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By De Simone A., Mundici D.

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Extra resources for A Cantor-Bernstein Theorem for Complete MV-Algebras

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Since −(1 + x) ∈ T and 1 − (1 + x) = −x ∈ T , it follows that −(1 + x) ∈ O1 . By Step 2 and Step 3, 1 + x = (−1)(−(1 + x)) ∈ O1 as required. Case 2: 1 + x ∈ T . Observe first that −x = (−1)x ∈ O2 (Step 2 and Step 3). Moreover, we saw in the proof of Case 2 of Step 2 that for each y ∈ O1 it follows that −y(1 + y)−1 ∈ O1 . Then (−x) −y(1 + y)−1 ∈ O1 , and so 1 + (1 + x)y = (1 + y) 1 + (−x) −y 1+y ∈T . Consequently, (1 + x)y ∈ O1 and 1 + x ∈ O2 , as claimed. Final Step. O is a valuation ring of K satisfying O× ⊆ T , and so the result is proved in this case: By Step 2, O · O ⊆ O.

Then h = cg1 /g2 ∈ K(X). 3. Suppose v : K −→ → Γ ∪ {∞} is a valuation of the field K, Γ is an ordered subgroup of an ordered group Γ , and γ ∈ Γ has the property: if n ∈ Z satisfies nγ ∈ Γ , then n = 0 . Under these conditions, there is exactly one valuation w on K(X) extending v, with w(X) = γ. For this w, we have K(X) = K and w(K(X)× ) = Γ ⊕ Zγ with the ordering induced from Γ . 2 Constructions of Valuations 35 Proof. 1. To prove uniqueness, let w be any such extension. Consider an f ∈ K[X], say, f = a0 + a1 X + · · · + an X n , with ai ∈ K.

5) as promised. The map w : K(X) −→ Γ ∪ {∞} is well defined: if f1 /g1 = f2 /g2 , then f1 g2 = f2 g1 . 5) implies w(f1 ) + w(g2 ) = w(f2 ) + w(g1 ) and so w(f1 ) − w(g1 ) = w(f2 ) − w(g2 ). 5) from the case of f, g ∈ K[X] \ {0} to the case of arbitrary elements h1 , h2 of K(X) \ {0}. For this, let g be a common denominator of h1 and h2 : hi = fi /g, where f1 , f2 , g ∈ K[X] \ {0}. Then w(h1 + h2 ) = w f1 + f2 g = w(f1 + f2 ) − w(g) ≥ min{w(f1 ), w(f2 )} − w(g) = min{w(f1 ) − w(g), w(f2 ) − w(g)} = min{w(h1 ), w(h2 )} .

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