Most of our work from now on depends on the following simple, but very useful observation:

Theorem 1Let be a field extension. Then with its usual addition is avector spaceover where multiplication of elements of by elements of is just the usual product of .

*Proof:* Remember that the statement means that the elements of (the *vectors*) can be added in a way that:

- Addition is commutative: for all
- Addition is associative: for all
- There is an additive identity for all
- All vectors admit additive inverses: For any there is a such that

All of these are immediate consequences of being a field. The statement also means that we can multiply vectors by elements of (the *scalars*) in such a way that:

- for all and
- for all and
- for all and all
- for all

Again, these are all immediate consequences of being a field, and in fact hold for all in

From linear algebra we know that if then the dimension of as a vector space over is well-defined, and it is the size of any basis. Recall that a basis is a set of vectors such that for any there are vectors and scalars such that

(we say that * spans *), and this representation is unique (we say that * is linearly independent over *). Another way of stating linear independence is by saying that if and

then

Definition 2If then is thedimension of as a vector space over .

We will be interested in finite-dimensional extensions. They correspond to extensions by *algebraic numbers*. Recall:

Definition 3If and we say thatis algebraic overiff there is a nonzero polynomial such that(If we simply say that is

algebraic.)If is not algebraic over we say that it is

transcendental over

For example, is transcendental over but algebraic over as witnessed by the polynomial (That is transcendental is a deep theorem of Lindemann that we won’t prove here.)

is algebraic over as witnessed by the polynomial This is also witnessed by the polynomial although this is not particularly surprising, since

This is an instance of a general phenomenon:

Theorem 4Let and suppose that is algebraic over Then there is a polynomial with the following properties:

- is nonzero.
- is monic.
- is irreducible over
Moreover, satisfies that whenever and then and is the unique polynomial in with these properties.

*Proof:* Let

We claim that is an ideal. To see this, note that since If then since If and then since

As shown last lecture, is principal. Let be such that

Note first that This is because is algebraic, so there is at least one nonzero polynomial such that This shows that so

If is the leading coefficient of then and so and if then and is monic. Moreover, This is because if then there is some such that But then so (This shows that On the other hand, since then )

Let Then is monic and nonzero, and Moreover, whenever and then We claim that is irreducible over To see this, suppose that for some polynomials Since we must have or so one of is also in and therefore or but then (since and ), or

Finally, we argue that is the unique polynomial in that is monic (and therefore nonzero), irreducible, and such that For suppose and is monic and irreducible. Then so Since is irreducible, we must have that for some unit The only units in are the nonzero elements of so Since both are monic, we must have so

Definition 5Suppose that is algebraic over Theminimal polynomial of overis the unique monic irreducible polynomial such that

The main connection between this concept and the idea of field extensions as vector spaces is the following result:

- If contains a transcendental over then
- If and is algebraic over then

*Proof:* Suppose first that is transcendental over We claim that the numbers are linearly independent over This is because for any finitely many integers if and

then is a root of the polynomial

Since is transcendental, we necessarily have i.e.,

This shows that if contains any transcendentals over then

Conversely, suppose and that is algebraic over Let Then are elements of that are linearly dependent over since but This shows that

If are not linearly independent over then there are numbers not all of them zero, such that

But then is nonzero and This contradicts that since and It follows that are linearly independent, and therefore

Corollary 7If is algebraic over and then is algebraic over Moreover,

*Proof:* By the above, Since then is algebraic.

Moreover, since we have as is a vector subspace of But and

Corollary 8If is algebraic over and then

moreover, if then there is exactly one way to write with

*Proof:* This is because the argument of Theorem 6 shows that is a basis of over

We already knew that Corollary 8 holds in a few particular cases, for example for for See also lecture 4.5. The argument above is much more general, and reduces significantly the amount of computations involved in each case.

Lemma 9If then

*Proof:* Let be a basis of over and be a basis of over It is enough to show the following two facts:

- is a basis of over and
- if and then implies and

The second fact shows that This together with the first fact implies the lemma.

To see the second fact, note that if then

and use that and Recall that is linearly independent over If then it follows that contradicting that is a basis so, in particular, it consists of nonzero elements. Then Since and is a basis, Then and we must have so This proves the second fact.

To prove the first fact, we need to show that spans over and that is linearly independent.

* spans.* Let Since is a basis of over there are vectors and scalars such that

Since is a basis of over for each there are and scalars such that

Replacing each with the corresponding expression above in the displayed representation of and expanding, gives as a representation of as a linear combination of elements of

* is independent.* If and are such that

we need to prove that To this end, note that each for some For each let Then we can rewrite the expression above as

Note that for all but finitely many for these we use the convention that empty sums are zero, that is,

For any Since is a basis for over it follows that for all Since for all and for all and is a basis of over then for all This holds for all and we are done.

The argument above is written in a way that proves the result even if the extensions are infinite-dimensional. However, all our applications will be restricted to finite dimensional extensions, in which case the write up above can be simplified slightly.

Corollary 10If and are algebraic over then so are and and if then so is

*Proof:* Note that if is algebraic over then it is also algebraic over since

Thus, Since the result follows from Theorem 6.

Corollary 11The set of algebraic numbers is a field.

*Proof:* This is immediate from the corollary above.

Note that a simple counting argument shows that is countable, which means that almost every complex number is transcendental over

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