Simplifymg Nested Radicals and Solving Polynomials by Radicals in
Minimum Depth
Gwoboa Horng and Ming-Deh A. Huang*
Department of Computer Science
University of Southern California
1
Introduction
1.c.m. of the exponents of the derived series of G. If
there is a denesting of a such that each of the terms
Simplifying nested radicals is an important problem in
has depth no more than t , then there is a denesting of
the development of algebraic and symbolic manipula-
a over k ( C ) with each of the terms having depth no
tion systems. The following example is taken from the
more than t + 1 and lying in L(Cl). 0
work of Ramanujan [8]:
One problem with the first result that typically
(/mm=
W(l+fl-(")
we are given a nested radical over the rationals or more
is
The nested radical of depth 2 at the left hand side is
simplified to one of depth 1 at the right hand side.
Such denesting is useful for manipulating large formulas as well as understanding the final results.
Simplifying nested radicals involving square roots
was studied by Caviness and Fateman [2], Zippel [lo],
Borodin, et al. [l]. More recently, Landau [5] obtained
the following general results.'
Theorem (Landau) Suppose a is a nested radical over
k, where k is a field of chat 0 containing all roots of
unity. Then there is a minimal depth nesting of a
with each of its terms lying in the splitting field of the
minimal polynomial of a over k. 0
Theorem (Landau) Suppose a is a nested radical over
k, where k is a field of chat 0. Let L be the splitting
generally an algebraic number fields, and adding all
roots of unity to the ground field is not computationally feasible. The second result gurantees a near optimum solution. However, it remains an open question
to decide whether a nested radical can be denested and
to find a denested form when it erists. The following
theorem provides a complete solution to this problem.
Theorem 1: Let k be a number field and let A be a
nested radical over k. Let u(A) be a value of A over
E. Let L be the splitting field of u(A) over k. Let p,
denote the set of all roots of unity. Then a denested
form B of u(A) over k(p,)
can be computed in time
polynomial in the length of A, the degree of L over Q
and the discriminant of L over Q. 0
Nested radicals represent roots of polynomials
field of k(a)over k, with Galois group G. Let 1 be the
which are solvable by radicals. Landau and Miller
*Researchsupported under NSF Grant CCR 8957317
In her pmentation at '89 FOCS conference, Landau pointed
[6] gave a polynomial time algorithm for deciding if
out the main result stated in [SI was incorrect. The statements
a polynomial is solvable by radicals and constructing
here are the corrected version presented in the conference.
such a solution under a suitable encoding. A related
a47
CH2925-6/90/0000/0847$01
.OO 0 1990 IEEE
awhere p
problem is to solve a polynomial by radicals in min-
successiveadjunctions of roots of the form
imum depth. A more subtle issue is to solve an ir-
is prime. The resulting nested radical is rather restric-
reducible polynomial e z a d l y by radicals in the sense
tive and in general not a solution of minimum depth.
that the nested radical constructed should represent
In this paper the notion of pure nested mdicab
precisely all the roots of the polynomial, and one would
and its field theoretic counterpart, pure root ezten-
like construct such a nested radical of minimum possi-
sions, are defined for investigating exact radical so-
ble depth. In a classical work of Gauss [4], he consid-
lution. When primitive roots of unity are treated as
ered the problem of constructing primitive p t h roots
symbols of no nesting depth, a general solution com-
of unity for prime p by successively solving binomials
parable to Theorem 1 is obtained:
of lower degree. When p = 17 the solution yields a
Theorem 2: Let A be a number field and f be an irre-
method for constructing a regular polygon of 17 sides
ducible polynomial over A. Then a pure nested radical
by ruler and compass, a problem which had eluded
of minimum depth over k ( p o o ) which solves f exactly
mathematicians for about 2000 years until Gauss.
can be computed in time polynomial in the length of
In examing Gauss' solution, one realises that he
not only solved the cyclotomic polynomial dP(z) = 0
f , the degree of L over Q and the discriminant of L
over Q . 0
but solved it exactly in that all the solutions one gets
from successively solving the binomial in his method
A more precise version of Theorem 1 and Theorem
2 will be stated in section 3.
are precisely all the primitive p t h roots of unity.
Solving the cyclotomic polynomials &(z) = 0 ex-
In terms of solving equations by radicals, t&.(z) =
actly by radicals in minimum depth is an interesting
0 can be solved by solving z p = 1. However, this would
question by itself. In this case, we have
violate Gauss' idea since we are solving an equation
Theorem 3: Let n be a natural number and let &(z)
of degree p which is larger than the degree of q5p(z).
be the n-th cyclotomic polynomial over Q. Then a pure
Another problem, in view of solving equation exacfly,is
that the solution to
zp
= 1 contains one that is not
a
solution to &(z) = 0, namely 1.
