two exercises concerning fields: order of a field

Hi there

I don't know how to solve these exercises.

Let be a field containing m elements and be positive.

Exercise 1)

Let . Show:

Exercise 2)

Let g: , . Find all elements of that map to

Notes: Those fields are finite at least are we doing this topic actually. m does not have to be a prime we have to do the proof for 1) in general.

And thats my problem for **exercise 1)**

If . The order of x is a divisor of m ? At least is true? I don't understand why this is true at this point already... If m is not prime (and we can't assume that) the number of units in doesn't have to be m-1 yeah???

If so the equation could be multiplied by x and what results is after subtracting x on both sides and this direction is proved.

Here from follows that so by dividing by x we have thus x has order m-1 so it is contained in the subset which consists of elements with order m-1 ????

Let's have a look at **exercise 2)**

So is isomorphic to where ?

?? Why?

...have no idea

Do you know how this works? Could you please give me a hint?

Regards

Huberscher

Re: two exercises concerning fields: order of a field

Hi,

How much do you know about finite fields? In particular, up to isomorphism, there is exactly one field of order q for q a power of a prime p. So you can assume:

Now in a finite field F, the multiplicative group of non-zero elements of F is cyclic with generator, say a. For your first problem, then, use this fact and the fact that in any field a polynomial of degree m has at most m zeros.

For your second question,

with . Also .

Re: two exercises concerning fields: order of a field

Hello and thank you for the reply.

I think now it's all clear but the following:

direction for **exercise 1)**

Obviously if x=0 then x is in that field. So if the order of x is a divisor of m-1.

Now it comes in that but how? which is cyclic regarding multiplication but why can I conclude now that x is in ? Is is just because the order of x is a divisor of m-1 and the number of elements in (=the units of ) is m-1 and that's it?

Regards

Huberscher

Re: two exercises concerning fields: order of a field

There are at most m values that satisfy . Now each is a root of this polynomial and there are m such. So the elements of form the complete set of solutions to . So if satisfies , .