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DECREASING CHAINS OF ALGEBRAIC SETS
by
Harvey M. Friedman
[email protected]
www.math.ohiostate.edu/~friedman/
November 12, 1999
1. IDEALS IN k VARIABLE POLYNOMIAL RINGS.
An ideal in a commutative ring R with unit is a nonempty I Õ
R such that for all x,y Œ I, z Œ R, we have x+y and xz Œ I.
A set of generators for I is a subset of I such that I is the
least ideal containing that subset.
One main formulation of the Hilbert basis theorem asserts
that for every k ≥ 1, every ideal in the k variable
polynomial ring over any field is finitely generated.
An equivalent way of saying this is that for every k ≥ 1,
there is no strictly ascending sequence of ideals in the k
variable polynomial ring over any field.
The presentation degree of an ideal I in the polynomial ring
F[x1,...,xk] is the least d such that there is a set of
generators for I consisting entirely of polynomials of degree
at most d.
Note that by linear algebra, there is a number f(k,d) such
that for all fields F, any ideal of presentation degree at
most d in the k variable polynomial ring over F has a set of
generators of degree at most d, of cardinality at most
f(k,d).
THEOREM 1.1. For all k,p there exists n such that the following holds. Let F be a field. There is no strictly ascending
sequence of ideals in F[x1,..., xk] of length n, indexed from
1, indexed from 1 through n, such that the ith ideal is of
presentation degree £ p+i.
Proof: Fix k,p, and suppose that this is false for all n. For
each n, let T[k,p,n] be the following theory in first order
predicate calculus in the language of fields, augmented with
unary predicate symbols P1,...,Pn.
i) the field axioms; 2
ii) for 1 £ i £ n, Pi is an ideal in the k variable polynomial ring over F;
iii) for all 1 £ i £ n1, Pi Õ≠ Pi+1;
iv) for 1 £ i £ n, Pi has a set of ideal generators of presentation degree £ p+i, of cardinality £ f(k,d).
Clearly, for each n, T[k,p,n] has a model. Hence there is a
model of the union of the T[k,p,n] (compactness theorem for
predicate calculus with equality). This model gives us a
field F together with a strictly ascending sequence of
ideals. This contradicts the Hilbert basis theorem. QED
We define I(k,p) be the least n given by Theorem 1.1. We also
define I(F,k,p) to be the least n given by Theorem 1.1 for a
given field F. Obviously I(k,p) is the minimum over all F of
I(F,k,p).
We can read off some crude upper bounds for I(k,p) using
machinery from logic. Of course, more precise upper bounds
can be obtained by getting rid of the logic and going into
the guts of the situation.
Specifically, we use logic to show that the function I(k,p),
k fixed, is primitive recursive in p. Fix k ≥ 1.
Step 1. We have to put the statement “for all p, I(k,p)
exists” into a more concrete form. Note that I(k,p) has the
property that the theory T[k,p,I(k,p)] has no model. By the
completeness theorem for predicate calculus with equality,
this means that the theory T[k,p,I(k,p)] is inconsistent.
Thus we can put the statement “for all p, I(k,p) exists” in
the form
*) for all p, there exists n such that T[k,p,n] is
inconsistent,
where any n with this property is at least as large as
I(k,p). Note that this statement is of the form
**) "p $n such that R(p,n),
where R is purely existential.
Step 2. We observe that the statement “"p, I(k,p) exists” is
provable in a certain formal system WKL0. Familiarity with the 3
technology makes this an easy exercise, and such familiarity
can be obtained by looking at [Si99].
Step 3. We also observe that the statment *) is likewise
provable in WKL0.
Step 4. Finally, we apply my old metatheorem that any sentence of the form **) that is provable in WKL0 must be primitive recursively true. See [Si99] for a proof of this
metatheorem.
Step 5. Thereby conclude that the function I(k,p) of p is
bounded by a primitive recursive function.
WKL0 is the so called second principal system of reverse
mathematics:
RCA0, WKL0, ACA0, ATR0, ’11CA0.
Another way of stating this result is by defining the
Ackerman hierarchy of unary functions on N = the set of all
nonnegative integers. For each strictly increasing f:N Æ N,
define f’:N Æ N by f’(k) = ff...f(1), where there are k f’s.
Take f1 to be doubling. Take fk+1 = (fk)’.
A function is bounded by a primitive recursive function if
and only if it is bounded by some fk. Thus
THEOREM 1.2. For each k, the function of p, I(k,p), is
bounded by some fr.
