We want to define the notion of **group** that will be fundamental to determine which polynomials are solvable by radicals. This notion is very important and appears in every area of mathematics.

We motivate the definition through the example that most concerns us: automorphisms of fields. They are particular class of isomorphisms, so we begin with them.

Some of the arguments below have been discussed in previous lectures.

**1. Isomorphisms **

Recall that a field homomorphism is a map between fields and such that and for all This implies that and that for all

There are two possibilities for the value of because so either or else In the first case, for all We will assume from now on that to avoid this degenerate case. Note that this implies that for all nonzero

The following observation is obvious:

Lemma 1Suppose and Suppose that is a field homomorphism.

Then is a field homomorphism.If is the restriction of to then is a field homomorphism.

Let If then is a field homomorphism.

We are interested in which sets can be the images of homomorphisms.

Lemma 2Suppose that is a field homomorphism. Then is a field.

*Proof:* We need to check that is a subfield of First note that it is nonempty, and it is closed under addition and multiplication. This is because Also, if then there are such that and and therefore Similarly,

One also checks easily that is closed under additive inverses and under multiplicative inverses of nonzero elements.

A field homomorphism is an **isomorphism** iff is a bijection.

Lemma 3If is an injective homomorphism, then is an isomorphism.

We are mostly concerned with isomorphisms between subfields of The following observation shows that any isomorphism always has many fixed points:

Lemma 4Suppose is a homomorphism and that and are subfields of Let

Then is a field.

*Proof:* By our convention, Note that is closed under addition and multiplication, under additive inverses, and under multiplicative inverses of nonzero elements. For example, if and then so

Since every subfield of contains it follows that every homomorphism between subfields of fixes

We are interested in the case where

In what follows, even if I forget to state this explicitly, all fields we consider are subfields of unless stated otherwise.

Lemma 5Suppose that and that is algebraic over Suppose that is an injective homomorphism, and that Let Then

is an isomorphism, and it is given as follows: Any element of can be written in a unique way in the form

where and where is the minimal polynomial of over Then

*Proof:* Note first that since and then

Given we can write as stated above. Then

and this number clearly belongs to

This shows that Note also that if

where then so

This means that is a root of Since is irreducible over then It follows that

From this we deduce that any element of can be written in the form

for some and therefore is onto. It is also 1-1, by assumption. Therefore, is a bijection.

This shows that any isomorphic image of must have the form for some that is also a root of This also means that there are at most injective homomorphisms with domain where By results from lecture 7.3, if is any root of the map defined as above is an isomorphism. We conclude that there are **exactly** injective homomorphisms with domain thanks to the following theorem:

Theorem 6If is irreducible, then all the roots of are simple. In particular, all the roots of the minimal polynomial of any algebraic over are distinct.

*Proof:* Suppose that is a double root of and that is irreducible. Then, over for some polynomial Note that since is not necessarily in then is not necessarily in

It follows that the derivative so is also a root of Note also that Since and have a common root, they cannot be relatively prime, because then there would be polynomials such that but this equation gives a conradiction when evaluated at

It follows that there is a nonconstant polynomial that divides both and But then and this contradicts that is irreducible.

For example, if is an injective homomorphism with domain , then or In the first case, is the identity. In the second, is the map

for any rational numbers

As another example, the only fields isomorphic to are and

The following follows from the same arguments as above:

Theorem 7Suppose that and that is algebraic over with minimal polynomial Let be any injective homomorphism with domain with Then is also a root of and

**2. Automorphisms **

Definition 8If is an isomorphism, then we say that is anautomorphsimof The collection of all automorphsism of is denoted by

For example, where denotes the identity map.

As another example, as well.

On the other hand, consists of precisely two elements: The identity, and the only homomorphism that sends to

In the next lecture we will study automorphisms in some detail, and introduce the notion of group.

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