For another example, consider z6 - 22'
nested radical A, of minimum depth over Q which
solves &(z) = 0 exactly can be computed in time
polynomial in n. Furthermore, let d(n) denote the
+ 4 = 0.
depth of A,.
Then d(1) = 4 2 ) = 0; d ( p ) = d(p-l)+l
+ 1 for p prime and
It can be solved exactly by first solving 'U = -3, then
for odd primes p; d ( p e ) = d ( p )
for all such U solve us = 1 + U . Symbolically, the set
e > 1; d(nm) = maz{d(n), d(m)} if (n,m) = 1. 0
When the input polynomial is abelian over Q , we
of solution to the equation can be exactly represented
by the formula
v m ' . However, the same equa-
tion can be solved, though not exactly, in one step by
solving the binomial
irreducible factor of
2'
2'
= 8, since 2' - 22'
+ 4 is an
= 8.
In [9] van der Wareden showed that an irreducibe
have
Theorem 4: Let f be an irreducible abelian polynomial over Q
a t most one more than the minimum depth can be
computed in polynomial time. 0
Because of space limitation, most proofs of lem-
polynomial sovable by radicals is solvable e z a d l y by
radicals. The exact solution he constructed involves
. Then a pure nested radical of depth
mas and theorems are omitted.
848
2
Definitions and Basic Results
Let K be a field and let K be the algebraic closure of
K. In the following definitions, let
* E {+, -, X , i}.
over K. We say B E RL is a denested form of A over L
w.r.t.
(1) An element a of K is a nested radical of depth
0. That is, d.@h~(o;)
= 0 for all a E K.
( 2 ) Let A, B be nested radicals. Then A
* B is a
nested radical of depth max(depthK(A),depthK(B)).
(3) Let A be a nested radical and n
> 1. Then
0is a nested radical of depth equal to depthK(A)+ 1.
We shall call f i a simple nested mdical of depth
if the following two conditions hold. (i) u(A) E
uaZue(B). (ii)VC E RL such that u(A) E uaZue(C),we
have depthL(C) 2 depthL(B).
Definition: The nested radicals and their depths over
K are defined as follows (cf [l, 51).
U
If B is a denested form of A over L w.r.t.
the minimum radical depth of A over L w.r.t.
then
U, de-
noted d o ( A / L ) ,is & p t h ~ ( B ) .
Definition: Given a nested radical A over K ,a radical valuation
U
over K and a field extension L I K , the
denesting problem is to find a denested form of A over
L w.1.t.
U.
0
Example: In the example given in the introduc-
depthK(A) + 1. 0
We shall use depth(A) instead of depthK(A) when
+
(
1G tion, B = m
form of A =
there is no ambiguity.
Definition: Let A be a nested radical over K. The
associated simple nested radicals of A, S(A), is defined
( f l ) 2is)a
denested
(/rm
over Q w.r.t.
radical valuation
U
the
> 0,
=as > 0,
such that u ( m ) =
> 0,
)-/((U
= a4 > 0,
u ( m ) = a2
u ( m )
as follows.
=
a6
>
a1
0.
Since
IuaZue(A)I = 125 and IuaZue(B)I = 25, there are radi-
(1) If A = a E K then S ( A ) = { a } .
cal valuations
(2) If A = B * C then S ( A ) = S ( B ) U S ( C ) .
( 3 ) If A =
U
K
0
U'.
We are mainly interested in the case when L =
Definition: Let R be the set of all nested radicals over
--t
over Q such that B is not a denested
form of A over Q w.r.t.
@ then S ( A ) = S ( B ) U { a}.
0
K. A radical valuation U over K is a map U : R
U'
K ( p o o )or L = K .
such that
(1) u(A) = A if A E K, and
Definition: Let k be a field and let a E k. Let n > 1
( 2 ) u ( A * B ) = u(A) * u ( B ) , and
be an integer. The radical
( 3 ) (U(
Let
U
f i is an irreducible mdical
of degree n over k if zn- a is an irreducible polynomial
a))"
= u(A). 0
be a radical valuation over K and let S is
over k. 0
a subset of nested radicals over K. We use u(S) to
Definition: Let k be a field. A field extension K over
denote the set { u ( a ) l a E S}.
k is a
Definition: Let A be a nested radical over K. Then
for all 1
uaZue(A) = { u(A)lu is a radical valuation over K }.
