Download No nontrivial Hamel basis is closed under multiplication

No nontrivial Hamel basis is closed
under multiplication
Richard D. Mabry
1
Introduction
Known since 1985 or earlier: there is no Hamel basis for
the reals over the rationals that is closed under multiplication [2, p. 365]. Here we prove the following, more
general
Theorem. Let F be a field and let P be a proper subfield
of F. Then the condition
h1 h2 ∈ H whenever h1 , h2 ∈ H
is impossible when H is a Hamel basis for F over P.
So the ≤ 1985 result is the special case that
F=R
and
P = Q.
It would be good to know what fields, proper subfields
and Hamel bases are, so let’s talk briefly about that and
return later to the theorem ...
1
2
Prelimiary ideas
Def ’n 1. A field is a set S along with binary operations
+ and ∗ such that all your favorite operations of arithmetic are valid.
•
+ and ∗ are commutative and associative.
•
∗ is distributive over +.
•
There is an additive identity 0 ∈ S and a multiplicative identity 1 ∈ S.
• Every element in S has an additive inverse in S and
every nonzero element in S has a multiplicative
inverse in S.
We recall that a binary operation on a set S takes
two elements in S and returns one. That is, a binary
operation is a function ? such that ? : S × S → S. We
usually like the so-called “infix” notation for this, and
we write a ? b in place of ?(a, b).
2
Familiar examples of fields include the following.
•
The field R of real numbers with the usual +, ∗.
•
The field C of real numbers with the usual +, ∗.
•
The field Q of rational numbers with the usual +
and ∗.
• The finite fields Zp = {0, 1, 2, 3, . . . , p−1} of integers
modulo a prime p with binary operators +p and
∗p being the usual + and ∗ , but modulo p.
• The field of rational functions in a variable x, with the
usual + and ∗ for such functions. This is also called
the “field of fractions” of the ring R[x] of polynomials in x.
We also keep in mind that the different +’s and ∗’s
used above are not really the same. In particular, the
+ and ∗ are different things in the cases of R and the
field of rational functions. (Why?) To be good boys and
girls, we usually include the + and ∗ along with the set of
elements being acted upon, and it is customary to define
a field as an ordered triple of a set with its two binary
operators having the properties listed previously.
So, the fields just described are more properly denoted
(R, +, ∗), (Zp , +p , ∗p ), etc.
3
Def ’n 2. Suppose (A, +A , ∗A ) and (B, +B , ∗B ) are fields
and that there is a function f : A → B such that
f (x +A y) = f (x) +B f (y)
and
f (x ∗A y) = f (x) ∗B f (y),
for all x, y ∈ A. Then f is called a homomorphism (or
a ring homomorphism) from A to B. We then call the
fields “homomorphic.”
(Among friends, we usually drop the subscripts on the
+ and ∗ because we know from the context what kinds
of elements are being “added” or “multiplied” and we
understand that the operations are different.)
To say that the fields A and B are homomorphic is to
say that they share a lot of structure in certain ways.
For instance, it isn’t difficult to show that the properties
above imply that
f (0A ) = 0B ,
f (1A ) = 1B ,
f (−x) = −f (x), ∀x ∈ A,
among other things.
An even greater level of similarity of structure happens
in the case that the function f above is a bijection. In
that case, we say that f is an isomorphism and that
the two fields are isomorphic to one another.
4
Def ’n 3. Suppose (A, +, ∗) and (B, +, ∗) are fields such
that A ⊆ B. We then say
A is a subfield of B
and
B an extension of A.
If A 6= B we call call A a proper subfield of B and
we call B a proper extension of A.
Examples from earlier:
Q is a proper subfield of R;
C is a proper field extension of R.
5
It is interesting that there are fields that lie “properly
between” Q and R. An easy one to describe is the field
√
denoted by Q( 2).
Def ’n 4. If F is a field, and α 6∈ F, then the
simple field extension of F
obtained by adjoining α
is defined to be the smallest field containing all the elements of F and the element α. This extension field is
denoted by F(α)
(We again suppress the writing of + and ∗.)
6
It is not too hard to show that
√
√
Q( 2) = {p + q 2 : p, q ∈ Q},
by simply showing that the object given by the right√
hand side above is a field containing Q and 2.
That it contains Q is clear, since it contains all num√
bers of the form q + 0 · 2, where q ∈ Q.
