Hey so I am having trouble with 3a) and the first part of 4.



For 3a) would I just start by checking all the field axioms for K and since F is a subset of K then it automatically gets the axioms for free?

For the first part of 4 I think that since I am working over Z_2 (i.e. modular arithmetic with mod 2) the only elements I have are 0 and 1 so is it enough to show that either of these do not satisfy the equation x^2 + x + 1 = 0 ?

I'm not too sure..

This post was edited by Bloo_Guardian on Jan 18 2015 06:34pm

3a) simply comes from the definition of a vector space.
See for example here :

http://en.wikipedia.org/wiki/Field_extension

You will have to check all vector space axioms for K, as a vector space over F. Every point is obvious actually.

4) is the construction of the unique finite field of size 4.
See for example here :
http://en.wikipedia.org/wiki/Finite_field
(Control F : Field with four elements)

You are right on the first part, since p(x) is a degree 2 polynomial (if it was possible to break it into (ax+b).(cx+d), then it should have roots).
You can even make a list of all polynomials of lesser degree :
0 ; 1 ; X ; X + 1
and show that no product of those equals X²+X+1 (they always differ for X=0 or for X=1).

For 3a is it enough to say that all axioms are inherited because both K and F are fields? I am having trouble figuring out what properties are inherited from what.

If not, how would I show that (K,+) is an abelian group and for 1,a,b in F and u,v in K, that 1v=v, (a+b)v=av+bv, a(u+v)=au+av, and that a(bv)=(ab)v ?

Thank you for the help

The givens : K is a field, F is a sub-field.
Gives you :
(K,+,x) is an abelian ring, in which every non zero element have a reciprocal.
F is included into K and (F,plus,times) is an abelian ring with all non-zero having a reciprocal (in F), with plus and times being the restriction of + and . to F, hence denoted simply the same : + and .
0 and 1 are automatically the same for K and F.

The demand : prove that K is a vector-space over F.
In other words, prove that :

(1) (K,+) is an abelian group
(2) check that the multiplicative law
F x K -> K
(f;k) -> f.k
verifies that :
(2a) 1.k = k , for every k in K
(2b) (f+g).k = f.k + g.k , for every f,g in F and k in K
(2c) f.(k+l) = f.k + f.l , for every f in F and k,l in K
(2d) (f.g).k = f.(g.k) for every f,g in F and k in K

(1) is a consequence of (K,+,.) being a ring : a ring (R,+,*) is an abelian group with * associative, distributive over +, and showing a neutral element "1" (unity, or multiplicative identity).
(2a) comes from the fact that unity in F is the same as unity in K
(2b), (2c) and (2d) comes from the axioms of rings : it is true for every f,g,k,l in K, a fortiori for every f,g in F and k,l in K.

That makes a lot more sense, we have not discussed this notion of rings yet so I guess being a field gives you all that

Thanks a lot

All fields are rings. Not all rings are fields.

Usually rings are taught before fields in my experience. You will find that rings obey weird properties that fields do not.