H. Normal Extensions.

An extension field E of a field F is called a normal extension if the group G of automorphisms of E which leave F fixed has F for its fixed field, and (E/F) is finite.

Although the result in Theorem 13 cannot be sharpened in general,there is one case in which the equality sign will always occur, namely, in the case in which a1, a 2, • • • , is a set of automorphisms which form a group. We prove

THEOREM 14. If <7j, <j2,• • • , On is a group of automorphisms of a field E and if F is the fixed field of q q . . , a , then (E/F) = n.

 

 

If at, a2,... T rrn is a group, then the identity occurs, say, c1 = I. The fixed field consists of those elements x which are not moved by any of the cr's, i.e., <7.( x) = x, i = 1, 2, . . • n. Suppose that (E/F ) > n. Then there exist n + 1 elements a a2, ■ ■ ■ > of EJ which are linearly independent with respect to F. By Theorem 1, there exists a non-trivial solution in E to the system of equations

 

We note that the solution cannot lie in F, otherwise, since a1 is the identity, the first equation would be a dependence between a,, . . . , a n .

 

Among all non-trivial solutions x1( x ■ ■ ■ , xn+1 we choose one which has the least number of elements different from 0. We may sup­pose this solution to be a2>. . . , a,, 0, . . . , 0, where the first r terms are different from 0. Moreover, r / 1 because a, <t1 (a ( ) = 0 implies al = 0 since <7t (a., ) = a^ ^0. Also, we may suppose ar = 1, since if we multiply the given solution by ar_1 we obtain a new solution in which the r-th term is 1. Thus, we have

 

 

for i = 1, 2, . . • , n. Since at, . . . , ar^ cannot all belong to F, one of these, say a 1? is in E but not in F. There is an automorphism crk for which ow{ a, ) ^ at . If we use the fact that ax, o2,. . . , on form a group,

 

is a permutation of

 

 

we see

 

which is a non-trivial solution to the system (' ) having fewer than t elements different from 0, contrary to the choice of r.

Corollary 1. If F is the fixed field for the finite group G, then each automorphism a that leaves F fixed must belong to G.

(E/F) = order of G = n. Assume there is a o not in G. Then F would remain fixed under the n + 1 elements consisting of <j and the elements of G, thus contradicting the corollary to Theorem 13.

Corollary 2. There are no two finite groups G1 and G2 with the same fixed field.

This follows immediately from Corollary 1.

Applying <jk to the expressions in ( *) we obtain

 

for j = l, 2, . . • , n, so that from crkcrj - ai

 

and if we subtract ( * * ) from ( * ) we have

 

If fl[x) is a polynomial in F, then fl[x) is called separable if its irreducible factors do not have repeated roots. If E is an extension of

the field F, the element a of E is called separable if it is root of a separable polynomial f(x) in F, and E is called a separable extension if each element of E is separable.

THEOREM 15. E is a normal extension of F if and only if E is the splitting field of a separable polynomial p(x) in F.

Sufficiency. Under the assumption that E splits p (x) we prove that E is a normal extension of F.

If all roots of p(x) are in F, then our proposition is trivial, since then E = F and only the unit automorphism leaves F fixed.

Let us suppose p(x) has n > 1 roots in E but not in F. We make the inductive assumption that for all pairs of fields with fewer than n roots of r>(x) outside of F our r)ror>osition holds.

 

be a factorization of p(x)

 

into irreducible factors. We may suppose one of these to have a degree greater than one, for otherwise p(x) would split in F. Suppose deg p,(x) = s > 1. Let aI be a root of p,(x). Then (F(a, )/F) = deg p,(x) = s. If we consider F ((7 ) as the new ground field, fewer roots of p( x) than n are outside. From the fact that p(x) lies in F(a, ) and E is a split­ting field of p(x) over F( ), it follows by our inductive assumption that E is a normal extension of F (a., ). Thus, each element in E which is not in F(a, ) is moved by at least one automorphism which leaves F(a, ) fixed.

p (x) being separable, the rootsa^,a2, - - - , as of p} (x) are a distinct elements of E. By Theorem 8 there exist isomorphisms

mapping respectively, which are each the identity on F and map a on a ,a ... a respectively. We now apply Theorem 10. E is a splitting

1 2' ' s

field of p(x) in F(a, ) and is also a splitting field of p(x) in F(a; ), Hence, the isomorphism uir which makes p( x ) in F ( ) correspond to the same p(x) in F( a ), can he extended to an isomorphic mapping of E onto E, that is, to an automorphism of E that we denote again by ait Hence, , (/2, . . . , cs are automorphisms of E that leave F fixed and map u, ontO a j, a2 , . • • a where the c. are in F. If we apply <7. to this equation we get, since

 

 

 

The polynomial

Now let 0 be an element that remains fixed under all automor­phisms of E that leave F fixed. We know already that it is in F (a ) and hence has the form has therefore the S distinct roots at, a2, ■ ■ ■ , a&. These are more than its degree. Sd all coefficients of it must vanish, among them c - 0. This shows $ in F.