> 1. It is a pure root eztension if in addition, for
5 i 5 m, 2''' - cli is an irreducible binomial
..,ai-1, a i + l , ...,am). Each ai is called a
over k(a1,.
5i 5
m, a
:
'
if K = k ( a 1 , . ..,a,) where,
=
ai E
k for some integer
ni
all 1
U
Let LIK be a field extension. Let R K , resp. RL,
be the set of all nested radical over K, resp.
RK
root eztension'
L. Then
RL. Let A E RK and let v be a radical valuation
'In
some books, ki is called a radical cztcnrion. However, w e
choose to called it a root cztcnrion to avoid confusions.
849
denoted d ( K / k ) , is the smallest integer m such that
genemting radical of degree Q.
A (pure) root tower over k is a tower of extensions
k = 60 C k i C
C kn such that kilki-1 is a (pure)
* *
- C k,,,
= K is a (pure) root tower.
If no such tower exists then d ( K / k ) = 00. 0
n. If a is a generating
For example, Q(cp) is a root extension of depth 1
klki-1 of degree m then a is called a gen-
for all p > 2. However, it is not a pure root extension
root extension for 1
radical for
k = Eo C kl C
5
5
i
of depth 1 for prime p
erating radical of degree m at level i 0
>
3. In fact, Q(cP) where
p > 3 is not a pure root extension, though it is always
There is a natural way to associate elements in a
contained in a pure root extension.
We are interested in a special subclass of nested
root tower with nested radicals.
Let k = ko
c
c
kl
c k,, be
a root
radicals, the pure nested radicals.
tower. Let a be a generating radical at level 1 of
Definition: Let A be a nested radical over a field k.
mfi
is
A is called a pure nested radical over k if for all radical
= a E ko. Then
degree m such that a"
k the root tower for A over k with
the nested radical associated with a. Inductively,
valuations v over
let a be a generating radical a t level i of degree m.
respect to v is a pure root tower. 0
Then a"
= g(&,
...,pl)
where /3; are the generat-
Definition: Let k be a field. Let f E k[z] be ir-
and g E k ( z 1 , ...,21). Let
reducible over k. Then f is said to be solvable b y
pi Then
pure radicals over k if there exists a pure root tower
..,Bl) is the nested radical associated with a.
Similarly, let a = g ( a l , l , al,,, ...,an,,=) E kn for some
E k(zi,ilzi,3,. . .,zn,,,) where ai,i,.-. ai,,; are all
k = ko C kl C . - . k n = K over k such that f has a
the generating radicals at level i . Then the nested rad-
other radicals in that they represent algebraic numbers
ing radicals a t level
<i
Bi be the nested radical associated with
"Jg(B1,.
ical associated with a is A
= g(A1,1, AI,^, ...,A,,,,)
root in K . 0
Irreducible radicals distinguish themseves from
uniquely upto coqjugacy. For example,
where Ai,j is the nested radical associated with the
z
roots of the irreducible polynomial '
generating radical ai,,.
can either be roots to
2'
- 2 or to
2'
~
represent
- 2 while fi
- 22' + 4. We
prove this is also true for pure nested radical of any
Conversely, there is a natural way t o associate
depth.
We summarbe the basic results about nested rad-
with a nested radical over a field k a root tower w.r.t.
icals in the following theorems,
a radical valuation over k.
Let A be a nested radical over a field k with
Theorem 2.1 Let K be a field. Let A be a nested
depth(A) = d and let v be a radical valuation over
radical over K Let G = G ( K / K ) . Then g a l , .
k.
K such that value(A) = uin,l{a:luE G}. 0
Let Si
S ( A ) be the set of all the depth i
simple nested radicals associated with A. Let
ki-l(v(Si)). Then E = Eo C kl
c
the root tower for A with respect to
..,an E
k =
C k d is called
U.
Definition: Let k be a field and K an extension over
Theorem 2.2 Let A be a pure nested mdical over a
field K . Let v be a mdical valuation over K and let
a = v ( A ) . Then value(A) = {aelu E G ( a / K ) } . 0
k. Then the minimum (pure) root depth of K over k,
850
Theorem 2.5 Let k be a field. Let k = ko C kl C
- c k,, be a pure root tower. Let A be the associated
polynomial in the length of f , the degree of L over Q
nested mdical of some element a E kn. Then A is a
and the discriminant of L over Q. U
k . The pure nested radical can be computed in time
pun nested radical over k . 0
Definition: Let G be a group and let
Theorem 2.4 Let k be a field. Let f be an irreducible
polynomial over k . Then f is solvable by radicals over
k e f is solvable by pun? radicals e f can be exactly
z-'y-'zy.
y
E G . The
The commutator subgroup of G is the sub-
group generated by all the commutators of G . 0
solved by radicals. 0
3
2,
commutator of z and y, denoted [z,y], io the element
Let G be a group. We shall use G' to denote the
Proof of Theorem 1 and 2
commutator subgroup of G and use
G' to denote the
commutator subgroup of Gibl for i
> 1.
