By Mary Gray

ISBN-10: 020102568X

ISBN-13: 9780201025682

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**Extra resources for A Radical Approach to Algebra (Addison-Wesley Series in Mathematics) **

**Example text**

6 32 CHAPTER 4 it follows that (iii) holds. Setting u = v in (iii), we obtain (i). iii, φ is identically zero. 3. ✷ Thus (ii) holds. 10. 11) for all v ∈ L. 12) for all a ∈ X and all v ∈ L. 14) for all a ∈ X, all v ∈ L and all t ∈ K. Proof. Choose a ∈ X and v ∈ L. 4) π(av) = π(a)σ q(v) + θ(a, v)σ f (1, v) + f (θ(a, v), 1)v σ + φ(a, v) · 1. ii. i. 12 holds. 14 holds (by C1 and C2). 18. Let char(K) be arbitrary. Then φ(a + b, u) = φ(a, u) + φ(a, v) + g(au, bu) − g(a, b)q(u) for all a, b ∈ X and all u ∈ L.

31 PROPER QUADRANGULAR ALGEBRAS Thus f (h(a, a), 1) = 0. 7 that g(a, a) = 0. 6. Thus (i) and (ii) hold. 6 for all v, w ∈ L. By C4 and (i), therefore, φ is identically zero. Hence (iii) holds. 8. When char(K) = 2, the map θ and the axioms C1–C4 in the deﬁnition of a quadrangular algebra are superﬂuous. i as the deﬁnition of θ and deduce C1–C4 from the other axioms as follows. Axioms C1 and C2 hold by A1 and B1. By A1, B1 and B2, we have h(a + b, (a + b)u) = h(a, au) + h(b, bu) + h(a, bu) + h(b, au) = h(a, au) + h(b, bu) + 2h(a, bu) + f (h(b, a), 1)u for all a, b ∈ X and all u ∈ L.

2 below. 9. Let char(K) be arbitrary. Then (i) f (θ(a, v), v) = f (π(a), 1)q(v); (ii) φ(a, u + v) + φ(a, u) + φ(a, v) = f (θ(a, u), v) + g(au, av); and (iii) f (θ(a, u), v) = −f (θ(a, v), u) + f (π(a), 1)f (u, v) for all a ∈ X and all u, v ∈ L. Proof. Choose a ∈ X and u, v ∈ L. 19, we can assume that char(K) = 2. 1. 6 32 CHAPTER 4 it follows that (iii) holds. Setting u = v in (iii), we obtain (i). iii, φ is identically zero. 3. ✷ Thus (ii) holds. 10. 11) for all v ∈ L. 12) for all a ∈ X and all v ∈ L.

### A Radical Approach to Algebra (Addison-Wesley Series in Mathematics) by Mary Gray

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