Necessity. If E is a normal extension of F, then E is splitting field of a separable polynomial p(x). We first prove the

Lemma. If E is a normal extension of F, then E is a separable extension of F. Moreover any element of E is a root of an equation over F which splits completely in E.

be the group G of automorphisms of E whose id be fixed field is F. Let a be an element of E, ar the set of distinct elements in the sequence Since G is a group,

 

 

Therefore, the elements a,a2, . . . , af are permuted by the automorphisms of G. The coefficients of the polynomial fl(x) = (x-«)( X-U 2). . . (x-af) are left fixed by each automorphism of G, since in its factored form the factors of fl(x) are only permuted. Since the only elements of E which are left fixed by all the automorphisms of G belong to F, f(x) is a polynomial in F. If g(x) is a polynomial in F which also has a as root, then applying the automorphisms of G to the expression g (a ) = 0 we obtain g(a.) = 0, so that the degree of g(x) > s. Hence f[x) is irre­ducible, and the lemma is established.

To complete the proof of the theorem, let olt oj2, . . . , cjt be a gen­erating system for the vector space E over F. Let f.(x) be the separable polynomial having co. as a root. Then E is the splitting field of

 

 

If f(x) is a polynomial in a field F, and E the splitting field of f (x ), then we shall call the group of automorphisms of E over F the group of the equation f(x) = 0. We corne now to a theorem known in algebra as the Fundamental Theorem of Galois Theory which gives the relation between the structure of a splitting field and its group of automorphisms.

THEOREM 16. (Fundamental Theorem). If p(x) is a separable polynomial in a field F, and G the group of the equation p(x) = 0 where E is the

splitting field of p(x), then: (1) Each intermediate field, B,is the fixed field for a subgroup GB of G, and distinct subgroups have dis­tinct fixed fields. We say B and GB "belong" to each other. (2) The intermediate field B is a normal extension of F if and only if the sub- group GB is a normal subgroup of G. In this case the group of automor- phisms of B which leaves F fixed is isomorphic to the factor group (G/G ). (3) For each intermediate field B, we have (B/F) = index of GB and (E/B) = order of G,.

The first part of the theorem cornes from the observation that E is splitting field for p(x) when p(x) is taken to be in any intermediate field. Hence, E is a normal extension of each intermediate field B, so that B is the fixed field of the subgroup of G consisting of the automor­phisms which leave B fixed. That distinct subgroups have distinct fixed fields is stated in Corollary 2 to Theorem 14.

Let B be any intermediate field. Since B is the fixed field for the subgroup GB of G, by Theorem 14 we have (E/B ) = order of G,. Let us call o(G) the order of a group G and i(G) its index. Then o(G) =: o( G B) • i(G B). But (E/F) = o(G), and (E/F) = (E/B)-(B/F) from which (B/F) = i (G, ), which proves the third part of the theorem.

The number i( GB ) is equal to the number of left COSetS of G,. The elements of G, being automorphisms of E, are isomorphisms of B; that is, they map B isomorphically into some other sub field of E and are the identity on F. The elements of G in any one coset of GQ map B in the same way. For let q . <j1 and &. a2 be two elements of the coset crGn. Since a, and on leave B fixed, for each a in B we have

 

Elements of different cosets give

different isomorphisms, ior it <j and r give the same isomorphism, a (a) = r{a) for each a in B, then o "!r(a) = a for each a in B. Hence, 01t --- ox, where (7j is an element of GB. But then r ~ oo^ and rGB (70-jG^ <jGb so that o and r belong to the same coset.

Each isomorphism of B which is the identity on F is given by an automorphism belonging to G. For let a be an isomorphism mapping B on B" and the identity on F. Then under a, p(x) corresponds to p(x), and E is the splitting field of p(x) in B and of p( x) in B ' , By Theorem 10, o can be extended to an automorphism a* of E, and since a* leaves F fixed it belongs to G. Therefore, the number of distinct isomorphisms of B is equal to the number of cosets of GB and is there­fore equal to (B/F).

The field <tB onto which a maps B has obviously oGB<r 1 as cor­responding group, since the elements of <jB are left invariant by precisely this group.

If B is a normal extension of F, the number of distinct automor­phisms of B which leave F fixed is (B/F) by Theorem 14. Conversely, if the number of automorphisms is (B/F) then B is a normal extension, because if F 1 is the fixed field of all these automorphisms, then F c F' c B, and by Theorem 14, (B/F ') is equal to the number of automorphisms in the group, hence (B/F ') = (B/F). From ( B/F) = (B/F")(F7F) we have (F'/F) = 1 or F = F \ Thus, B is a normal extension of F if and only if the number of automorphisms of B is (B/F).

