Linearly Independent Integer Roots over the Scalar Field Q

3 Pages·2007·0.087 MB·English

by  Eric Jaffe

Q
Linearly Independent Integer Roots over the Scalar Field
Eric Jaffe
July 12, 2007
√ √
It is easy to show that certain integer roots are irrational; the numbers 2 and 3 4 are good
√ √
examples. An equivalent statement is that the sets {1, 2} and {1, 3 4}, respectively, are linearly
independent over the scalar field Q.
√ √ √ √
Furthermore, it can be shown that 3+ 2 is irrational, and that q 3+r 2 is irrational for
√ √
all q,r ∈ Q. In fact, we can say that {1, 2, 3} is linearly independent over Q.
In this paper we will generalize the above notions. First, we aim to determine for which integers
√
ρ and n > 0 the set {1, n ρ} is linearly independent over Q. We can do this quickly by employing
Lemma 2, which is a critical insight concerning the prime numbers due to Euclid. Theorem 3 shows
√
that {1, n ρ} is linearly independent exactly when ρ is not the nth power of some integer.
Lemma 1. Let a be an integer, and p a prime. If p does not divide a then gcd(p,a) = 1.
Proof. We have gcd(p,a)|p, so gcd(p,a) = 1 or gcd(p,a) = p since p is prime. But by assumption
p does not divide a, and gcd(p,a) does, so we must have gcd(p,a) = 1.
Lemma 2. Let a ,a ,...,a be integers, and p a prime. If p|a a ···a then there is some i,
1 2 n 1 2 n
1 ≤ i ≤ n, such that p|a .
i
Proof. The Lemma is clear for n = 1, so assume that it holds for n−1.
Suppose that p does not divide a . Then gcd(p,a ) = 1 by Lemma 1. Hence there exist integers
1 1
r,s such that
1 = ps+a r.
1
It follows that
a a ···a = pa a ···a s+a a a ···a r.
2 3 n 2 3 n 1 2 3 n
But p divides both terms on the right, so p|a a ···a . In particular, p divides one of a ,...,a by
2 3 n 2 n
the inductive hypothesis. This completes the proof.
√
Theorem 3. Let ρ 6= 0 and n > 0 be integers. The set {1, n ρ} is linearly independent over Q if
and only if ρ is not the nth power of some integer.
Proof. Suppose first that ρ = σn,σ ∈ Z (if n is even then let σ be positive). Then
1 √
1− n ρ = 0
σ
√ √
is a nontrivial linear combination of 1 and n ρ with rational coefficients, so {1, n ρ} is linearly
dependent.
√
Conversely, suppose {1, n ρ} is linearly dependent over Q. Then there exist integers a and
b > 0, with gcd(a,b) = 1, such that ρ = (a/b)n. Hence
ρbn = an.
1
In particular, this shows that b|an. Therefore if p is some prime that divides b, then p also divides
an. It follows from Lemma 2 that p|a, which is impossible since gcd(a,b) = 1 < p. Therefore b
must have no prime divisors, so it must be that b = 1 and ρ = an.
This question of linear independence generalizes to larger sets of numbers, and in this paper
we will answer a broader question involving sets of square roots. For example, the claim that
√ √ √
{1, a , a ,..., a } is linearly independent over Q immediately implies that all numbers of the
1 2 n
form
√ √ √
x +x a +x a +···+x a
0 1 1 2 2 n n
are irrational whenever x ,x ,...,x ∈ Q. The next theorem provides a class of a ,a ,...,a for
0 1 n 1 2 n
which this is the case.
Theorem 4. The set
√
S := { n : n is a squarefree positive integer}
is linearly independent over Q.
Note that an integer is squarefree if its prime factorization contains no prime more than once.
The sequence of squarefree integers is
1,2,3,5,6,7,10,11,13,14,15,17,19,21,...
In order to prove Theorem 4, we shall use the concept of field extensions. If F is a subfield of F ,
1 2
written F ≤ F , then we shall say that F over F is a field extension. We use the shorthand
1 2 2 1
F /F to refer to F as a field extension over F , although such notation has nothing to do with
2 1 2 1
quotient groups.
When F /F is a field extension one can consider F as a vector space over the scalar field F .
2 1 2 1