Before giving the proofs of Theorem 1 and Theorem 2,
Lemma 3.1 Let G be a group and let H be a normal
we restate them using the technical terms defined in
subgroup of G .
section 2.
Theorem 1: Let k be a number field and let A be a
nested radical over k. Let
U
H.
> 0.
n be a natural number which is divisible by [ L : Q ]
and the discriminant of L over Q. Then the tower of
field extensions corresponding to the derived series for
Gal(L((n)/k((n))determines a denested form E of A
U.
Furthermore, E can be computed
in time polynomial in the length of A, the degree of L
over Q and the discriminant of L over Q . 0
0
Let k be a field and K be a finite solvabe G&
be a radical valuation over
k and let L be the splitting field of v ( A ) over k. Let
over k(poD)w.r.t.
If G / H is abelian then G'
(ii) ( G / H ) ' Y G'HIH for all i
(i)
lois extension over k . Let G be the Galois group
of K over k.
Then G" = 1 for some n, G'+' is
a normal subgroup of G' for all i and the chain of
group G 3 G1 3
..- 3
G" = 1 is called the de-
rived series of G . The length of the derived series,
denoted Z(G), is n. The corresponding tower of subfields k = ko
c kl c
- - - C k,
= K has the property
that ki+l is the maximum abelian subextension of K
over ki by Lemma 3.1 (i). In particular, when k con-
Theorem 2: Let k be a number field and f be an
irreducible polynomial over E . Let L be the splitting
field o f f over k . Let n be a natural number which is
divisible by [L: Q ] and the discriminant of L over Q .
Then the tower of pure root extensions corresponding
tains a primitive N-th root of unity where N is the
5 n,
The maximality of k+l
lcm of the exponents of G ( k i + l / k i ) for all 1 5 i
the tower is a pure root tower.
over ki within K implies that any pure root tower from
k to K has depth at least n.
to the derived series for Gal( L(cn)/k(Cn))determines
Theorem 5.1 Let A be a nested radical over a field
a pure nested radical of minimum depth over k(p,)
k . Let a = v ( A ) for some radical valuation v over
which represents precisely all the roots of f over k
k . Let 1; be the splitting field of a over k . Let B
when (,, is viewed as a symbol which can assume all
be a denested form of A over k(p,) w.r.t. U . Then
conjugate values of a primitive n-th root of unity over
L ( A l k ( l . 0 ~ )=) l ( G ( L b m ) / k b m ) ) ) .0
85 1
a set of generating radicals. This can be done by using
Proof of Theorem 1:
Let K
=
Lagrange resolvents. From this we get a nested rad-
L n k(pOO).
G ( L ( h ) / k ( b ) ) . Suppose K
Then G ( L / K ) S
ical for A of minimum depth. All the computations
c
involved can be done by Galois theoretic algorithms
k ( & ) for some n
divisible by the exponent of G ( L / k ) . Since K =
L n k(Cn), G ( L / K )
2
involving L*-algorithm [7].
G'(L(L)/k(C,,)). Hence
G(~(Cn)/K(CTb)) 2 G(L(POO)/k(POa)). Therefore, a
denested form of a over k ( p O O w.r.t.
)
U
can be found
by constructing the tower of subfields corresponding
The overall time complexity is polynomial in the
degree of the splitting field of the input nested A and
the discriminant of L over Q . This proves Theorem 1.
0
to the derived series of G(L(C,,)/k(<,,))by Theorem
3.1. Our goal thus is to show that such n exists and
Proof of Theorem 2:
The above proof can be refined to yield a proof for
to bound it in terms of k and L.
Let F = k n Q ( b )Then
.
G(k(h)/k)Y
G ( Q ( h ) / F ) .Let M = K n Q ( p L O OThen
) . K = kM
and G ( K / k ) Y G ( M / F ) . Let m be the conductor of
M over Q . It follows from class field theory that M
is contained in the ray class field of modulus mm over
Q , which is precisely Q(Cm). Hence M = K n Q(&,,)
and it follows that K = k M C k(Cm).
