SOME PROPERTIES OF EXTENSIONS OF SMALL DEGREE OVER Q. 1. Quadratic Extensions

SOME PROPERTIES OF EXTENSIONS OF SMALL DEGREE OVER Q TREVOR ARNOLD Abstract This aer demonstrates a few characteristics of finite extensions of small degree over the rational numbers Q It comrises attemts at classification of all such extensions of degrees, 3, and 4, as well as mention of how one might begin the rocess of classifying all finite fields with degree a ower of Some toics discussed are discriminants of certain fields, how to reresent a given extension as a simle extension, and some alied Galois theory 1 Quadratic Extensions It is fairly clear why quadratic extensions are in most resects the easiest to understand: they all have the same Galois grou (Z ), their degree over Q is as small as ossible without being trivial, and the number is in many ways very easy to work with 11 The Make-u of a Quadratic Number Ring We begin by comletely classifying the elements of the number ring associated with any number field of the form Q( m), where m 1 or 0, is squarefree (ie a quadratic number field) This ring is by definition simly A Q( m), where A is the set of algebraic integers, defined here to be the set {α C : f(x) Z[x] f(α) 0, f(x) monic and irreducible} 111 Preliminaries We know that [Q( m) : Q], and so any element of A Q( m) is going to have a minimum olynomial f(x) with deg f(x) Thus we know that the algebraic numbers in Q( m) will simly be those elements α C such that f(α) 0 for some f(x) Z, monic and irreducible, with deg f(x) Our goal in section?? will be to show that any root of such a olynomial can be exressed in one of the following ways: as a + b m, where a, b Z and m or 3 (mod 4), for some squarefree m or as a+b m, where a, b Z, a b (mod 4), and m 1 (mod 4), for some squarefree m Date: May 15, 000 1

TREVOR ARNOLD This is trivial to show if deg f(x) <, and so we will only give a treatment of the case where f is a genuine quadratic 11 Roots of quadratic olynomials Take any f(x) x + cx + d Z[x] By an elementary theorem of arithmetic, we know that f(x) 0 x c ± c 4d Choosing t and m aroriately, we can write c 4d t m where m is squarefree Since t 0, 1,, or 3 (mod 4), it follows that t 0 or 1 (mod 4) This gives us two cases If t 0 (mod 4), we have 4 c 4d 4 c c c ± t m Z, and hence we know that any root of f(x) has the form a + b m, a, b Z This is what we want for m or 3 (mod 4), and for m 1 (mod 4), we simly note that have a + b m a + b m where 0 a b (mod ) If t 1 (mod 4), consider the ossibility that m or 3 (mod 4) We would c 4d c 1 or 1 3 or 3 (mod 4), which is an imossibility for the square of an integer (see above) Thus we must have m 1 (mod 4) since m is assumed to be squarefree, and therefore c 4d c t m 1 1 1 (mod 4) Hence t and c are both odd (or else t c 0 (mod 4)) From all of this it follows that when t 1 (mod 4), c ±t (mod ) and m 1 (mod 4) Thus we have shown exlicitly that all roots of a monic quadratic olynomial f(x) Z[x] have the form stated above, from which it directly follows that any α A Q( m) also has this form for some m 1 The Discriminant Now that we have shown what an element of A Q( m) has to look like, we want to know if everything of this form has to be in A Q( m) for some m; more secifically, we want to show A Q( {a + b m} m or 3 (mod 4) m) { } a+b m : a b (mod ) m 1 (mod 4)

SOME PROPERTIES OF EXTENSIONS OF SMALL DEGREE OVER Q 3 Once we know this, it will be clear that the following will be a set of generators for A Q( m) as a Z-module (ie an integral basis): { 1, m } when m or 3 (mod 4), { 1, 1 + 1 } m when m 1 (mod 4) From this we will be able to directly comute the discriminant of Q( m) 11 Preliminaries Although we have shown exlicitly what the elements of a number field look like, we have as yet to show that if a+b m 1 A Q( m ) then necessarily m 1 m We do this here in rearation for demonstrating that the generators are exactly what we want them to be It is clear that the result will follow immediately if we can show so this how we shall roceed Q( m 1 ) Q( m ) Q, From elementary field theory, we want to show that one of the two conditions Q( m 1 ) Q( m ) Q Q( m 1 ) Q( m ) must hold, which we will be able to conclude from the following: [ Q( m 1 ) : Q ] [ Q( m ) : Q ] both Q( m 1 ) and Q( m ) contain Q( m 1 ) Q( m ) [ Q( m 1 ) Q( m ) : Q ] 1 or Thus all that we will need to show is that if there is some β / Q with β Q( m 1 ) Q( m ), then it must be the case that m 1 m There will be such a β a 1 + b 1 m1 a + b m with a i, b i Q if and only if we can find rational numbers a and b such that a + b m 1 m But then we would have the following: m a + ab m 1 + b m Z a 0 or b 0 If a 0 then m b m 1 Say b c d (where c, d Z are relatively rime, c 0, and d 0) We then have d m c m 1 First off, we may assume that c 1, because if so, we would either have m 1 not squarefree (in the case that d 1), or we would have m 1 m (if d 1), which we assumed wasn t the case But since (c, d) 1, we know that (c, d ) 1, and hence that for any dividing c, it is also the case that m, contradicting the assumtion that m is squarefree On the other hand, if b 0, then m a, again contradicting the fact that m is

