Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

Can someone help me get started on the following problem.

Determine the degree over of and of

Peter

Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

So the degree of is 2 since it cannot be 1

Besides, satisfies the irreducible polynomial

Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

. Are you sure that is what you want? Or did you mean ? Assuming it is, . Then, and generate that field extension. So, is reducible over ? If not, then has degree 3 over . Try to figure out why the degree of over equals the degree of over times the degree of over .

Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

To see that is irreducible over , assume that it is reducible. Then with . Multiply that out and try to solve for . You find that (by equating coefficients). You find that any solution must have , and since , is indeed irreducible over .