L. Kummer's Fields.


If F contains a primitive nth root of unity, any splitting field E of a polynomial


i = 1,2, . . . , r will be called a Kummer extension of F, or more briefly, a Kummer .field,


xn - 1

cannot have more than q distinct roots. But we assumed that F has a primitive n root of unity and would be If a field F contains a primitive nth root of unity, the number n is not divisible by the characteristic of F. Suppose, to the contrary, F has characteristic p and n = qp. Then yp _ 1 = (y - 1 )p since in the expansion of (y - 1 )p each coefficient other than the first and last is divisible by p and therefore is a multiple of the p-fnld of the unit of F and thus is equal to 0. Therefore n distinct roots of It follows that n is not divisible by the characteristic of F. For a Rummer field E, none of the factors has are separable, 90 that_E has repeated roots since the derivative only the root 0 and has thereto re no roots m common with Therefore, the irreducible factors of is a normal extension of F.


Let ce ^ be a root of are the n distinct nth roots of unity in will be n distinct 90 that and hence will be the roots of roots of


Let o and t be two automorphisms in the group  G of E over F. For each a., both q and j map aL on some other root of  

where fi(J and * are nth and Thus roots of unity in the basic field F. It follows that

Since oand r are commutative over the generators of E, they commute over each ele­ment of E. Hence, G is commutative. If q f G, then a(a- } = (■„a , a2(a.) = f. 2a., etc. Thus, «-"'(a.) = a. for n. such that

iff 1 v i/ lG i     1          i           i

e. * =1. Since the order of an n th root of unity is a divisor of n, we have a a divisor of n and the least common multiple mofnI,n2,...,nr is a divisor of n. Since q ) = a{ for i = 1( 2, . • • , r it follows that m is the order of ex. Hence, the order of each element of G is a divisor of n and, therefore, the least common multiple r of the orders of the ele­ments of G is a divisor of n. If { is a primitive nth root of unity, then fn/r is a primitive r th root of unity. These remarks can be summarized in the following.

contains a primitive nth root of unity, then: (a) E is a normal extension of F; (b) the group G of E over F is abelian, (c) the least common multi­ple of the orders of the elements of G is a divisor of n.

Corollary.. If E is the splitting field of xp - a, and F contains a primitive pth root of unity where p is a prime number, then either E = F and xp ~ a is split in F, or xp a is irreducible and the group of E over F is cyclic of order p.

The order of each element of G is, by Theorem 23, a divisor of p and, hence, if the element is not the unit its order must be p. If a is a root of xp - a, then a,«z, . . . ,fp'!a are all the roots of xp - a 90 that F(a ) = E and ( E/F ) < P • Hence, the order of G does not exceed p 90 that if G has one element different from the unit, it and its powers must constitute all of G. Since G has p distinct elements and their behavior is determined by their effect on a, then a must have p distinct images. Hence, the irreducible equation in F for a must be of degree p and is therefore xp a = 0.

The properties (a), (b) and (c) in Theorem 23 actually characterize Kummer fields.

Let us suppose that E is a normal extension of a field F, whose group G over F is abelian. Let us further assume that F contains a primitive rth root of unity where r is the least common multiple of the orders of elements of G.

.THEOREM 23. If E is a Kummer field, i.e., a splitting field of whete a, lie in F, and F

The group of characters X of G into the group of r th roots ofunity is isomorphic to G. Moreover, to each a t G, if o é 1, there exists a character C fX such that C (a) ^ 1. Write G as the direct product of the cyclic groups

?f orders m1 may be written m2 . . . m,. Each g ( G Call C. the character sending ai  into ( ., a primitive m,ih root of unity and g into 1 for j ^ i. Let C be Conversely, C ' 1 . . .           defines a character. Since the order of C. is any character.


then we have


Thus at least one vsay i/ , is not divisible by ml-

Let A denote the set of those non-zero elements a of E for which a' f F and let F^ denote the non-zero elements of F. It is obvious that A is a multiplicative group and that Fj is a subgroup of A. Let Ar de­note the set of rth powers of elements in A and FJ the set of r powers of elements of F1. The following theorem provides in most applications a convenient method for computing the group G.

THEOREM 24. The factor groups (A/F, ) and (Ar/F j) are iso- morphic to each other and to the groups G and X.