4 TREVOR ARNOLD squarefree (since we know in this case that a, and hence a, is necessarily in Z since a Q) Thus we know there can be no such β, and hence Q( m 1 ) Q( m ) Q 1 A comlete descrition of A Q( m) What we have shown so far (combining sections?? and??) is that A Q( {a + b m} m or 3 (mod 4) m) { } a+b m : a b (mod ) m 1 (mod 4) What we want to show now is, which will rove to be relatively straightforward Once we have done this, all that we need to do to find the discriminant will be a short, simle calculation Let a+b m be in one of the sets on the right hand side of the above equation (where a b (mod ) and where a and b are odd only if m 1 (mod 4)) This is clearly a root of the following olynomial: x ax + a b m 4 Thus all we need to show is that a b m 4 Z We consider two cases If a and b are even then 4 a and 4 b so 4 a b m If a and b are odd, then (by assumtion) m 1 (mod 4) Since a and b are odd, we have a b 1 (mod 4) a b m 1 1 1 0 (mod 4) 4 a b m x ax + a b Z[x] We can therefore conclude that A Q( m) is in fact equal to what we want it to be 13 The discriminant: summing u Now that we have our hands on exactly what the makeu of a quadratic number ring is, we can take the integral basis that we have found and lug them into the defining formula for the discriminant For m or 3 (mod 4), we know that {1, m} is an integral basis, and that the only conjugate of m is m Hence: disc ( Q( m) ) 1 m 1 ( m m ) ( ) m 4m m { For m 1 (mod 4), a set of generators over Z is 1+ m is 1 m Hence: disc ( Q( m) ) 1 1 1+ m 1 m ( 1 m 1, 1+ m } ; the conjugate of 1 + ) m ( m ) m

SOME PROPERTIES OF EXTENSIONS OF SMALL DEGREE OVER Q 5 And so we have established the following formulae for the discriminant of any quadratic number field: disc ( Q( m) ) 4m if m or 3 (mod 4) m if m 1 (mod 4) Degree 3 Extensions Degree 3 extensions are nearly as routine as those of degree, and for many of the same reasons, eg their Galois grous are always isomorhic to Z 3 In this section, we will show which cubic olynomials have slitting fields of degree 3 over Q and what the discriminant of a simle radical extension of degree 3 over Q is, following the outline on 45 50 of [?] (and Ex 41 in articular) 1 Cubic Polynomials In this section, it will be necessary to introduce the concet of the discriminant of a olynomial Say f(x) F [x], F a field, has roots x 1,, x n in its slitting field K F The discriminant D f of f is defined by the formula D D f (x i x j ) 1 i<j n It is clear that for any φ G(K : F ), which acts by necessity as a ermutation on the roots, then φ(d) D Since D is fixed by every element of the Galois grou, it must be the case that D F We roceed to find necessary and sufficient conditions for a cubic olynomial to have a slitting field of degree 3 over Q (as outlined on 13 of [?]) Assume that f(x) Q[x] is monic and irreducible of degree 3, say f(x) x 3 + ax + bx + c It can be shown by lengthy and tedious calculation (or the command [> discrim(x^3 + a*x^ + b*x + c, x); in Male) that in fact D f 7c +18cab+a b 4a 3 c 4b 3 It will be convenient to define the value d to be a square root of D, ie d ± (x i x j ), 1 i<j n where we may choose the sign arbitrarily It is clear that d K, the slitting field for f over Q Also convenient will be the variable change x u a/3, which yields f(x) g(u) u 3 + u + q for some, q Q We note that D g D f since our variable change simly translates the roots of f, and hence, for our uroses, we may assume that a 0 to

