MICHAEL J. NEELY, UNIVERSITY OF SOUTHERN CALIFORNIA, FALL 2007 1EE 441: Axioms and Lemmas for a Field FI. AXIOMS FOR A FIELDLet F be a set of objects that we call “scalars.” We assume that F has at least two distinct elements. We say that F is a fieldif there are rules for addition and multiplication of the elements of F such that for any α ∈ F, β ∈ F, we have:• α + β ∈ F• αβ ∈ FFurther, the addition and scalar multiplication must satisfy the following properties (for any α, β, γ ∈ F):1) α + β = β + α , αβ = βα (Commutative)2) (α + β) + γ = α + (β + γ) , (αβ)γ = α(βγ) (Associative)3) α(β + γ) = αβ + αγ (Distributive)4) (Existence of 0) There is an element “0” such that α + 0 = α for any α ∈ F.5) (Existence of 1) There is an element “1” such that α1 = α for any α ∈ F.6) (Additive inverse) For every α ∈ F, there exists an additive inverse −α such that α + −α = 0.7) (Multiplicative inverse) For every α ∈ F such that α 6= 0, there exists a multiplicative inverse α−1such that αα−1= 1.II. LEMMASSuppose we have a field F. Recall that the field must contain at least two elements. We have the following arithmetic factsfor any field (the proofs are on the back page).Note: For any values a ∈ F, b ∈ F, the following fact holds: If a = b, then for any c ∈ F we have a + c = b +c and ac = bc.That is, we can add any value c to both sides of a true equation to yield another true equation, and we can multiply any valuec to both sides to yield another true equation. The proof is obvious (For addition: If a = b then a + c = b + c, done). (Formultiplication: If a = b then ac = bc, done).Lemma 1: α0 = 0 for any α ∈ F.Lemma 2: 0 6= 1.Lemma 3: For any element a ∈ F, the additive inverse −a is unique.Lemma 4: For any element a ∈ F such that a 6= 0, the multiplicative inverse a−1is unique.Lemma 5: For any element a ∈ F, we have that −a = (−1)a.Lemma 6: (−1)(−1) = 1.Lemma 7: For any elements a, b in F, we have:a(−b) = −(ab) = (−a)b , (−a)(−b) = abMICHAEL J. NEELY, UNIVERSITY OF SOUTHERN CALIFORNIA, FALL 2007 2Definition 1: (Subtraction) If a ∈ F and b ∈ F, then the subtraction operator a − b produces another element in F, definedas follows:a − bM=a + (−b)From this subtraction rule, it immediately follows that a(b − c) = ab − ac.Definition 2: (Integer Powers) For any element a ∈ F, we define the square of a (written a2), to be aa. Similarly, for anypositive integer k, we use the notation akto represent a multiplied by itself k times (so that a1M=a, a2M=aa, a3M=aaa, etc.). Forany a 6= 0 and for any positive integer k, we define a−kto be the product of a−1with itself k times (so that a−2= a−1a−1).For any a 6= 0, the power a0is defined to be 1. The power 0−kis not defined for any integers k ≥ 0.Lemma 8: For any two elements a, b in F, if ab = 0 then either a = 0 or b = 0 (or both).Lemma 9: For any element a ∈ F, if a2= 1, then a ∈ {−1, 1}.Lemma 10: (Polynomial Factoring) Suppose a ∈ F and b1, b2, . . . , bnare n additional elements of F. Suppose the followingequation holds: Πni=1(a − bi) = 0. Then a ∈ {b1, b2, . . . , bn}.The proof of Lemma 1 (that α0 = 0) is given below. The proofs of the remaining lemmas are given on the next page. Forfun, you are encouraged to prove the lemmas yourself and then check the proofs on the back if you need help or want tocompare proof methodologies.Proof: (Lemma 1: α0 = 0.) Note that 1 + 0 = 1. Therefore, for any α ∈ F, we have:α = α1= α(1 + 0)= α1 + α0= α + α0Thus, we have:α = α + α0Adding −α to both sides of the above equation yields:−α + α = −α + α + α0and therefore we have 0 = 0 + α0 = α0, and we are done.MICHAEL J. NEELY, UNIVERSITY OF SOUTHERN CALIFORNIA, FALL 2007 3Proof: (Lemma 2: 0 6= 1.)Let a ∈ F. If 0 = 1, then 0 = a0 = a1 = a, and hence a = 0. Thus, if 0 = 1, then all elements a of F are equal to 0,contradicting the fact that F has at least two elements.Proof: (Lemma 3: −a is unique.)Suppose that a + b = 0. Then adding −a to both sides, we get a + (−a) + b = −a and hence 0 + b = −a. That is, b = −a.In other words, the only element that can be added to a to get 0 is the element −a.Proof: (Lemma 4: If a 6= 0, then a−1is unique.)Suppose a 6= 0 and there is a b ∈ F such that ab = 1. Because a 6= 0, a−1exists. Multiplying both sides of the equationab = 1 by a−1yields: a−1ab = a−11 = a−1. But a−1a = 1, and hence b = a−1.Proof: (Lemma 5: −a = (−1)a.)Note that 0 = a(1 + −1) = a1 + a(−1) = a + a(−1). Adding −a to both sides yields −a = 0 + a(−1), and hence−a = (−1)a.Proof: (Lemma 6: (−1)(−1) = 1.)0 = (−1)(1 + −1) = (−1)1 + (−1)(−1) = −1 + (−1)(−1)Adding 1 to both sides of the equation yields:0 + 1 = −1 + 1 + (−1)(−1) = (−1)(−1)Therefore, 1 = (−1)(−1).Proof: (Lemma 7: a(−b) = −(ab) = (−a)b and (−a)(−b) = ab.)We have by the previous lemmas and by commutativity of multiplication:a(−b) = a(−1)b = (−1)ab = −(ab)Similarly, (−a)b = (−1)ab = −(ab). Finally: (−a)(−b) = (−1)a(−1)b = (−1)(−1)ab = ab.Proof: (Lemma 8: If ab = 0, then either a = 0 or b = 0.)Suppose ab = 0, but that both a and b are nonzero. We reach a contradiction. If a 6= 0 and b 6= 0, then there existmultiplicative inverses a−1and b−1. Thus, multiplying the equation ab = 0 by the product a−1b−1yields:aba−1b−1= 0and hence: 0 = aa−1bb−1= (1)(1) = 1. Thus, 0 = 1. This is a contradiction, as we know that 0 6= 1.Proof: (Lemma 9: If a2= 1, then a ∈ {−1, 1}.)Suppose a2= 1. Then we have:(a + 1)(a − 1) = a2− a + a − 1 = 1 − 1 = 0Therefore, either (a + 1) = 0 or (a − 1) = 0. Equivalently, either a = −1 or a = 1.Proof: (Lemma 10: Πni=1(a − bi) = 0 implies a ∈ {b1, . . . , bn}.)This fact follows by using induction together with Lemma 8.III. EXAMPLESBasic examples of fields are R, Q, and C (the …