Fields vs. vector spaces
Definition N18.1 (Field). A field is a triple with (distributive law), where is abelian with identity and is abelian with identity . A field with , , contains .
(Margin, next to “”: wth does characteristic mean again? See below.)
Definition N18.2 (Vector space). A vector space over comes with
where is abelian with identity , and scalar multiplication has identity .
Definition N18.3 (Characteristic of a field). The characteristic of a field is such that, given ’s multiplicative identity , we add
where is the additive identity of . The for MUST be the smallest such that ( times) holds. This is the general case for a ring; however for a field there exist only 2 cases:
Note that for C1 we have , i.e. it must be prime. If instead as in case 2, it means that there does not exist an such that , so in .
Theorem N18.4. The size of any finite vector space is a prime power.
Proof. Let be a finite vector space over . Then is finite too ( is injective for ), so is prime and . So is a vector space over the field ; being finite, it has a finite basis, say of elements, and therefore . (For , .) ∎
Subspaces, span and linear independence
Definition N18.5 (Subspace). A subspace is a subset containing such that is closed under and .
Definition N18.6 (Span).
Let be a list of vectors of . We say the span if .
Definition N18.7 (Linear independence). The are linearly independent if only if for all .
(Note: NO non-linear functions.)
Note. The span is the set of all linear combinations. Also, for all gives the zero vector, so the zero vector is in every span.
Example N18.8. In , let , , . Then .
Example N18.9. In consider . Then , which equals , i.e. the line . And if we take , then if are linearly independent.
Exercises on fields
Exercise N18.10 (Is a field?). Consider with and . Is this a field? If not, which are fields? My guess: primes (intuition).
Show solution
For to be a field, must be abelian with identity , must be abelian with identity , and must hold.
abelian with identity is obvious. seems suspicious: is not even a group ( has no inverse), so it cannot be an abelian group. In order for to be a field, every nonzero must be coprime with .
Exercise N18.11 (Zero divisors). Consider a field . If , then either or . Why does this fail in ?
Show solution
Proof. , and adding to both sides gives . ∎
Let , but and . Then
Conversely, using the same method (the lemma) for , and : if then ; if then ; and if then . No matter which case we have, we contradict our assumption.
As to why fails this: there does not exist an inverse under multiplication for in , which can be seen from .
Exercise N18.13. Suppose is a field with . What is in this field? ? ? Show that in general, for , every element satisfies .
Show solution
, and given , . So , , and .
Now for the general case: means , so any multiple of is as well, since
Exercise N18.14. If is a field, either or .
Show solution
Suppose with . Then , i.e. in . (Note is , but .) A field has no zero divisors (previous exercise), so or . But and is the smallest number with , so and : a contradiction. Thus must be either or , and cannot be anything else.
Note. For , , and for , . We have only if and .
as a vector space
Exercise N18.15. Let be a field. Show that is a vector space over . What is ? What about over as a vector space? What are the operations?
Show solution
We need operations satisfying the following: for there is with , and for there is with , for all , with no product of two vectors required. This also needs to satisfy the distributive law. Our main issue is that , so it seems the structure cannot be simple multiplication. But if then , and we can say that vector multiplication is defined in . Vector spaces don’t require that not be valid, so we can consider a vector space over with the additional inclusion of scalar-by-scalar multiplication. Since scalars and vectors are the same things in (), , which is (maybe) why we can have . However, in we have , so we cannot have , since has for its basis.
Exercise N18.16. cannot be a vector space over (show)!
Show solution
Consider , with . Then taking , for must again be written in a basis of . Consider a basis ; then , but for all , , so , and thus scalar multiplication is not defined in over . So cannot be a vector space over (with scalar multiplication given by multiplication of complex numbers).