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

and

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.

Proof: Let and be elements in
. Then

Also,

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 .

Proof: For all
, we have

and

- 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 .

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

and

Multiply the first equation by and the second by to get

If we add these equations, we get

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:

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 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

Now

and

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

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

- 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 .

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