√
That it contains 2 is also clear, since
√
it contains 0 + 1 · 2.
√
Next, we can verify that Q( 2) is indeed a field by
checking that it satisfies all of the field properties we
gave earlier. We note only a few of the more interesting
ones.
√
It contains 0, since 0 = 0 + 0 · 2.
It has additive inverses, since
√
√
−(p + q 2) = (−p) + (−q) 2.
It has multiplicative inverses, since
√
1
1
p−q 2
√ =
√ ·
√
p+q 2
p+q 2 p−q 2
√
p−q 2
= 2
p − 2q 2
=
−q √
p
+
2.
p2 − 2q 2 p2 − 2q 2
7
You can similarly show that
√
√
√
3
3
3
Q( 2) = {p + q 2 + r( 2)2 : p, q, r ∈ Q}.
√
3
Notice that extending Q by 2 is slightly more com√
plicated than extending it by 2, as it adds more complication to the forms of the new elements. But whenever α
is an algebraic number, the extension Q(α) extends Q
via finitely many different forms√(which look suspiciously
3
like polynomials evaluated at 2), as above. Such an
extension is called a simple finite field extension or
an algebraic field extension. (A number is algebraic
if it is the root of a polynomial with integer coefficients.)
Something dramatically different occurs when we add
the number π, however, because π is transcendental,
i.e., it is not algebraic.
It turns out that infinitely many elements must be
added to Q to form Q(π); this is a simple infinite field
extension or transcendental extension.
8
For a field F and a variable x, we let F[x] denote the
set of all polynomials in the variable x whose coefficients
are in F. So our usual set of polynomials is R[x]. We
then denote by F(x) the set of all fractions of elements
of F[x] (just don’t divide by zero). Our field of rational
functions from earlier, which is just the set of fractions
whose numerator and denominator are members of R[x],
is R(x).
Now here is something cool: for a variable (or “indeterminate”) x, the field extension Q(π) and the field
Q(x) are isomorphic! What is interesting about this (to
me) is that one is a set of numbers, the other a set of
functions, the “rational functions over Q”. The different
+’s and ∗’s in these fields really are quite different.
Here is another isomorphism: Let f : C → C be given
by f (z) = z̄. That is, f is the complex conjugation
function. It is easy to show that f is an isomorphism. It
is actually called an automorphism.
Def ’n 5. When f : F → F is an isomorphism, then f is
called an automorphism of F.
Put another way, an automorphism of a field F is an
isomorphism from F into (actually, onto) itself.
Def ’n 6. When f : F → F is a homomorphism, f is
then called an endomorphism of F.
So an endomorphism is a homomorphism from a field
into itself. Also, an automorphism is a bijective endomorphism.
9
√
Look again at Q( 2).
√
√
Q( 2) = {p + q 2 : p, q ∈ Q}
Let’s write is another way:
√
√
Q( 2) = {p · 1 + q · 2 : p, q ∈ Q}
Compare this now with C, which we can write as
C = {x · 1 + y · i : x, y ∈ R},
√
where i = −1. (You might notice that C = R(i).)
One more comparison: the vector space R2 can be
expressed as
R2 = {(x, y) : x, y ∈ R} = {x(1, 0) + y(0, 1) : x, y ∈ R}.
Each of the structures above is a vector space. In
fact, each of these has dimension two. Most interest√
ing to us is Q( 2), which is expressed above as a twodimensional vector space over the rationals.
√
Using rationals for the “scalars”, we see that Q( 2)
is spanned by the two basis vectors given by 1 and
√
2. Not only that, these basis vectors are independent
— there is only one way to reperesent each member of
√
Q( 2) in this way.
10
This is a different way of thinking about a subset of the
reals, which we usually consider a one-dimensional vector
space (when we think of it as a vector space at all)! The
difference here is that we are restricting the multipliers,
the scalars, the coefficients of the basis vectors, to be
members of Q.
√
We also notice that Q( 2) is not all of R, in fact, far
from it, as it is only a countable set of reals, while the
reals are uncountable.
Def ’n 7. We say that a subset H of reals is a Hamel
basis for R over Q if H is a linearly independent spanning set (over Q) for R. That is, for every nonzero x ∈ R
there is a unique, finite set
{h1 , h2 , . . . , hn }
of members of H and a unique set of rationals
{q1 , q2 , . . . , qn }
such that
x = q1 h1 + q2 h2 + · · · + qn hn .