Theorem 2. The splitting field for u(A) is now replaced by the splitting field L for the input irreducible
polynomial f . The tower of extensions corresponding
to the derived series for G(L(C,,)/k((,,)) is in fact a
pure radical tower. Computationally we then have to
compute for each extension in the tower a set of irreducible radicals which are disjoint and we omit the
details here.
The conductor of M over Q divides the discriminant D ( L / Q ) of L over Q .
Let n
Let a be a root o f f in L(<,,). Let B be the as-
=
D ( L / Q ) [ L : k1. Then G(L(Cn)/k(Cn))2 G ( L / K ) Y
G ( L ( h ) / k ( b ) ) .Hence the corresponding subfields
ciated nested radical of a over k(C,,). In the following
we show that B represents exactly all the solutions to
C,
of the derived series of G(L((,,)/k(C,,)) is a tower of
f over k when
pure root extensions and determines an equivalent
sume all conjugate values of a primitive n-th root of
nested radical of minimum depth for the given nested
unity over k .
radical A.
To compute the minimum depth nested radical
is viewed as a symbol which can as-
Let g E k(Cn)[z] be a irreducible factor o f f over
k(C,,) such that g ( a ) = 0 and let G = G(k(C,,)/k).
Then
gQ E k [ z ] and any root o f f over k is a
L over k . We then compute an irreducible polynomial
noEG
root of noEG
ge E k [ z ] and vice versa. Let A , be a
equivalent to u(A), we first compute the splitting field
g for L over Q and its discriminant D ( g ) . We ob-
pure nested radical over k ( t ) with a E
serve that D ( g ) is divisible by the discriminant of L
Then uuZue(A,) = {XIg(X) = 0). Let
over Q . Let n be the product of D(g) and the degree
be the nested radical obtained by replacing all
of L over k . We then compute the derived series for
A, by
G ( L ( b ) / k ( h ) )and the corresponding tower of ex-
UQEGUdUe(A:)is the set of all the solutions to f over
tensions. For each extension in the tower, we compute
1. 0
852
CE.
U
UUZw(Aa).
E G . Let A:
c,,
in
Then vuZw(AZ) = {p1gU(p)= 0). Hence
Proof of Theorem 3 and 4
4
p t h root mod p. The Galois group G(Q(a)/Q)is
generated by the element U such that U ( Q ) = ag. Let p
Throughout this section, let K be a number field.
be a primitive (p- 1)-st root of unity and let t(p,a)=
We shall only consider pure nested radicals in this
section. Let
U
a
be a pure nested radical over K . Let
V be the set of all the radical valuations over K . By
Theorem 2.2, & ( a / K ) and & ( a / K ) have the same
Let
= t(/T,a)then t!-' and
under
U.
Therefore, once
are invariant
e-'
is known
, we
have
+. - - + t p - l )is a description of a
in terms of known quantities ifp is known. For detail,
pure nested radical depth of a over K and denote it
please see [3]. 0
as d ( a / K ) or &(a).
Let L / K be a finite extension. We say that L
We will ilx a radical valuation
U
over K and adopt
has the unique subfield property (wp) if, for every
the following convention throughout thm section. Let
E K,l 5 i 5
U.
m,...,
.J.';)).
U(
We have u(t) = P"t.
a = (p- l)-'(tl +t2
value for all U, U' E V. We call this value the minimum
K(v(
+ pa9 + - + pP-2ag'-'.
Then K(*
,..., W
d I [L: K], there is exactly one subfield M of L such
=
EM
that K
L. Clearly, every cyclic extension has
the unique subfield property by Galois theory.
Lemma 4.1 Let (,, E K and let L / K be a field ezten-
Lemma 4.2 Let p be a prime and let
sion of degree n. Then L / K is cyclic iff L = K ( ~)
(4
E K if p =
2. Let L = K ( a ) where a i s a root of the irreducble
for some A E K . 0
polynomial zP*
- a.
Then L / K has the w p . 0
Lemma 4.3 Let L be an eztension of K generated
Remark: It is well known that such A can be found
by an a with am" = a E K and (m,n) = 1. Then
using Lagrange resolvents as follows.
L = K(am,a"). 0
Let K ( a ) / K be a cyclic extension of degree n and let U
be a generator of the Galois group G ( K ( a ) / K ) .Let C
Lemma 4.4 Let K ( a ) / K be a Galow eztension of de-
be an n-th root of unity, then the associated Lagrange
gree p' where p i s a prime and
resoluent is t ( ( ,a)= a0
ai
0
+Cal + - - +
("-l a,+1
where
= uia. Let ti(c,a) = u i t ( ~ , a=
) c-'t(c,a), for
K(a)
5 i 5 n -- 1. We have to((, a)" = tl((,a)" = -.-=
tn-l(<,a)" = A. Note that A is invariant under
C E K then K = K(C). Therefore, K ( a ) = K ( 6).
