Math (abstract algebra)?
Suppose that L is an extension of k and a,b in L. If a is algebraic over k of degree m and b is
algebraic over k of degree n, where m and n are relatively prime, show that [k(a, b) : k] = mn.
Suppose that L is an extension of k and a,b in L. If a is algebraic over k of degree m and b is
algebraic over k of degree n, where m and n are relatively prime, show that [k(a, b) : k] = mn.
Comments
Note that [k(a,b):k] = [k(a,b):k(a)][k(a):k] = [k(a,b):k(a)]m, which shows that m divides [k(a,b):k]. Symmetrically [k(a,b):k] = [k(a,b):k(b)][k(b):k] = [k(a,b):k(b)]n, showing that n divides [k(a,b):k]. Since m and n are relatively prime then mn divides [k(a,b):k].
Now note that if f(x) = the irreducible polynomial of b over k, and g(x) = irreducible polynomial of b over k(a), then g(x) divides f(x). So [k(a,b):k(a)] = deg g <= deg f = [k(b):k] = n. From [k(a,b):k] = [k(a,b):k(a)][k(a):k] <= nm.
Since [k(a,b):k] <= mn and mn divides [k(a,b):k] then [k(a,b):k] = mn.
that is not authentic. case in point, take F to be the sphere with p^n components (p best, n > a million). Then F is a commutative ring, that could't be isomorphic to Z_{p^n}, using fact Z_{p^n} incorporates 0 divisors and F does of direction no longer.
cant write it here