Most of our work from now on depends on the following simple, but very useful observation:
Theorem 1 Let be a field extension. Then with its usual addition is a vector space over 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
Definition 2 If then is the dimension of as a vector space over .
We will be interested in finite-dimensional extensions. They correspond to extensions by algebraic numbers. Recall:
Definition 3 If and we say that is algebraic over iff 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 4 Let 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.
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 5 Suppose that is algebraic over The minimal polynomial of over is 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 7 If 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
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 9 If 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 10 If 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 11 The 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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