Bourbaki

Exercise 1, § 2

Algebra, III, 618

Let E be an A-algebra with a basis of two elements e1,e2e_1, e_2 and a unit element e=λe1+μe2e = \lambda e_1 + \mu e_2.

(a) Show that the ideal aa of A generated by λ\lambda and μ\mu is the whole ring A (in the contrary case, note that E/aE=E⊗A(A/a)={0}E/aE = E \otimes_A (A/a) = \{0\} would hold contrary to the fact that A/a≠{0}A/a \neq \{0\} and E is a free A-module).

(b) If α,β\alpha, \beta are two elements of A such that αλ+βμ=1\alpha \lambda + \beta \mu = 1, show that the elements ee and u=βe1−αe2u = \beta e_1 - \alpha e_2 also form a basis of E and deduce that E is a quadratic A-algebra.

Solution

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Transcribed from the printed volume. Not translated.