We write [F : F ] to denote the dimension of this space; this number is also called the degree of
2 1
F /F . It can be shown that degrees are“multiplicative in towers”—that is, if F ≤ F ≤ F then
2 1 1 2 3
[F : F ] = [F : F ][F : F ].
3 1 3 2 2 1
Finally, if F /F is a field extension and K ⊂ F , then F (K) is the smallest subfield of F which
2 1 2 1 2
contains K and is an extension of F . For example, when Q is considered as a subfield of R,
1
√ √
Q( 2) = {a+b 2 : a,b ∈ Q};
√ √ √ √ √ √ √
Q({ 2, 3}) = (Q( 2))( 3) = {a+b 2+c 3+d 6 : a,b ∈ Q}
√ √
= {a+b 3 : a,b ∈ Q( 2)}.
We shall now prove the following Lemma. Observe that Lemma 5 implies Theorem 4, since for any
finite subset T ⊂ S of squarefree positive integers, we can find a suitable set A such that T = B
n n
(A and B are defined below).
n n
Lemma 5. Suppose that A := {ρ ,ρ ,...,ρ } ⊂ Z is a set of positive integers such that no
n 1 2 n +
ρ ∈ A is the square of any integer, and every pair of elements in A is relatively prime. Then
i n n
the set
√
B := { σ σ ···σ : 0 ≤ k ≤ n;each σ is a distinct element of A }
n 1 2 n i n
√ √ √
is a basis of the space Q( ρ , ρ ,..., ρ ) over the scalar field Q. (Note that B has exactly 2n
1 2 n n
elements, corresponding to the power set of A .)
n
2
Proof. The proof is by induction on n. Suppose that ρ is a positive integer that is not a perfect
√ √ √
square. Then{1, ρ}certainlyspansQ( ρ), sinceeveryelementofthelatterisoftheforma+b ρ
for a,b ∈ Q. Linear independence follows from Theorem 3. Hence the Lemma holds for n = 1. The
Lemma also holds for n = 0, since {1} is a basis of Q/Q.
Now suppose the Lemma holds for n−1 and n−2 and define the fields
F := Q
0
√
F := Q( ρ )
1 1
√ √ √
F := F ( ρ ) = Q( ρ , ρ )
2 1 2 1 2
.
.
.
√ √ √
F := F ( ρ ) = Q( ρ ,..., ρ ).
n n−1 n 1 n
By the induction hypothesis we have [F : F ] = 2n−1 since B is a basis of F /F . Let
n−1 0 √n−1 n−1 0
β ,β ,...,β be the 2n−1 distinct elements of B . Since {1, ρ } spans F /F , and then
1 2 2n−1 n−1 n n n−1
since B spans F /F , every element x ∈ F can be written as
n−1 n−1 0 n
√
x = a +a ρ , a ,a ∈ F
1 2 n 1 2 n−1
2n−1 2n−1
X √ X
= b β + ρ b β , b ,...,b ∈ F
k k n 2n−1+k k 1 2n 0
k=1 k=1
2n−1 2n−1
X X √
= b β + b (β ρ ).
k k 2n−1+k k n
k=1 k=1
√ √
But the 2n numbers {β ,...,β ,β ρ ,...,β ρ } are exactly the elements of B , so we
1 2n−1 1 n 2n−1 n n
conclude that B spans F /F and [F : F ] ≤ 2n.
n n 0 n 0
It remains to show that B is linearly independent. This will now follow immediately if we can
n
show that [F : F ] = 2n (since B spans F /F , if it were linearly dependent then we could discard
n 0 n n 0
elements to obtain a basis of < 2n elements, which would be a contradiction). And degrees are
multiplicative in towers, so it suffices to show that [F : F ] = 2.
n n−1
Suppose not. Then we must have [F : F ] = 1, which means that F and F are the same
√ n √n−1 n n−1
field; in particular, ρ ∈ F . Since {1, ρ } spans F /F , there exist scalars a,b ∈ F
n n−1 n−1 n−1 n−2 n−2
such that
√ √
a+b ρ = ρ .
n−1 n
That is,
√
a2+2ab ρ +b2ρ = ρ .
n−1 n−1 n
√
Now if ab 6= 0, then this would give an expression for ρ in terms of scalars in F . But
n−1 √ n−2
[F : F ] = 2 6= 1 by the inductive hypothesis, so we must have ρ ∈/ F and therefore
n−1 n−2 n−1 n−2
ab = 0. Since F is a field this means that either a = 0 or b = 0.
n−2 √ √
Ifa = 0thenwehavebρ = ρ ρ ,whichimpliesthat ρ ρ ∈ F . Nowρ ρ isnot
n−1 n n−1 n n−1 n−2 n n−1
thesquareofanyintegersinceρ andρ arerelativelyprime,sothiscontradictstheinductivehy-
n n−1 √
pothesis when applied to A = {ρ ,ρ ,...,ρ ,ρ ρ }. Similarly, if b = 0 then we have ρ ∈
n−1 1 2 n−2 n n−1 n
F , which contradicts the inductive hypothesis when applied to A = {ρ ,ρ ,...,ρ ,ρ }.
n−2 n−1 1 2 n−2 n
Therefore B is linearly independent over Q, so it is a basis of F /F , as desired.
n n 0
3