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# 2.4 Some Consequences of the Field Axioms.

2.59   Theorem (Cancellation laws.) Let be a field, let be elements in , and let . Then
 (2.60) (2.61) (2.62) (2.63)

(2.60) and (2.61) are called cancellation laws for addition, and (2.62) and (2.63) are called cancellation laws for multiplication.

Proof: All of these results are special cases of the cancellation law for an associative operation (theorem 2.19).

2.64   Theorem. In any field

Proof: These are special cases of the remark made earlier that an identity element is always invertible, and is its own inverse.

2.65   Theorem (Double inverse theorem.) In any field ,

Proof: These are special cases of theorem 2.17.

I will now start the practice of calling a field . If I say  let be a field" I assume that the operations are denoted by and .

2.66   Theorem. Let be a field. Then

Proof: We know that , and hence

Also, , so

By the cancellation law for addition, .

2.67   Corollary. Let be a field. Then for all , .

2.68   Theorem. Let be a field. Then for all in
 (2.69)

Proof:

Case 1:
Suppose . Then (2.69) is true because every statement implies a true statement.
Case 2:
Suppose . By theorem 2.66, , so

Since , we can use the cancellation law for multiplication to get

and hence

Thus (2.69) holds in all cases.

2.70   Remark. We can combine theorem 2.66, corollary 2.67 and theorem 2.68 into the statement: In any field ,

2.71   Exercise. Let be a field. Prove that has no multiplicative inverse in .

2.72   Theorem (Commutativity of addition.) Let be any field. Then is a commutative operation on .

Proof: Let be elements in . Then since multiplication is commutative, we have

By the distributive law,

Since is the multiplicative identity,

and hence

By the cancellation law for addition

By commutativity of multiplication and the distributive law,

and

Since is the multiplicative identity and addition is associative

and hence

Since multiplication is commutative

and by the cancellation law for addition,

Hence, is commutative.

2.73   Remark. Let be a field, and let . Then
 (2.74)

 (2.75)

Proof:

2.76   Theorem. Let be a field. Then

Proof: Let . By (2.74) it is sufficient to prove

Well,

2.77   Exercise. Let be a field, and let . Prove that
1. a)
2. b)
3. c)

2.78   Exercise. A Let be a field and let be non-zero elements in . Prove that

2.79   Definition (Digits.) Let be a field. We define

I'll call the set

the set of digits in . If are digits, I define
 (2.80)

and
 (2.81)

Here should not be confused with .

2.82   Example.

In general, if , I define

Then for all digits

so, for example

2.83   Remark. The set of digits in may contain fewer than ten elements. For example, in ,

and you can see that .

2.84   Theorem. In any field , and .

Proof:

Also,

2.85   Exercise. Prove that in any field , and .

2.86   Exercise. Prove that in any field , .

2.87   Remark. After doing the previous two exercises, you should believe that the multiplication and addition tables that you learned in elementary school are all theorems that hold in any field, and you should feel free to use them in any field.

2.88   Exercise. Let be a field and let . Prove that

Next: 2.5 Subtraction and Division Up: 2. Fields Previous: 2.3 The Field Axioms   Index