Bourbaki

Corollary 1

Tag 00MX, permanent. Cite it as [Bourbaki, Tag 00MX].

Algebra, VIII, p. 386, § 20. Linear Representations of Algebras

Let E and F be semisimple A-modules of finite dimension over K, and let A\mathscr{A} be a generating subset of the K-vector space A. Suppose that for every a∈Aa \in \mathscr{A}, the characteristic polynomials of the endomorphisms aEa_E and aFa_F of the K-vector spaces E and F are equal. Then the A-modules E and F are isomorphic.

Let aa be an element of A\mathscr{A}. The characteristic polynomials of aEa_E and aFa_F have the same degree, so the dimension of E is equal to that of F; we denote it by nn. Let PcE(a;T)\mathrm{Pc}_E(a; T) be the characteristic polynomial of aEa_E. In K(T)K(T), we have the equalities

χE(a;T)=det⁡(1+aET)=(−T)ndet⁡(−1T−aE)=(−T)nPcE(a;−1T), \chi_E(a; T) = \det(1 + a_E T) = (-T)^n \det \left( \frac{-1}{T} - a_E \right) = (-T)^n \mathrm{Pc}_E \left( a; \frac{-1}{T} \right),

and χF(a;T)\chi_F(a; T) is given by an analogous formula. Because of our assumptions, we have χE(a;T)=χF(a;T)\chi_E(a; T) = \chi_F(a; T). By Theorem 2, we have [E] = [F], which implies that E and F are isomorphic (VIII, p. 190, Corollary of Proposition 7).

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.