0
5
i
5
n
- 1.
When t , ( ( , a ) ,O 5 i
.
known, we can evaluate ao,.. , a,-l
follows, ai =
5 n - 1, are
very simply as
ti((, a))/=
When a is a primitive p t h root of unity. Then
Q ( a ) / Qis cyclic
>
0.
If K ( Q ) w
K(
~ ' f..i, , .
or
(4
where oi E K . Fur-
E K when p = 2 then
e>ei>O,lSiSs. 0
Therefore, it must be an element of the field K(C). If
in n. Once A is known, we can evaluate t i ( C , a ) for
C
thermore, i f p is odd
U.
Note the time required to compute A is polynomial
e
contained in a pure root tower of depth 1 over K then
Lemma 4.5 Let
(2p
E K where p i s a pn'me and let
e be a positive integer. Let L / K be a finite eztension
such that L is a splitting field of the polynomial (zP*' a1)(zP"
for 1
- a2)- .- (zp" - a,) where 0 < ei 5 e , ai E K
5 i 5 U.
Let M / K be cyclic of degme p' and K c
M c L. Then M is a splitting field of the irreducible
polynomial
of degree p- 1. Let g be a primitive
zP'
- a for some a E K .
Now we prove the following
853
0
Theorem 4.1 Let K ( a ) be a cyclic eztension of de-
>0
gree p' over K , where p is a prime and e
and let
K . Let L I K be a finite field eztenswn of degree 1. If
(I:
m) = 1 then
L n M N = L n N. 0
K ( a ) be contained in a pure root tower of depth 1 over
K . Then
CP
E K . Furthermore, assuming
>
when p = 2 and e
CPU
(4
E K
1, i f K ( a ) n K(Cpe) = K then
Proof of Theorem 4.2 :
Since &(a)
EK.
root extension K (
p e e , . .
.,
*'fi,...,
~'*-m
and, w.l.o.g.,
Let L = K (
For d = 1, we have
and L ( a ) C_ L(
we have L ( a ) = L(
E L. Suppose
p
e
a
By Lemma 4.2,
.
pm. Therefore,
Cp
E L
-K
cPd
E L. Hence
then K C K(Cp) C
< p , a contradiction.
Hence, we have Cp E K.
and 1 < [K(CP) : K]
Cp @ L - K.
L
Therefore,
*
- (zp.'
- a,) and
K ( a ) is the splitting field of
Therefore,
(pa
cp E K, by Lemma 4.5,
zPo
E K by Theorem 4.1.
Assume the theorem hold for d = 1,.
Let d ~ ( a=
)d=k
1
<
..,k 2 1.
+ 1. If p = 2 then d ~ ( < , =)
d. Therefore, we may assume p is odd. Since
d~(a=
) d, there is a pure root tower K = KO c
K1 C
... C Kd
such that a E Kd - Kd-1, KilKi-1 is
a pure root extension Let M = K1 n K(a,Cp.).
(i) If [M : K ] 2 p then there is a cyclic extension
Since K ( a ) is contained in the splitting field of
(d'
-01)
c,,
Therefore, d ~ ( ( , )= 0.
as-
sume a @ L. Then L ( a ) is cyclic of degree pd for some
> 0 over L
+ 1. We do this by
induction on d = dK(a).
where aj E K
and e 2 ei 2 1.
Cp
Lemma 4.1, we only
) dK(C,)
have to show that d ~ ( a2
Proof: By Lemma 4.4, K ( a ) is contained in a pure
d
5 d~((,,+
) 1 by
- a for some a E K.
CPU E K ( a ) . If K ( a ) n K(CPu)= K
then
E K. 0
of K of degree p has depth 1. Therefore,
E K by
Theorem 4.1 We have &(CPp.) = 1 < d.
(ii) If [M : K] = 1 then K r ( a ) n K1(cp.) = K1 by
Lemma 4.6.
Therefore, dK,(a) = d - 1 = k. By
5 d-
induction hypothesis,
Next, we extend Theorem 4.1 to the following
Cp
2+ 1 = d - 1 = k.
