Bourbaki

Exercise 15, § 9

Algebra, I, 177

Bourbaki marks this one as harder than the rest.

Let A be a non-commutative ring with no divisors of 0. A is said to admit a field of left fractions if it is isomorphic to a subring B of a field K such that every element of K is of the form x−1yx^{-1}y, where x∈B,y∈Bx \in B, y \in B.

(a) Let A′A' be the set of elements ≠0\neq 0 in A. For A to admit a field of left fractions, it is necessary that the following condition be fulfilled: (G) for all x∈A,x′∈A′x \in A, x' \in A', there exist u∈A′u \in A' and v∈Av \in A such that ux=vx′ux = vx'. (b) Suppose conversely that condition (G) is fulfilled. Show that in the set A×A′A \times A' the relation R between (x,x′)(x, x') and (y,y′)(y, y') which states "for every ordered pair (u,v)(u, v) of elements ≠0\neq 0 such that ux′=vy′,ux=vy"ux' = vy', ux = vy" is an equivalence relation.

Let (x,x′),(y,y′)(x, x'), (y, y') be two elements of A×A′A \times A', ξ\xi and η\eta their respective classes (mod. R). For every ordered pair (u,u′)∈A×A′(u, u') \in A \times A' such that u′x=uy′u'x = uy', show that the class (mod. R) of (uy,u′y′)(uy, u'y') depends only on the classes ξ\xi and η\eta; if it is denoted by ξη\xi \eta, a law of composition is defined on the set K=(A×A′)/RK = (A \times A')/R. If K′K' is the set of elements of K distinct from the class 0 of elements (0,x′)(0, x') of A×A′A \times A', K′K', with the law induced by the above law, is a group.

For every element x∈Ax \in A, the elements (x′x,x′)(x'x, x'), where x′x' runs through A′A', define an isomorphism of A (with multiplication alone) onto a subring of K. Identifying A with its image under this isomorphism, the class (mod. R) of an ordered pair (x,x′)∈A×A′(x, x') \in A \times A' is identified with the element x′−1x{x'}^{-1}x.

This being so, if ξ=x′−1x\xi = {x'}^{-1}x is an element of K and 1 denotes the unit element of K′K', then ξ+1\xi + 1 denotes the element x′−1(x+x′){x'}^{-1}(x + x'), which does not depend on the representation of ξ\xi in the form x′−1x{x'}^{-1}x. We then write ξ+0=ξ\xi + 0 = \xi and, for η≠0\eta \neq 0, ξ+η=η(η−1ξ+1)\xi + \eta = \eta(\eta^{-1}\xi + 1). Show that the addition and multiplication thus defined on K determine on that set a field structure which extends the ring structure on A; in other words, condition (G) is sufficient for A to admit a field of left fractions.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.