Page 75 - Textos de Matemática Vol. 45
P. 75

6.3. Schreier internal structures in semirings 67 (c) the reflexive graph is an internal groupoid.
Proof. We already proved (see Proposition 6.3.8) that, if a Schreier reflex- ive graph is an internal category, then the kernels of the domain and of the codomain cooperate. Hence (b) implies (a).
Conversely, suppose that we have a Schreier reflexive graph
d0 // K[d0]   // X1 oo  s0  // X0,
ooq k
d1
with the Schreier split epimorphism associated with the codomain d1 given by
oo t oo s0 K[d1] // X1
// X0,
such that k and l cooperate (we recall that (d1,s0) is a Schreier split epimor-
phism thanks to Proposition 3.1.13 and Lemma 6.0.8). This means that x·y=y·x=0 forall x∈K[d0], y∈K[d1].
We already know from Proposition 3.3.5 that the unique possible multiplication is the map defined by
m(a′, a) = q(a′) + a,
and that it is a morphism of monoids (since the additive monoids involved are all commutative). We only have to show that it is a morphism of semirings, i.e. that m(a′1 · a′2, a1, a2) = m(a′1, a1) · m(a′2, a2). We have that:
m(a′1 ·a′2,a1,a2)=q(a′1 ·a′2)+a1 ·a2 =
= q(a′1) · q(a′2) + q(s0d0(a′1) · q(a′2)) + q(q(a′1) · s0d0(a′2)) + a1 · a2,
while
m(a′1, a1) · m(a′2, a2) = (q(a′1) + a1) · (q(a′2) + a2) = =q(a′1)·q(a′2)+q(a′1)·a2 +a1 ·q(a′2)+a1 ·a2,
so it suffices to show that
q(s0d0(a′1) · q(a′2)) = a1 · q(a′2)
and
q(q(a′1) · s0d0(a′2)) = q(a′1) · a2.
a1 · q(a′2) = (t(a1) + s0d1(a1)) · q(a′2) = t(a1) · q(a′2) + s0d1(a1) · q(a′2) =
We have:
l d1


































































































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