We map A on A' by making a ( A correspond to aT ( A'. If ar ( FJ, where a f F then b ( A is mapped on e? if and only if br = ar, that is, if b is a solution to the equation xr - ar = 0. But a, ea, f2a, . . . , a are distinct solutions to this equation and since ( and a belong to Fj, it follows that b must be one of these elements and must belong to F1. Thus, the inverse set in A of the subgroup FJ of Ar is Fj , so that the factor groups (A/F ) and (Ar/FJ ) are isomorphic.

m^, the character group X of G is isomorphic to G. If o ^ 1, then in


If a is an element of A, then

Hence,  a/<j{a) is an rth root of unity and lies in F . By Theorem 22, a/(j (a ) defines a character C (a) of G in F. We map a on the corresponding character C. Each character C is by Theorem 22, image of some a. Moreover, a . a 1 is mapped on the character C * (<7) =so

that the mapping is homomorphism. The kernel of this homomorphism is the set of those elements a for which a/o(a) = 1 for each o, hence is Fj. It follows, therefore, that (A/F1 ) is isomorphic to X and hence also to G. In particular, (A/Ft ) is a finite group.

We now prove the equivalence between Kummer fields and fields satisfying (a), (b) and (c) of Theorem 23.

THEOREM 25. If E is an extension field over F, then E is a Kummet field if and only if E is normal, its group G is abelian and F contains a primitive rth root ( of unity where r is the least common multiple of the orders of the elements of G.


theorem. Tb this end it suffices to show that

The necessity is already contained in Theorem 23.. We prove the)e the cosets sufficiency. Out of the group A, let of Fj. Since ß ( A, we have

Thus, Q; is a root of the md since are also roots, equation must split in E. We prove that E is the splitting field of vhich will complete the proof of the


Suppose that

Then intermediate field between F and E, and since E is normal over which


F(a,,... , a ) there exists an automorphism

Moreover, A C F(a,, . . . ,a, ) since all the cosets «jFj are Contained in F(a,, . . . ,a,). Since F(a,, . . .,a() is by assumption left fixed by a, a(a) = a which contradicts a/a(a) ^ 1. It follows, therefore, that Corollary. If E is a normal extension of F, of prime order p, and if F contains a primitive pth root of unity, then E is splitting field of an irreducible polynomial xp — a in F.

E is generated by elements dj, . . . , an where n? f F. Let at be not in F,. Then xp - a is irreducible, for otherwise F(a ^ ) would be an intermediate field between F and E of degree less than p, and by the product theorem for the degrees, p would not be a prime number, Con­trary to assumption. E = F ( ax ) is the splitting field of xp - a. M. Simple Extensions.

We consider the question of determining under what conditions an extension field is generated by a single element, called a primitive. We prove the following is an leaves F(dj, . . . , a,) fixed. There exists a character C of G for which C(a) jL 1. Finally, there exists an element a in E such that


->y Theorem 22, hence a t A.

THEOREM 26. A finite extension E of F is primitive over F ifand only if there are only a finite number of intermediate fields.

Let E = F(a) and call f(x) = 0 the irreducible equation for a in F. Let B be an intermediate field and g(x) the irreducible equa­tion for a in B. The coefficients of g(x) adjoined to F will generate a field B 1 between F and B. g ( x ) is irreducible in B, hence also in B ' . Since E = B'(a) we see (E/B) = ( E/B ' ). This proves B' = B. So B is uniquely determined by the polynomial g(x). But g(x) is a divisor of fl(x), and there are only a finite number of possible divisors of fl[x) in E. Hence there are only a finite number of possible B's.

Assume there are only a finite number of fields between E and F. Should F consist only of a finite number of elements, then E is generated by one element according to the Corollary on page 53. We may therefore assume F to contain an infinity of elements. We prove: To any two elements a, /3 there is a y in E such that F (a, fi ) = F (y). Let y ~ a + a^S with a in F but for the moment undetermined. Con­sider all the fields F(y) obtained in this way. Since we have an infinity of a's at our disposal, we can find two, say a, and a„, such that the corresponding and yield the same field

Since both y and y2 are in F ( y )7  their difference (and therefore y3) is in this field. Consequently also tention is proved. Select now v in E in such a way that


Since our con as large as possible. Every element e of E mustbe in F( r/) or else we could find an element § such that F(6) contains both r/ and e, This






is a finite extension of


the field F, and a a2> • • • , an are separable elements in E, then there exists a primitive 0 in E such that

Proof: Let L(x) be the irreducible equation of aj in F and let B be an extension of E that splits fi(x) f2( x) . . . f,(x). Then B is normal over F and contains, therefore, only a finite number of intermediate fields (as many as there are subgroups of G). So the subfield E contains only a finite number of intermediate fields. Theorem 26 now completes the proof.