6 TREVOR ARNOLD begin with Thus we have D f 7c 4a 3 c The Galois grou G of K over Q can be viewed as a subgrou of S 3, the symmetric grou on 3 letters, acting on the three roots of f (which must be distinct since f is irreducible, and hence searable, over Q) Furthermore, G must act transitively on the roots of f The only transitive subgrous of S 3 are S 3 itself and A 3 Z3, the alternating subgrou comrising all even ermutations of S 3 It is clear that if φ Q is even, then φ(d) d (indeed, this is one way to define the term even ), ie G A 3 Z3 if and only if φ(d) d for each φ G, ie if and only if d Q Since the Fundamental Theorem of Galois Theory tells us that the degree of a Galois extension is equal to the order of its Galois grou, we have shown that [K : Q] 3 if and only if d Q, ie D is a square in Q Discriminants Following the sketch in [?], we will now show how to find the discriminant of Q( 3 m) where m is a cubefree integer We will need to use a standard theorem from algebraic number theory (taken from [?]), stated here without roof: Theorem If R is a number ring and α R has degree n over Q, then there is a basis for R over Z of the form 1, f 1(α),, f n 1(α) d 1 d n 1 where d i Z with d 1 d d n 1 and f i Z[x] is monic and of degree i The idea, of course, is to use this to find an integral basis for Q( 3 m) in this form, after doing which it will be a straightforward calculation to find the discriminant of this field We begin by verifying some needed facts 1 A few rearations Let K be an extension of Q of degree n and choose any two subsets of K {η 1,, η n } and {θ 1,, θ n } which generate the same Z- submodule Then there is a matrix M over Z such that η 1 M θ 1 η n θ n If σ 1,, σ n are the n isomorhisms between K and other subfields of C, then we may aly each σ i to the system of equations above to yield the matrix equation [ σj (η i ) ] M [ σ j (θ i ) ], which yields, according to the definition of the discriminant of an n-tule, disc (η 1,, η n ) M disc (θ 1,, θ n )

SOME PROPERTIES OF EXTENSIONS OF SMALL DEGREE OVER Q 7 Since the determinant of M must be an integer, we may conclude that in fact disc (η 1,, η n ) disc (θ 1,, θ n ) Equality follows from the fact that our argument is symmetric It is an easy alication of this result that disc (α) disc ( 1, f 1 (α),, f n 1 (α) ) for α R and it follows from the definition of the discriminant that ( ) disc (α) (d 1 d d n 1 ) disc ( Q( 3 m) ) since 1, f 1 (α),, f n 1 (α) is an integral basis for Q( 3 m) Now choose two ositive integers i and j such that i + j < n Then we know that f i (α)f j (α) is monic of degree i + j and hence f i (α)f j (α) d i d j d i+j + i+j k1 m kd 1 d k f k (α) d i+j m k Z, imlying that d i d j d i+j Inductively then we can see that for i < n, d i 1 d i, and hence by (??), we have that d n(n 1) 1 disc (α) An integral basis The general manner in which we will roceed to find our integral basis for Q( 3 m) is by considering and rejecting various ossibilities In order to kee notation consistent with the receding section, we set K Q( 3 m) and α 3 m Furthermore, we choose two relatively rime, squarefree integers h and k such that m hk To start things off, we will show that disc (α) 7m To do this, we use the formula for the discriminant given by disc (α) ±N K/Q ( m α,q (α) ), where m α,q (x) is the minimum olynomial for α over Q and + holds if and only if n 0 or 1 (mod 4) Since m α,q (x) x 3 m, disc (α) N K/Q (3α ) 3α 3ζα 3ζ α 7m, where ζ is a rimitive 3 rd root of unity Thus, by the fact that d 6 1 disc (α) 7m, we know that d 1 1 or 3, and can only equal 3 when 9 m However, if 9 m we have β (α + a)/3 R for some integer a But then Tr(β 3 ) Z, and ( 1 Tr(β 3 ) Tr 7 m + 1 9 aα + 1 9 a α + 1 ) 7 a3 m + a3 Z 9 imlies that 3 a (since 9 m), which is imossible since then α/3 R and hence m/7 R, contradicting the fact that m is cubefree (the only rational numbers in a number ring are integers) Thus we have shown that d 1 1 in any case, and hence we may choose f 1 (α) α