(iii) If 1 < [M : K ] < p then M = M n K(a,Cp.) =
= pe for some
M n K(CPe)= K1 nK(Cp.) by Lemma 4.7. Therefore ,
> 1. If K ( a ) is a
we can apply Lemma 4.6 and use the same argument as
cyclic eztension of K of degree n and K ( a )nK(C,) =
(ii) to show d ~ ( ( p ) k. Thus the proof is completed.
K then d ~ ( a=
) d ~ ( c+~1.) 0
0
Theorem 4.2 Let p be a prime. Let
e 2 1. Assume
(4
E K i f p = 2 and e
n
We need two more preparatory lemmas to prove
<
Lemma 4.8 Let p be a prime and let g = [ Q ( ( l ,
Theorem 4.2.
Q(C)]. Then
Lemma 4.6 Let K ( a ) and K ( P ) be finite abelian ez-
pe-1
if (Z,fl)
g
= ( p - l)pe-' i f (1, p') = 1 and
=p.
:
g
=
0
tensions of K and K ( a ) n K ( P ) = K . Let L I K be
any finite field eztension and let M = L n K ( a , P ) . If
M = K or M = L n K ( a ) or M = L n K ( p ) then
~ ( a )Ln( p ) = L .
0
n
(i) For each n,we shall construct T,, a pure root tower,
Q = Q o , C~ Ql,, C
Lemma 4.7 Let M and N be Galois extensions of K
with degree m and
Proof of Theorem 3:
respectively. Assume
M nN =
that
C,
C Q4, where d
E Qd,a and, for 1 5 i
following properties:
854
a . .
5
= d(n) such
d(n), Qi,, has the
P1. Qi,,/Q is the composite field of Q((
01)
Pi
Q ( a )be the subfield of L of degree
for
+
cp. E
cp E Qi-l,,.
Qi,,
then
+
plying Theorem 4.2, dQ(a) >_ d(qe) 1 = d(Z) 1.
some primes pj and e j 2 1.
P2. Let e > 1 if
over Q . Ap-
Therefore, d ( p ) 2 d ~ ( a2) d(Z)+ 1. By induction
hypothesis, d(Z) = d(1m) = d ( p - 1). Therefore,
We use induction to show the second part of The-
we have d ( p ) = d ( p - 1) + 1.
orem 3
It's easy to see d(1) = 4 2 ) = 0 and TI= Ti = Q.
The associated pure root tower Tp is Q = Q o ,C~
We also have d(3) = 1 and the associated tower T3 is
Q I , ~C
.
C
Q4p-i),p
C
Qi,p
=
1) and Q4p),p
=
Q4p),p
where
Q = Qo,a C 91,s = Q0,3(It's
m
easy
. to see TI,Ti,
Qi,p-l for 0
and Ta satisfy properties P1 and P2.
Qqp--l),p(
'-6)
with X E Q4p-1),p. Note X can
Assume for n = 1,2,. . .,k, Theorem 3 holds and
5
i
5 d(p -
be found by using Lagrange resolvent (see the re-
there is a T, satisfies properties P1 and P2.
mark after Lemma 4.1). Since inductively Tp-l
+
For n = k 1 > 3, we consider the following cases.
satisfies P1 and P2, it's easy to
n = pe where e > 1 and p is a prime.
properties P1 and P2.
If p = 2 then it's easy to see d(2') = 1 and Tp is
n
Q = Qo,i= C Qi,i*= Qo,i-(
If p # 2 then since
have d(p')
5
d(p)
cp. =
+ 1.
'*-m.
n;=,pzi where
pi
1, we
cn
We can apply Theorem
=
n;=,
&,;i.
2 m ~ = l { d ( p ~ i )since
}
In the following, we show
d(n) 5 max&l{d(p;i)}.
+ 1. Therefore, d ( p e ) =
+ 1. We omit the details here.
Tp satisfies
# p, for i # j.
It's easy to see d(n)
?*-z
>
for e
=
see
Thus complete the proof
4.1 to prove d ( p e ) 2 d ( p )
of d(mn) = max{d(m), d( n)} for (m, n) = 1 and
d(p)
the second part of Theorem 3.
n =p
We only have to show the composite of Tp:i, 1 5
> 3, a prime.
5 8 , is a pure root tower and satisfies properties
Let p- 1 = lm where 1 is odd and m is a power of
i
2. We first consider the case when 1 = 1, i.e., p =
P1 and P2. This can be done by using Lemma
2b 1 for some b > 1. Since d ( p - I) = ~ i ( 2 =
~ )1,
4.8. Again we omit the details.
we only have to show d ( p ) # 1. This can be done
(ii) The following procedure is based on the above
+
with the help of Lemma 4.4. We omit the details
discussion.
here.