It is an amazing fact that Hamel bases for R exist.
However, one must invoke the axiom of choice (AC) to
prove that Hamel bases exist. We will assume AC with
no hesitation (the Banach-Tarski paradox notwithstanding).
11
Unfortunately, we have no time to dwell on the implications of AC. (Ask your professors about it in class.)
Hamel bases are commonly used to provide various “pathological” examples in analysis.
3
A few more remarks
Here again is the theorem we wish to prove.
Theorem. Let F be a field and let P be a proper subfield of F. Then the condition
h1 h2 ∈ H whenever h1 , h2 ∈ H
(1)
is impossible when H is a Hamel basis for F over P.
The most commonly encountered Hamel basis for a
field F over a subfield P of F occurs when P is the field
Q of rationals and F is either the field C of complex
numbers or the field R of reals. We note, however, that
Hamel bases for F over P exist, for instance, if F is C or
R, and P is a subfield of F having cardinality less than
the continuum ([3, p. 220]).
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We also note that the assumptions of our theorem exclude the situation of F = P, and in particular, that
of F = Q; in these cases one can have the trivial basis
H = {1}, which is indeed closed. (This is the only reason
that the word “nontrivial” is in the title.)
Otherwise, we make no assumptions about the sizes
of F and P, so F could be, for example, a field properly containing C. We require only the usual definition
of representation by Hamel bases, namely, that of elements of a vector space (F) whose nonzero elements are
uniquely expressible as finite linear combinations (over
P) of basis elements (from H). (That is, this theorem is
very general.)
13
Recall a definition.
An endomorphism of the field F is any function
f : F → F satisfying
f (x + y) = f (x) + f (y)
and
(2)
f (xy) = f (x)f (y),
∀x, y ∈ F.
Here’s a well-known theorem that we’ll need to deploy.
If f is an endomorphism of R, then either f is the identity function or f is the zero function.
(We’ll prove it later, if we have time.)
It follows that the only automorphism of R is the identity. (Of course, the identity is always an automorphism,
for any field.)
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An interesting role is played by the function τH : F →
F defined by
à n
!
n
X
X
τH
pi h i =
pi ,
(3)
i=1
i=1
which we note is not identically zero (since τH (h) = 1 for
h ∈ H), is P-valued (and therefore not onto F), and is,
as can be easily checked, an endomorphism of F in the
event that (1) holds.
The proof sketched in [2] for the case of our theorem in
which F = R and P = Q is based on the fact mentioned
above, that there are no nontrivial endomorphisms on R
([2, p. 356]). That is, there are no functions f : R → R,
other than f (x) = x and f (x) = 0, for which (2) holds
when F = R. But if (1) holds, then τH is a nontrivial
endomorphism of R, which is impossible. This proves
that a Hamel basis H for R over Q cannot be closed
with respect to multiplication.
On a related note, J. Smı́tal proved that a weaker sort
of closure is possible, namely that a Hamel basis H for
R over Q does exist for which hn ∈ H whenever h ∈ H
and n is an integer [4].
15
On the other hand, here is a really wild fact: there
do exist nontrivial endomorphisms of the complex plane,
these being functions f : C → C, other than f (z) = z,
f (z) = z̄ and f (z) = 0, satisfying (2). See [1] or [2, p.
358].
This would seem to leave open the question of the existence of a Hamel basis H for C over Q satisfying (1).
It turns out that the proof for the real case can be easily adapted to answer this question. This was pointed
out by a referee, who noted that the only real-valued endomorphism of C is the zero function. (To see this, let
z ∈ C and let u be a square root of z. Then if f is any
real-valued endomorphism of C, we have f (z) = f (u2 ) =
(f (u))2 ≥ 0, since f (u) is real. Likewise, −f (z) =
f (−z) = f ((iu)2 ) = (f (iu))2 ≥ 0, so f (z) must be identically zero.) But assuming that F = C and Q ⊆ P ⊆ R,
we see that τH is real-valued and nonzero on C. If (1)
holds, then τH is a real-valued nonzero endomorphism of
C. Hence (1) is impossible when F = C and Q ⊆ P ⊆ R.
Our main result is therefore of most interest when P
is not a subset of R or when F differs from R and C.