8 TREVOR ARNOLD Now we attack f and d We will need the fact that α /k R, which is shown simly by the fact that it is a root of x 3 h k Now let m ±1 (mod 9) and let β (α 1) /3 It is easy to comute that β 3 β + 1 ± m (m 1) β 0, 3 7 which tells us that β R (which follows from our assumtion that m ±9 (mod 9)), and so too is αβ/3 (α ± k α + k )/3k Note that this imlies that when m ±1 (mod 9), 3k d (we have shown before that k d when m ±1 (mod 9)) By the revious section, we know that d disc (α) 7m, imlying that d 3m In order to work with the coefficients of f, it will be convenient to say f (α) α + aα + b for some integers a and b In order to start determining which rimes can divide d, we now choose a rime number such that 3, m, and m If d then it is clear that (α + aα + b)/ R Taking the trace of this yields ( α ) + aα + b Tr 3b Z, and hence b, imlying that (α + aα) R But then cubing this yields ( m + 3mα + 3a α + a 3 ) m Tr 3 3(m + a 3 m) 3 Z, and hence 3 m(m + a 3 ), imlying that m + a 3 (since m) But then (again since m) a 3, and it follows that a 3 and hence m, a contradiction Thus no such rime can divide d So suose 3 and m We know from above that k d Since m, it must be that k, and hence d If d, by the same reasoning as above we obtain 6 m(m + a 3 ) and hence 3 m, another contradiction Thus d for such rimes At this oint we know that d 3 i k for some i If we square f (α)/d, we get the following: mα + a m + a α + abα + b d R, which imlies that d divides all of a + b, m + ab, and b + am, since no ratio in R can have a denominator greater than d To find the ower of 3 dividing d, first consider the case where 3 m We have already shown that d 3m, and so it is easy to conclude that 9 d When m ±1 (mod 9), we already know that d is divisible by 3 since it is divisible by 3k If m ±1 (mod 9), suose 3 d If a or b 0 (mod 3), then m 0 (mod 3) since 3 divides m + ab However, we know that a + b 0 (mod 3) and so a 1 b (mod 3), imlying that m ab a (mod 3) All of this

SOME PROPERTIES OF EXTENSIONS OF SMALL DEGREE OVER Q 9 imlies that (α + mα + 1)/3 is in R If m (mod 3), then we can see that (α 1) /3 R Hence ( ) (α 1) 4 Tr 8m 56m + 1 Z, 3 7 and so m(m ) 1 (mod 9) It is easy to check that the only way this can haen is when m 1 (mod 9), contradicting our assumtion A similar contradiction arises if we ick m (mod 9), and hence our initial assumtion that 3 d must have been incorrect Now we consider the case where 3 divides m but 9 does not Assume 3 d Then 3 also divides a + b and b + am Hence 3 b and also 3 a Thus α /3 is in R, and so m /7 is in Z, contradicting our assumtion that 9 m Our final case is when 9 m Assume now that 9 d Then 9 b + am imlies 9 b Then at least 3 b Then since 9 a + b, we have 9 a, which imlies that 9 b as well But then we recall from above that (α + aα)/3 R, and so 9 3 m(m + a 3 ), which, if we stare at it long enough, leads to 9 m, a contradiction 3 The discriminant finally Although the last section was a bit technical an somewhat obscure, it none the less got the job done Though at first it might not be aarent, we have in fact shown that 3k if m ±1 (mod 9) d, k if m ±1 (mod 9) which also means that an integral basis for Q( 3 m) has the form } {1, α, α when m ±1 (mod 9), k {1, α, α ± k α + k } when m ±1 (mod 9) 3k Now that we have found an integral basis, calculation of the discriminant amounts to calculating the square of the determinant of a 3 3 matrix We will not erform this calculation here, but the result is disc ( Q( 3 m) ) 3kh if m ±1 (mod 9) 7kh if m ±1 (mod 9) To sum u, the amount of work required to find the discriminant of even a small class of cubic extensions surasses that which is required to determine the discriminant of all quadratic extensions, affirming the large increase in comlexity associated with even a small increase in degree

