So the degree of is 2 since it cannot be 1
Besides, satisfies the irreducible polynomial
. 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 .
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 .