Question

Q: Asking for assistance in understanding and solving this example on Modern Algebra II with the steps of the solution to better understand, thanks.

**Please give the step by steps with details to completely see how the solution came about.

1) Determine all elements in an integral domain that are their own inverses under multiplication.

2) Let F be a finite field with n elements. Prove that x^(n-1 )= 1 for all nonzero x in F. Hint: Use the fact that the nonzero elements of F form a group under the multiplication operation.

Answer 1

1)

Definition: An integral domain is a commutative ring R with identity element 1, such that multiplication of any two non zero elements of R is non zero.

That means if R is an ID (ID stands for integral domain) , 'a' and 'b' is in R such that , then either a=0 or b=0.

Let, S be the set of all such elements in an ID 'R' which are self inverse.

i.e. S={ }

Let, 'a' belongs to S. Then,

implies, (Composing 'a' on both side)

As,

Now since R is an ID so,

either i.e.

or   i.e.  

This shows that the only possibilities are either a=1 or a=-1.

also observe that 1.1=1 and (-1).(-1)=1

Hence S={1,-1}.

2)

Definition: A field F is a commutative ring with identity 1 such that every non zero element of F has an inverse.

i.e. for any such that an element such that .

So look at {} i.e. all the non zero elements of F under the multiplication.

(i) Elements of F* are closed under multiplication:

let then from definition of F*  .

Then from definition of field both a and c have inverses. Let if possible , then   , Which is clearly a contradiction since .

Hence

So proved that F* is closed under multiplication.

(ii) F* contains 1 ( Since ) ,such that for all .

(iii) By definition of F* , every element in F* has an inverse.

(i), (ii), (iii) implies that F* is a group under multiplication. Also F* contains all the elements of F except 0. So order of the group F* is (n-1).

We know (Recall Group Theory) that order of an element in a group is divisor of the order of the group.

Let x be any element of F* such that O(x)=m. Then m divides |F*|=n-1.

So n-1=d*m for some natural no d.

Since O(x)=m so,

Since x was chosen arbitrarily from F* so this is true for any non zero element of F.

Hence proved that for all non zero x in F x^(n-1)=1.