10 TREVOR ARNOLD 3 Quartic Extensions Degree 4 extension are, dans un certain sens, the first interesting case, at least when it comes to the determination of Galois grous This section will be entirely devoted to the classification of the Galois grous of extensions of degree 4 over Q There are only two grous of order 4: V, the Klein 4-grou, and Z 4, the cyclic grou of order 4 31 Biquadratic Extensions We first consider the easiest case, ie when K (finite and Galois over Q) has G(K : Q) V The grou V has two distinct subgrous of order, which, by the Fundamental Theorem of Galois Theory (FTGT), corresond to two distinct subfields of degree over Q These two subfields can be written in the form Q( m) and Q( n) where m n and both are squarefree integers The smallest field containing both of these is, of course, Q( m, n), which has degree 4 over Q Thus we can conclude that every extension K with Galois grou over Q isomorhic to V is in this form Going the other way, if we have two squarefree integers m n, then every element of G ( Q( m, n) : Q ) has at most order, since the only conjugate of m (or n) is m (or n) This tells us that the Galois grou must be isomorhic to V, as Z 4 has two elements of order 4 3 Cyclic Extensions Now the more interesting case We aroach cyclic extensions in the natural way: namely, we will view them as degree extensions of quadratic extensions Thus they will all have the form Q( m)( α), where α Q( m) and hence α a + b m for some a, b Q We know that if b 0, then we have a Galois extension with Galois grou isomorhic to either V or Z, deending on whether Q( a) Q( m) or not If a 0, then the extension looks like Q( 4 d) (where d b m) The conjugates of 4 d over Q are the other roots of x 4 b m: ζ 4 d, ζ 4 d, and ζ 3 4 d, where ζ is a rimitive 4 th root of unity Then it is easy to see that if we are working with a Galois extension, then i Q( 4 d), which means that Q(i) Q( 4 d), and so this must be the unique quadratic subextension of degree (unique by the FTGT) Hence we may assume that m 1 to begin with, and so our extension is nothing more than Q( ib) Since i 1 + 1 i, we know that ib b + i b, and hence Q( ib) Q(i, b), which means that the Galois grou of Q( ib) is isomorhic to either Z or V, since the Galois grou of Q(i, b) has already been shown to be one of these Hence we know that if we are looking for a cyclic extension, we need only consider the cases where a 0 b 31 Some necessary and sufficient conditions So let us choose a, b Q, not both zero, such that Q ( a + b m ) is a Galois extension of Q (m, as usual, is

SOME PROPERTIES OF EXTENSIONS OF SMALL DEGREE OVER Q 11 taken to be squarefree) Our field must contain, then, not only a + b m, but also a b m, since they are conjugates (from now on we will call them α and ᾱ, resectively) The other two conjugates differ by only a negative sign, and so we can correctly claim that our field is a Galois extension if and only if it contains ᾱ We now roceed to rove a rather general statement which can be alied to our secific case This statement is: for a field K with char K and any elements a, b K, not squares in K, K( a) K( b) if and only if a x b for some x K Let us first assume the equality of the two fields Then a y + z b for some y, z K We then get a y + yz b + z b Now we can see that either y 0 or z 0 (but not both), or else a / K If z 0, then a is a square, contra our assumtion Hence y 0 and we get a z b where z K Now assume that a x b for some x K Then a ±x b K( b) and b ± 1 x a K( a) Hence K( a) K( b) We can aly this directly to our situation We have Q(α) Galois if and only if Q(α) Q(ᾱ), if and only if α c ᾱ for some c Q( m), and this holds if and only if a + b m a b m (a + b m) a b m c, ie if and only if a b m is a square in Q( m), say a b m (d + e m) d + ed m + e m As before, either d 0 or e 0 In the latter case, we have a b m d We will show that in this case, our extension does not have a Galois grou isomorhic to Z 4 (which is unfortunate since it is rather easy to find all solutions of this equation for a and b) First we note that a b m ᾱα d, and so α d/ᾱ Thus if we consider any element φ of the Galois grou, the order of φ can be at most since this is certainly the case if φ takes α to α, and if φ takes α to ᾱ, we have ( φ (α) φ ± d ) ± d α φ(α) ± d ± d α, α and hence φ has order Therefore, it is a necessary and sufficient condition for K to be Galois and cyclic over Q that a b m e m for some e Q We can simlify this condition by dividing through by e and absorbing it into a and b, yielding a /m b 1 3 Some more on this note There are two basic aroaches (as there are for most roblems) to trying find all fields of degree 4 over Q with cyclic Galois grou : the clever aroach and brute force We will discuss the latter aroach first The idea is that there is an easy solution to our equation, viz a 0, b i, but we are not allowed to use it However, what we are allowed to do is use this solution to