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4
Proof of the result
In what follows, when a fixed nonzero element z of F is
such that
n
X
z=
pi hi ,
(4)
i=1
where the hi ∈ H are all distinct and
each pi ∈ P\{0}, we occasionally indicate the uniqueness
H P
of such a representation by writing z = ni=1 pi hi .
To prove our theorem, we first establish a few useful
facts via a sequence of lemmas.
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Lemma 1. Suppose that
n
X
i=1
pi h i =
m
X
rj kj , where the
j=1
hi and kj are in H and the pi and rj are in P. Suppose
further that the pi are all nonzero and the hi are all distinct. Then each of the hi must occur among the kj , i.e.,
for each i = 1, 2, . . . , n, it must be that hi = kj for some
j = 1, 2, . . . , m.
Proof. This is clear using the independence (over P) of
the elements of H.
It is important to note that the reverse is not true in
the above lemma, i.e., the kj need not appear among the
hi , because when the kj are not assumed distinct, the
various terms rj kj could cancel.
18
For the remainder of the paper we assume (until we
obtain a contradiction) that H is a Hamel basis for F over
P (with the conditions on F and P as already noted) such
that (1) holds.
Lemma 2. If h ∈ H and k ∈ H, then h/k is in H.
n
h HX
Proof. Let =
pi hi . Then
k
i=1
h=
n
X
pi h0i ,
(5)
i=1
where h0i = khi ∈ H for each i. But the h0i are all themselves distinct, for if h0i = h0j then we have khi = khj ,
implying hi = hj , which contradicts the uniqueness of
the representation for h/k. But then it must be that
n = 1, because the lefthand side of the representation (5)
is certainly unique. This means that p1 = 1 and h01 = h,
hence h/k = h1 ∈ H.
Letting h be any member of H and letting k = h in
Lemma 2 gives us the following useful fact.
Lemma 3. 1 ∈ H.
19
Lemma 4. If hn = 1 for h ∈ H and a positive integer
n, then h = 1.
Proof. Assume, to the contrary, that h ∈ H and that
h 6= 1. We may assume that n is the smallest positive
integer for which hn = 1. Clearly, n > 1, so we have
n
0 = h − 1 = (h − 1)
n−1
X
hi ,
i=0
and therefore,
1=
n−1
X
−hi .
i=1
Since each hi is in H, at least one of the hi must equal 1,
by Lemma 1 and Lemma 3. But this means that hi = 1
for some i strictly between 0 and n, contradicting the
minimality of n.
Lemma 5. Suppose h and k are in H and that n is a
nonzero integer such that hn = k n . Then h = k.
Proof. The assumptions imply that (h/k)n = 1, so Lemma 2
and Lemma 4 give us the proof.
We’re now ready to prove the main result.
20
Proof of Theorem. Let k1 and k2 be distinct members of
H and observe that k1 + k2 6= 0. Writing the Hamel
representation for 1/(k1 + k2 ), i.e.,
n
X
1
H
=
pi hi ,
k1 + k2
i=1
we have
1=
n
X
pi h0i
+
i=1
n
X
(6)
pi h00i ,
i=1
where
the h0i and h00i are elements of H
with h0i = k1 hi and h00i = k2 hi .
It is also clear that h0i 6= h00i for each i, and the h0i are
all distinct, as are the h00i .
By Lemma 1 and Lemma 3, one of the h0i or h00i must
equal 1, so without loss of generality (by re-indexing , if
necessary) we may let h01 = 1.
We note that if n = 1, then 1 = p1 h01 + p1 h001 . Because
h01 and h001 are distinct with h01 = 1, it would then follow
that the coefficient of h01 is 1 while the coefficient of h001 is
zero. This is clearly nonsensical, so we must have n > 1.
Writing
(1 −
p1 )h01
+
n
X
−pi h0i
i=2
=
n
X
j=1
21
pj h00j ,
(7)
the observations about the distinct elements, along with
Lemma 1, show that each h0i on the left is equal to exactly one h00j on the right. (Our attention is, however,
first drawn to the fact that the coefficient of h01 might be
zero, but this cannot be the case. For if so, then the pigeonhole principle puts two distinct h00j equal to some h0i ,
and therefore equal to each other, which is impossible.)
