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# 4.1 Construction of .

Throughout this chapter, will represent a field in which is not a square. For example, in an ordered field is not a square, but in , so is a square. In ,

so is not a square in .

Let be a field in which is not a square. I am going to construct a new field which contains (a copy of) and a new element such that . The elements of will all have the form

where and are in . I'll call the complexification of . Before I start my construction, note that if are in and , then by the usual field axioms
 (4.1)

and
 (4.2)

Let be a field in which is not a square. Let denote the Cartesian product of with itself (Cf. definition 1.55). I define two binary operations and as follows (cf. (4.1) and (4.2)): for all ,

and

We will now show that is a field.

4.3   Theorem (Associativity of .) The operation is associative on .

Proof: Let and be elements in . Then

 (4.4)

Also,
 (4.5)

Now by using the field properties of , we see that the (4.4) and (4.5) are equal, and hence

Hence, is associative on .

I expect the multiplicative identity for to be .

4.6   Theorem (Multiplicative identity for .)The element is an identity for on .

Proof: For all , we have

and

4.7   Exercise.
a)
Show that is associative on .
b)
Show that there is an identity for on .
c)
Show that every element in has an inverse for .
d)
Show that is commutative on .
e)
Show that the distributive law holds for .
f)
Show that the additive and multiplicative identities for are different.

As a result of exercise 4.7 and the two previous theorems, we have verified that satisfies all of the field axioms except existence of multiplicative inverses. Note that up to this point we have never used the assumption that is not a square in .

4.8   Theorem (Existence of multiplicative inverses.)Let be a field in which is not a square and let be an element in . Then has an inverse for .

Proof: Let . I want to find a point such that

Since multiplication is commutative, this shows that and hence that is a multiplicative inverse for . I want

so I want
 (4.9)

and
 (4.10)

Multiply the first equation by and the second by to get

If we add these equations, we get
 (4.11)

In the next lemma I'll show that if is not a square then for all , so by (4.11), . Now multiply (4.9) by and (4.10) by to get

If we add these equations, we get

so

I've shown that if , then . A direct calculation shows that this works:

4.12   Remark. The above proof shows that for all ,

4.13   Lemma. Let be a field in which is not a square. Let be an element in . Then .

Proof: Since , either or .

Case 1:
Suppose , then , so

Since is not a square in , .
Case 2:
Suppose . Repeat the argument of Case 1 with the roles of and interchanged.
We now have verified all of the field axioms so we know that is a field. Hence we can calculate in using all of the algebraic results that have been proved to hold in all fields.

4.14   Notation () Let be a field in which is not a square. We will denote the pair by , and if we will denote the pair by .

We have is the additive identity for , and is the multiplicative identity for . If , then . Also

so is a square root of .

If , then

and hence every element can be written in the form . We have

Hence contains a  copy of ". Each element in corresponds to a unique in in such a way that addition in corresponds to addition in and multiplication in corresponds to multiplication in . We will henceforth drop the tildes, and we'll denote by and by as is usual in fields. Then every element in can be written uniquely as where and .

We consider to be a subset of . An element of is in if and only if . If , then

4.15   Examples. I will find the square roots of in . Let . Then

Now

and

which is impossible. The only possible square roots of are . You can easily verify that these are square roots of .

4.16   Example. I can solve the quadratic equation
 (4.17)

in by using the quadratic formula for . (by the previous example). Since is a square, the equation has the solution set

4.18   Exercise. Check that and are solutions to (4.17).

4.19   Exercise. A
a)
Write in the form where .
b)
Find all solutions to

in . (You may want to use the result of example 4.15.) Write your solutions in the form where .

4.20   Entertainment. We noted earlier that is not a square in , so has a complexification, which is a field with 9 elements. Show that if , then the 9 elements in are

Can you figure out before you make any calculations which of these elements is ?

Next: 4.2 Complex Conjugate. Up: 4. The Complexification of Previous: 4. The Complexification of   Index