1 TREVOR ARNOLD find other solutions It is clear that any rational solution to this equation will lie on a line connecting the oint (0, i) C with our solution, and that this line will have comlex rational sloe Likewise, if we take any line with comlex rational sloe through the oint (0, i), we will get another comlex rational solution (this is a simle algebraic comutation), ie all solutions can be found in this way Thus we take b ( c d + i e f )a + i for some c, d, e, f Z We may assume that none of these integers are zero, since if c 0 then b is ure imaginary, and if e 0 then Imb i if a is real If we lug this into our original equation (a /m b 1) and ask a symbolic engine to solve for a, we get a m( e f c d ( ) i) 1 mc d me f + i mce df Again asking a symbolic engine, we find that if we require a R, then md 3 f 4 m df 4 c 3 md 3 f ce 0, an equation which can be whittled down to d f mf c md e 0 by using the fact that we are dealing with nonzero numbers The quadratic formula then tells us that this equation holds if and only if m d ±fc f me Since d is an integer and m is squarefree, the only way that this can haen is if f me m At first it may seem that we are right back where we started, but then we realize that f and e are integers, ie we have just shown that if there is a solution the equation a /m b 1 over the rationals, then there must also be one over the integers This is as far as the brute force method can take us The clever aroach leads to a different though likewise incomlete result We again rewrite our equation as a m(1 + b ) We can show that if ζ is a rimitive m th root of unity, then ( ) m Qζ of a,, as follows: c ( ) c Now we take the Gauss sum of ζ, and let For a rime, define the Legendre symbol 1 if a is a square mod 1 otherwise 1 ( ) c g ζ c Q( m) c1 Looking at the comlex conjugate ḡ of g, we see that 1 ( ) c ḡ ζ c c1 1 c1 ( ) c ζ c ( ) 1 g,

SOME PROPERTIES OF EXTENSIONS OF SMALL DEGREE OVER Q 13 but we also have (summing over equivalence classes mod, which we are clearly justified in doing) ( ) 1 ( ab gḡ a,b ( ) 1 ( 1 ( 1 1 + a ) 1 + ( 1 a ) ζ a+b ( ) 1 b 1 ) ) 1 b1 ( 1 ) ζ a( 1+1) + ( ) 1 b 1 ( b ) ( a b a ab ( ) 1 ( b a,b 1 ) ( ζ b+1 ) a a ( ) b ( 1), ( ) 1 ) ζ a(1+b) ( 1 ( ) b ζ a(1+b) ( 1 ) + b 1 and hence g The fact that ḡ ±g tells us then that g ± ) a,b ( ) b ( ) b Since we will only be interested in ( rimes ) equivalent to 1 mod 4, we assume now that 1 (mod 4) and thus 1 (the easiest way to see that this is 1 true is to note that in this case Z 1 Z has an element of order 4) So now we have g, imlying that Q(ζ) Moreover, the (cyclic) Galois grou of Q(ζ) has degree divisible by 4, and so by the FTGT we know that there is a unique degree 4 cyclic subextension, necessarily containing Q( ) since this is the unique subextension of degree If m 1 n where i 1 (mod 4), then if ζ is a rimitive m th root of unity the Galois grou of Q(ζ) is isomorhic to Z 1 1 Z n 1 Let us consider the subgrou of order generated by the element ( ( 1 1)/,, ( n 1)/ ) We can choose ζ to be the roduct ζ 1 ζ n where ζ i is a rimitive th i root of unity It is clear that we can view each factor of Z i 1 in the Galois grou of Q(ζ) as corresonding to the subfield of Q(ζ i ) Doing this also makes it clear that each subfield generated by ( 0,, ( i 1)/, 0, ) contains i Hence the subfield in consideration contains m, and is, in fact, Q( m) This allows us to find a large class of examles of cyclic extensions of degree 4, which can be quite a useful, and at least tells us that there infinitely many such extensions 33 Further ossibilities The main oint to kee in mind when looking for extensions of degree a higher ower of is that most of the work is done in characterizing the next lowest ower Grous of order n will always have subgrous of order n 1, and will often be largely determined by them Thus, once all fields of degree n 1 have been classified, it becomes much easier to classify fields of the next highest

14 TREVOR ARNOLD degree For examle, the work done in this aer can be used directly to find all fields of degree 8 whose Galois grou is isomorhic either to Z Z Z or to Z 4 Z Adding a factor of 3 into the degree of the field is likewise relatively straightforward References [1] Larry Grove, Algebra, Academic Press, San Diego, 1983 [] Serge Lang, Algebraic number theory, Addison-Wesley, Menlo Park, CA, 1970 [3] Daniel A Marcus, Number fields, Sringer-Verlag, New York, 1977