Lemma 2

Let M and $M'$ be A-modules that are finite-dimensional vector spaces over K. The supports (VIII, p. 190) of M and $M'$ are disjoint if and only if there exists an $a\in A$ such that $a_M= 0_M$ and $a_{M'}= 1_{M'}$.

Suppose that there exists an $a\in A$ such that $a_M= 0_M$ and $a_{M'}= 1_{M'}$. Let S be a simple A-module. If cl(S) belongs to the support $\mathscr{S}_M$ of M, then the A-module S is isomorphic to one of the quotients of a Jordan–Hölder series of M and we have $a_S= 0_S$. Likewise, if cl(S) belongs to the support $\mathscr{S}_{M'}$ of M’, then we have $a_S= 1_S$. It follows that $\mathscr{S}_M$ and $\mathscr{S}_{M'}$ are disjoint.

Conversely, suppose that the sets $\mathscr{S}_M$ and $\mathscr{S}_{M'}$ are disjoint. They are finite because M and $M'$ are finite-dimensional over K. Every simple A-module S whose class belongs to $\mathscr{S}_M\cup \mathscr{S}_{M'}$ is finite-dimensional over K and a fortiori over the field End$_A(S)$. By Corollary 1 of Proposition 4 (VIII, p. 83), there exists an element $b\in A$ such that we have $b_S= 0_S$ (resp. $b_S= 1_S$) for every simple A-module S whose class belongs to $\mathscr{S}_M$ (resp. $\mathscr{S}_{M'}$). Let $(M_i)_{0\leqslant i\leqslant n}$ be a Jordan–Hölder series of M. By the above, we have $bM_i\subset M_{i+1}$ for $0\leqslant i < n$ and therefore $(b^n)_M= 0_M$. The existence of a natural number $m$ such that $((1-b)^m)_{M'}= 0_{M'}$ is proved in the same way. Set $P(X) = 1-(1-X^n)^m$ and $a= P(b)$. The polynomial P(X) is a multiple of $X^n$, so we have $a_M= 0_M$, while the polynomial $1-P(X)$ is a multiple of $(1-X)^m$, so we have $a_{M'}= 1_{M'}$. This concludes the proof.

Let L be an extension of K. If M is an A-module that is finite-dimensional over K, then $M_{(L)}$ is an $A_{(L)}$-module that is finite-dimensional over L. Moreover, for every exact sequence

of A-modules, the sequence of $A_{(L)}$-modules

deduced from it by extension of scalars is exact (II, §7, No. 7, p. 308, Proposition 14). Hence there exists a unique ring homomorphism $u: R_K(A)\rightarrow$ $R_L(A_{(L)})$ such that $u([M]) = [M_{(L)}]$ for every A-module M that is finite-dimensional over K (No. 5).