B is a normal extension of F if and only if each isomorphism of B into E is an automorphism of B. This follows from the fact that each of the above conditions are equivalent to the assertion that there are the same numberof isomorphisms and automorphisms. Since, for each

B = tfB is equivalent to <tGb(7 C GB, we can finally say that B is a normal extension of F and only if GB is a normal subgroup of G.

As we have shown, each isomorphism of B is described by the effect of the elements of some left Coset of G,. If B is a normal exten­sion these isomorphisms are all automorphisms, but in this case the cosets are elements of the factor group (G/GB ). Thus, each automor­phism of B corresponds uniquely to an element of ( G/GB ) and con­versely. Since multiplication in ( G/GB ) is obtained by iterating the mappings, the correspondence is an isomorphism between (G/Gfi ) and the group of automorphisms of B which leave F fixed. This completes the proof of Theorem 16.

Finite Fields.

It is frequently necessary to know the nature of a finite subset of a field which under multiplication in the field is a group. The answer to this question is particularly simple.

THEOREM 17. If S is a finite subset (^0 ) of a field F which is a group under multiplication in F, then S is a cyclic group.

The proof is based on the following lemmas for abelian groups. Lemma 1. If in an abelian group A and B are two elements of orders a and b, and if c is the least common multiple of a and b, then there is; an element C of order c in the group.

Proof: (a) If a and b are relatively prime, C = AB has the re­quired order ab. The order of C 3 = Ba is b and therefore c is divisible by b. Similarly it is divisible by a. Since Cab = 1 it follows c = ab.

If d is a divisor of a, we can find in the group an element of order d. Indeed Aa/d is this element.

 

 

Call tj the larger of the two numbers n. and ra . Then

 

Now let us consider the general case. Let pir p2, • • • , Pr be the prime numbers dividing either a or b and let

According to (b) we can find in the group an element of order p.1 and one of order p.1"1. Thus there is one of order p. i. Part (a) shows that the product of these elements will have the desired order c.

Lemma 2. If there is an element C in an abelian group whose order c is maximal (as is always the case if the group is finite) then c is divisible by the order a of every element A in the group; hence x = 1 is_ satisfied by each element in the group.

Proof: If a does not divide c, the greatest common multiple of a and c would be larger than c and we could find an element of that order thus contradicting the choice of c.

We now prove Theorem 17. Let n be the order of S and r the largest order occuring in S. Then xr - 1 = 0 is satisfied for all ele-ments of S. Since this polynomial of degree r in the field cannot have more than I roots, it follows that I > n. On the other hand f < n be­cause the order of each element divides n. S is therefore a cyclic group consisting of 1, f, f 2, . . . , en_1 where fn = 1.

Theorem 17 could also have been based on the decomposition theorem for abelian groups having a finite number of generators. Since this theorem will be needed later, we interpolate a proof of it here.

Let G be an abelian group, with group operation written as f. The element gj, . . . , gk will be said to generate G if each element g of G can be written as sum of multiples of g,, . . . , gk, g = n^gj + . . . + nkgk< If no set of fewer than k elements generate G, then g,, . . . , g^ will be called a minimal generating system. Any group having a finite genera­ting system admits a minimal generating system. In particular, a finite group al.ways admits a minimal generating system.

 

From the identity

 

 

it follows that if

 

 

generate G, also

 

 

generate G.

 

 

will be called a re­

 

An eouation lation between the generators, and nij, . . . , m^ will be called the co­efficients in the relation.

We shall say that the abelian group G is the direct product of its if each g f G is uniquely representable as a subgroups where

Decomposition Theorem. Each abelian group havini ber of generators is the direct product of cyclic subgroups

ig a finite num and n is where the order of G. divides the order of the number of elements in a minimal generating system.

may each be infinite, in which case, to be precise, we assume tne tneorem true tor all groups naving minimal genera­ting systems of k-1 elements. If n = 1 the group is cyclic and the theorem trivial. Now suppose G is an abelian group having a minimal generating system of k elements. If no minimal generating system satis­fies a non-trivial relation, then let g^ t g^, . . . , gfc be a minimal generating system and G1?G2, . . . , Gk be the cyclic groups generated by them. For each g ( G, g = l^gj + . . . + nkgk where the expression is unique) otherwise we should obtain a relation. Thus the theorem would be true. Assume now that some non-trivial relations hold for some mini­mal generating systems. Among all relations between minimal genera­ting systems, let be a relation in which the smallest positive coefficient occurs. After an eventual reordering of the generators we can suppose this coefficient to be mj, In any other relation between g . . . , g,. we must have n^/n^ Otherwise ^ = qcr^ + r, 0 < r < and q times relation (1) subtracted from relation (2) would yield a relation with a coefficient r < irij. Also in relation (1) we must have nij/m., i = 2,. . . , k.

