\displaystyle  R^\times = \mathrm{GL}_1( R) \to K_1( R)

which obviously splits the determinant.



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