We will repeatedly use the fact that when h0i = h00j we
have
hj
k1
= .
k2
hi
Now it cannot be that h01 = h001 , so we must have h01 =
h00j for some j > 1. We may let (by reindexing the h00j ,
if necessary) h01 = h002 . Considering the possibility that
h02 = h001 , we would then have
k1
h2
h1
=
= .
k2
h1
h2
But this implies that
µ ¶2
k1
h2 h1
=
= 1,
k2
h1 h2
so k12 = k22 . But then k1 = k2 by Lemma 5. This is
impossible, as is the remaining possibility that h02 = h002 .
We therefore must have n > 2.
This process continues and shows that no finite sum
such as (7) is possible. The proof is by induction. If (7)
is impossible for n = 1, 2, . . . , N (which we have shown
22
explicitly for N = 1, 2), it must be that n > N , so we
relabel when appropriate and necessary to obtain the
pairings
h01 = h002 ,
h02 = h003 ,
...,
h0N = h00N +1 .
If we now consider the possibility that h0N +1 = h001 , we
obtain
k1
h2
h3
hN +1
h1
=
=
= ··· =
=
.
k2
h1
h2
hN
hN +1
It follows that
µ
¶ µ
¶
µ ¶N +1 µ ¶ µ ¶
h2
h3
hN +1
h1
k1
=
···
·
= 1.
k2
h1
h2
hN
hN +1
But then k1N +1 = k2N +1 , so again by Lemma 5, we have
the contradiction k1 = k2 .
This shows that it is impossible that h0N +1 = h001 and so
it must be that h0N +1 = h00i for some i > N + 1 (otherwise
h0N +1 = h00N +1 ). But this means n > N + 1 and shows
that (7) is impossible for n = N + 1, and inductively, for
all n.
We have now proved that no representation of the
form (6) is possible, and hence H cannot span F over
P. Thus H cannot be a Hamel basis for F over P.
23
References
[1] Kestelman, H., Automorphisms of the field of complex numbers, Proc. London Math. Soc., II. Ser., 53
(1951), 1-12.
[2] M. Kuczma, An introduction to the theory of
functional equations and inequalities, Panstwowe
Wydawnictwo Naukowe, Warszawa, 1985.
[3] J. C. Morgan II, Point Set Theory, Marcel Dekker,
New York, 1989.
[4] Smı́tal, J., On a problem of Aczél and Erdős concerning Hamel bases, Aequationes Math., 28 (1985),
no. 1-2, 135-137.
24
Sketch of proof of the endomorphism-on-R theorem. We
are trying to show that when f is an endomorphism of R,
then f is either the identity function or it is identically
zero.
We already know that f (0) = 0. Now suppose that
f (x) = 0 for some x 6= 0. Then for any other y 6= 0 we
have
³ y´
³y ´
f (y) = f x ·
= f (x)f
= 0.
x
x
That is, if f (x) = 0 for some x 6= 0, then f is identically
zero on R.
We may now suppose that this is not the case, and
we’ll show that f is the identity function.
Actually, we already have f (x) = x whenever x ∈ Q.
This follows easily from the additivity of f .
Next notice that if x ≥ 0, we have
√
√
√ √
f (x) = f ( x · x) = f ( x)f ( x) ≥ 0.
That is, f (x) ≥ 0 for all x ≥ 0. It follows from this
that f (x) ≤ 0 whenever x ≤ 0, since for x ≤ 0, we have
−x ≥ 0, so
−f (x) = f (−x) ≥ 0.
Next we use the additive property of f to prove that
the graph of f has a very nice property. Let’s put it this
way: if
(x1 , y1 ), (x2 , y2 ), (x3 , y3 )
are distinct points on the graph which are not collinear,
then the point (x4 , y4 ) that completes a parallelogram is
also on the graph.
25
With this, we can suppose that f (x0 ) 6= x0 for some
x0 6= 0. Of course, x0 will have to be irrational. It
suffices to suppose that either 0 < f (x0 ) < x0 or 0 <
x0 < f (x0 ). Both cases are similar, so we’ll consider only
the first. Draw the line L0 parallel to y = x and through
the point A = (x0 , f (x0 )). Notice that it passes through
quadrant IV. Choose any rational q > 0 and notice that
the point B = (q, q) is on the graph of f . Now, since
the rationals are dense in R, we can choose a rational
p > 0 so that the line through the point C = (p, p) and
parallel to AB intersects L0 in quadrant IV at a point
D = (x1 , y1 ). This point, as well as C, lie on the graph
of f , by the parallelogram property. But D cannot be on
the graph, since then we have f (x1 ) < 0, whereas x1 > 0.
26