Procedure primitive-root-of-unity (n)
Now we consider the case when 1
>
1. Since
Input: n is a positive integer, n 2 2,
Q ( c p ) / Qis cyclic of degree p - 1, by Lemma 4.1,
Output: a pure nested radical of minimum depth for
d ( p ) 5 d(p-1)+1. In the following, we shall prove
Cn.
I. If n = 2 then return(-1).
that d ( p ) 2 d(p-1)+1. Hence, d(p) = d(p-1)fl.
Let L and M be subfields of
[L : Q ] = 1 and [ M : Q ]
Q(cP)such that
=
m.
We have
2. If n = pel where p a prime, e
>
primitive-root-of-unity(p) and return(
1 then let t
=
.*-a.
M n Q ( C ) = Q since Q(Cp) n Q(C) = Q . BYin-
5. If n
duction hypothesis, there is an odd prime q such
let tl = primitive-root-of-unity($), let t i = primitive-
that qellZ for some e 2 1 and d(1) =
a($). Let
= p'm, where p a prime and ( p , m ) = 1 then
root-of-unity(m) and return(tlt2).
855
References
4. If n = p an odd prime then
0
let t = primitive-root-of-unity(p- I);
0
let g = a primitive root mod p;
[l] A. Borodin, R. Fagin, J. Hopcroft, and M.Tompa
Decreasing the nesting depth of expressions in-
o f o r i = l top-2
voling square roots.
let ti = 2 + tizg + t2'zga +
0
compute r, =
0
let
81
=
...+ t(p-l)'zg'-';
1:169-188, 1985.
for 1 5 i 5 p- 2;
and let
p - f i
J. Symbolic Computation,
[2] B. Caviness and R. Fateman. Simplification of
= r i / ~ f - ' - ~for 2 5
radical expressions. In Pm. SYMSAC 77,1977.
i5p-2;
[3]H. M. Edwards. Galois Theory. Springer-Verlag,
o l e t Cp=(p- l)-'(rl+rl+...+(-l));
0
1984.
return(CP).
[4]C. Gauss. Disquisitwnes Arithmeticae (English
5. End.
Tmnsl.). Yale, New Haven and London, 1966.
This completes the proof of Theorem 3. 0
Proof of Theorem 4:
[5] S. Landau. Simplification of nested radicals. In
Let a be a root of f. Let Q(a) =
Q1
- e -
that Qi/Q is cyclic of degree p:i and n =
Q, such
nhlp:'.
Note pi are not necessary distinct primes. By Theorem 4.1, d(Qi/Q)
2
d(pi) + 1. It's easy to see that
+ 1)
is in polynomial time. JCSS,30179-208, 1985.
[7]S. Lanndau. Factoring polynomial over algebraic
= D . As in the proof of The-
number fields. SIAM J. Comput., 14184-195,
Therefore, d Q ( a )
orem 3, there is a tower of pure radical extension
Q = KO C K1 C
[6]S. Landau and G. Miller. Solvability by radicals
2
d ~ ( a2
) mt%&{d(Qi/Q)}.
m+,(d(pi)
Proc. 30th IEEE FOCS, pages 314-319, 1989.
KD such that Cpti E KD
for 1 5 i 5 8. Let's decompose KD(IZ)/KD
into cyclic
subextensions Mi/Ko,l5 i 5 8 such taht K o ( a ) =
M I M, and Mi n Ml-..
Mi-lMi+l M, = KD.
Let qi = [Mi: KO].Since qiln, Cq6 E KD. Hence, we
can use Lagrange resolvent to find a, such that Mi =
K~(ai)where ai = t/x; for some Ai E K O . Since
MinMl.-.Mi-lMi+l...M,
=K~,a,,15
i<s,are
e . .
C
1985.
[8]S. RamanGan.
Problems and solutions. Col-
lected works of S. Ramanujan Cambridge University Press, 1927.
a - .
disjoint irreducible radicals over KO.
To find a, the exact solution of f(z) in radical
over Q, we can factor f(z) over
L = KD(a1,. . .,a,).
All operations (Lagrange resolvent and factoring polynomial) can be performed in polynomial time.
Note d ( L / Q ) = D + 1. Therefore, a found in this
way is at most one more than the minimum depth. 0
[9]B. van der Waerden. Modern Algebru. Frederick
Ungar, 1953.
[ 101 R. Zippel. Simplification of expressions involving radicals. J. Symbolic Computation, 1:189-210,
1985.