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Definition 8.2.1. Let f(x) be a polynomial in K[x], and let F be a splitting field for f(x) over K. If f(x) has the factorization
f(x) = (x - r1)m1 (x - r2)m2 · · · (x - rt)mtover F, then we say that the root ri has multiplicity mi.
Definition 8.2.2. Let f(x) be a polynomial in K[x], with f(x) = ak xk. The formal derivative f'(x) of f(x) is defined by the formula
f'(x) = k ak xk-1,where k ak denotes the sum of ak added to itself k times.
Proposition 8.2.3. The polynomial f(x) in K[x] has no multiple roots if and only if gcd(f(x),f'(x)) = 1.
Proposition 8.2.4. Let f(x) be an irreducible polynomial over the field K. Then f(x) has no multiple roots unless chr(K) = p 0 and f(x) has the form
f(x) = a0 + a1 xp + a2 x2p + · · · + an xnp.
Definition 8.2.5. A polynomial f(x) over the field K is called separable if its irreducible factors have only simple roots.
An algebraic extension field F of K is called separable over K if the minimal polynomial of each element of F is separable.
The field F is called perfect if every polynomial over F is separable.
Theorem 8.2.6. Any field of characteristic zero is perfect. A field of characteristic p>0 is perfect if and only if each of its elements has a pth root.
Corollary 8.2.7. Any finite field is perfect.
Theorem 8.2.8. Let F be a finite extension of the field K. If F is separable over K, then it is a simple extension of K.
8. Let f(x) be a polynomial in Q[x] be irreducible over Q, and let F be the splitting field for f(x) over Q. If [F:Q] is odd, prove that all of the roots of f(x) are real. Solution
9. Find an element a with Q( , i) = Q(a). Solution
10. Find the Galois group of x6-1 over Z7. Solution
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