For suppose nij does n°t divide one coefficient, say m2 . Then In the generating system we should have a relation where the coefficient

is minimal gen­

g,= g, i-q2g2+ • • • + qkgk, g2, . . •, gk

r contradicts the choice of m . Hence The system In any relation erating, and our presince m j is a coefficient in a relation between

vious argument yields nij nj , and hence

 

 

Let G be the subgroup of G generated by g2, . . . , gk and Gt the cyclic group of order nij generated by gj . Then G is the direct product of G. and G* . Each element g of G can be written implies hence so that the relation and also

The representation is unique, since

 By our inductive hypothesis, G 1 is the direct product of k-1 cyclic groups generated by elements g g . . . , gk whose respective orders t2, . . . , tk satisfy t. t.fl , i = 2, ... , k-1. The preceding argu­ment applied to the generators gj, g2, . . . , gk yields m1 t2, from which the theorem follows.

By a finite field is meant one having only a finite number of elements.

Corollary. The non-zero elements of a finite field form a cyclic

If a is an element of a field F, let us denote the n-fold of a, i.e.the element of F obtained by adding a to itself n times, by na. It is ob­vious that n • (m • a) = (nm ) - a and (n-a)(m>b) = nm • ab. If for one element a ^ 0, there is an integer n such that n • a = 0 then n • b = 0 for each b in F, since n. b = ( n . a) (a1 b) 0 - a"1 b = 0. < If there is a positive integer p such that p - a = 0 for each a in F, and if p is the smallest integer with this property, then F is said to have the charac- Jeristic.p. If no such positive integer exists then we say F has charac­teristic 0. The characteristic of a field is always a prime number, for if p = r • S thenpa = rs-a-r.(s-a).How ever, s . a = b ^ 0 if a ^ 0 and r b 0 since both r and S are less than p, so that pa / 0 contrary to the definition of the characteristic. If na = 0 for a ^ 0, then p divides n, for n = qp + r where 0 <r < p and na = (qp + r)a = qpa + ra. Hence na = 0 implies ra = 0 and from the definition of the characteristic since r < p, we must have r = 0.

If F is a finite field having q elements and E an extension of F such that (E/F) = n, then E has qn elements. For if , 6J2, . • • , oj is a basis of E over F, each element of E can be uniquely represented as

 

a linear combination

where the x. belong to

F. Since each can assume q values in F, there are qn distinct possible choices of x . . . , x and hence qn distinct elements of E. E is

 

(The non­

 

finite, hence, there is an element a of E so that

zero elements of E form a cyclic group generated by a).

If we denote by P s [0,1, 2 , ... , p-1] the set of multiples of the unit element in a field F of characteristic pf then P is a subfield of F having p distinct elements. In fact, P is isomorphic to the field of integers reduced mod p. If F is a finite field, then the degree of F over

P is finite, say (F/P) - n, and F contains pn elements. In other words, the order of any finite field is a power of its characteristic.

If F and F' are two finite fields having the same order q, then by the preceding, they have the same characteristic since q is a power of the characteristic. The multiples of the unit in F and F1 form two fields P and P' which are isomorphic.

The non-zero elements of F and F1 form a group of order q-1 and, therefore, satisfy the equation xq_1 1-0. The fields F and F 1 are splitting fields of the equation x l"1 = 1 considered as lying in P and P' respectively. By Theorem 10, the isomorphism between P and P can be extended to an isomorphism between F and F 1. We have thus proved

THEOREM 18. Two finite fields having the same number of ele- ments are isomorphic.

 

Differentiation.

nomial in a field F, then we define

 

 

 

The reader may readily verify that for each pair of polynomials f and g we have

THEOREM 19. The polynomial f has repeated roots if and only if in the splitting field E the polynomials f and f ' have a common root. This condition is eauivalent to the assertion that f and f ' have a where common factor of degree greater than 0 in F.

If a is a root of multiplicity k of f(x) then This gives then a is a root of f' of multiplicity at least k-1. If Thus, f and f' have a root a in common if and only if a is a root of f of multiplicity greater than 1.

If f and f' have a root a in common then the irreducible polynomial in F having a as root divides both f and £' . Conversely, any root of a factor common to both f and f1 is a root of f and f 1 .

Corollary. If F is a field of characteristic 0 then each irreducible polynomial in F is separable.

Suppose to the contrary that the irreducible polynomial fl[x) has a root a of multiplicity greater than 1. Then, f 1 (x) is a polynomial which is not identically zero (its leading coefficient is a multiple of the leading coefficient of f(x) and is not zero since the characteristic is 0) and of degree 1 less than the degree of f(x). But a is also a root of f (x) which contradicts the irreducibility of f(x).