Page 82 - Textos de Matemática Vol. 45

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and
Chapter 6. Semirings
l(x) · tg(b) = l(x) · tg′(b).
By hypothesis, these conditions are satisfied when b = z (because in this case
g(z) = d and g′(z) = d′). Then, it suffices to prove that the set Z of all ele- ments b ∈ B such that the conditions above are satisfied for any x ∈ X is a subsemiring of B, because in this case it coincides with B, since B is generated by z, and z ∈ Z.
So,letb1,b2 ∈Z.Then
tg(b1 + b2) · l(x) = (tg(b1) + tg(b2)) · l(x) = tg(b1) · l(x) + tg(b2) · l(x) =
= tg′(b1) · l(x) + tg′(b2) · l(x) = (tg′(b1) + tg′(b2)) · l(x) = tg′(b1 + b2) · l(x), and, similarly, l(x) · tg(b1 + b2) = l(x) · tg′(b1 + b2), so b1 + b2 ∈ Z. Obviously
0∈Z,soitremainstoprovethatb1·b2 ∈Z:
tg(b1 · b2) · l(x) = (tg(b1) · tg(b2)) · l(x) = tg(b1) · (tg(b2) · l(x)) =
= tg(b1) · (tg′(b2) · l(x)) = tg′(b1) · (tg′(b2) · l(x)) = tg′(b1 · b2) · l(x), where the first equality in the last row holds because tg′(b2) · l(x) ∈ l(X). This
concludes the proof. Theorem 6.5.4. For any semiring X, every object in SSplExt(X) is accessible. Proof. Given a Schreier split extension
0 //Xoo q //Aoo p //B //0,
we have to build a faithful one
k
s
// oo q′ r //
0 X //Coo D 0
lt
and a morphism of Schreier split extensions into it, the so-called index of the Schreier split extension in question. Let us define the following relation R on B:
bRb′ ⇐⇒ s(b)·k(x) = s(b′)·k(x) and k(x)·s(b) = k(x)·s(b′) for all x ∈ X.
It is immediate to see that R is an equivalence relation. It is a congruence on B, indeed, if b1Rb′1 and b2Rb′2, then (b1 + b2)R(b′1 + b′2) and (b1 · b2)R(b′1 · b′2), because
s(b1+b2)·k(x)=s(b1)·k(x)+s(b2)·k(x)=s(b′1)·k(x)+s(b′2)·k(x)=s(b′1+b′2)·k(x),
//