By Takashi Ono (auth.)

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**Example text**

2. 5) and let fa be the minimal polynomial of a over G b: fa E Gb[X]. 2(3), we have degfa = [K: G b]. For a E G, by the definition of G b, we have fa( aa) = O. As a is a generator of Kover k, we have aa =1= a" when a =1= 7:, a, 7: E G, which implies that degfa 2! [G]. On the other hand, since the polynomial g(X) = TIaEG (X - aa) belongs to Gb[X] and has a as a root, we must have fa Ig; therefore we find that degfa ~ [G]. 7). 4). 7). 5). D. 4. t In particular, we have F/k is a Galois extension~F# is a normal subgroup ofG(K/k).

Next, for any c E G and a EN, f(cac- 1) = f(c)f(a)f(c)-1 = f(c)e'f(c)-1 = f(c)f(c)-1 = e', which means that cac -1 EN. From this we find that cN = N c, i. e. , N is a normal subgroup of G. Now, for Na E G / N, put f(Na) = f (a); as f (N) = {e'}, f is well-defined and one verifies easily that f is a surjective homomorphism G / N ~ G'. , Na = Nb; therefore f is injective. D. We can do the same thing for rings. 16. A mapping f: R~R' of a ring R to a ring R' is a homomorphism if we have f(a+b)=f(a)+f(b) and f(ab) = f(a )f(b), a, b E R.

D. 6. Let K :::J L :::J k be fields such that K/ k is finite. Prove that NK1ky = NLlk(NKILY) and tK1ky = tLlk(tKILY) for Y E K. EXERCISE Now we shall consider the discriminant for a finite extension K/k of number fields. For Yl1 ... , Yn E K, n = [K: k], put d Klk ( Yl1 ... ,Yn ) ~(d - et ( Yj(i»))2 , 1:$ i, j:$ n. Since this equality can also be written as it follows that d K1k (Yl1 ... , Yn) E k. In particular, for a primitive element a for K/k: K = k(a), we have DKlk(a)~fdKlk(1, a, ... , ~-l)= n (a(i)-a(j))2.