A more detailed study should reveal a good r = r(k) so that
I(k,p) is eventually dominated by Fr, and how long you have to
wait for the domination. This can probably be done by avoiding the logic and going after the technical guts of the
problem. However, there are additional metatheorems from
logic that could be used to get some sort of reasonable
estimate for this, but perhpas not as good as really getting
your hands dirty.
And there is the question of comparing the function I(k,p) of
two variables with the function A(k,p) of two variables. We
make the same remarks as above. 4
Notice that we have avoided saying that the functions I(k,p),
k fixed, is primitive recursive. In fact, we don’t even know
if they are recursive.
CONJECTURE: Each I(k,p), k fixed, is primitive recursive.
I(k,p) is recursive.
We now come to lower bounds. We will even discuss the lower
bounds to the weaker theorem that requires the ith ideal to
have degree exactly p+i+1  not just £ p+i+1.
We can already get big lower bounds if we just look at ideals
consistting entirely of monomials with coefficient 1, no
matter what the field is.
We use N for the set of all nonnegative integers. For n in
Nk, write M(x) for the monomial x1n_1...xkn_k.
LEMMA 1.3. Let F be a field, A Õ Nk, and n Œ Nk. M(n) is in
the ideal generated by {M(m): m Œ A} if and only if there
exists m Œ A such that m £ n coordinatewise. Also the ideal
generated by {M(m): m Œ A} has presentation degree exactly
the maximum of the l1 norms of the elements of A.
LEMMA 1.4. Let F be a field. Let n1,...,nr Œ Nk have l1 norms
p+1,...,p+n, where for no i < j is ni £ nj coordinatewise.
Then L(F,k,p) > r.
In 3 dimensions, can have l1 norms p+1,...,p+2n. Then we can
start an induction through the Ackerman hierarchy of functions. Get L(F,k,p) > A(k1,p) for k ≥ 3. With more work, can
do better.
Gallo and Mishra, A solution to Kronecker’s problem,
Applicable Algebra in Engineering, Communication and
Computing, 5(6):343370, 1994,
consider such chains of ideals  only over the integers  in
connection with the polynomial ideal membership problem over
the integers. The latter problem was given an effective
procedure by
Ayoub, On constructing bases for ideals in polynomial rings
over the integers, J. of Number Theory, 17:204225, 1983. 5
Gallo and Mishra show that upper bounds for these chains give
upper bounds for the sizes coming up in this ideal membership
problem. Gallo and Mishra give upper bounds in terms of the
Ackerman function. In our setup, their upper bounds would
involve looking at the 8kth level of the Ackerman hierarchy
in dimension k. They do not discuss lower bounds, although
the above lower bounds would serve for their chains also.
However, there are no good lower bounds for the polynomial
ideal membership problem over the integers. Over the standard
fields, it is known to be at most double exponential
(Hermann, etcetera).
I suspect that one could read off primitive recursive bounds
from Ayoub’s paper using logic, thereby obtaining the main
conclusion of Gallo and Mishra immediately from Ayoub, but I
have to look.
2. ALGEBRAIC SETS IN K DIMENSIONS OVER A FIELD.
Let F be a field and k ≥ 1. An algebraic set in Fk is the set
of simultaneous zeros of a system of polynomials in k
variables over F.
The presentation degree of an algebraic set in Fk is the least
d such that there is a set of generators for I consisting
entirely of polynomials of degree at most d.
Note that by linear algebra, there is a number f(k,d) such
that for all fields F, any algebraic set in Fk of presentation degree at most d in the k variable polynomial ring
over any field is the zero set of a system of k variable
polynomials of degree at most d, of cardinality at most
f(k,d).
THEOREM 2.1. For all k,p there exists n such that the following holds. Let F be a field. There is no strictly decreasing
sequence of algebraic sets in Fk, indexed from 1 through n,
such that the ith set is of presentation degree £ p+i.
Notation: Let A be algebraic in Fk. Write I(A) for the ideal
of all k variable polynomials over F which vanish on A.
Theroem 2.1 follows immediately from 6
LEMMA 2.2. Let F be a field and A,B be algebraic in Fk, where
A ⊇≠ B. Then I(A) Õ≠ I(B). Also the degree of I(A) is at most
the presentation degree of A.
Proof: Assume hypotheses. Clearly I(A) Õ I(B). Now the
presenting polynomials for B do not vanish on all of A, and
so I(B) does not vanish on all of A. Hence I(B) is not I(A).
In fact, this argument shows that any upper bound for Theorem
1.1 is an upper bound for Theorem 2.1.
Now we discuss the lower bounds for Theorem 2.1. In fact,
define L(F,k,p) to be the longest strictly decreasing of
algebraic sets in Fk, indexed from 1, such that the
presentation degrees successively go up by 1, starting with
presentation degree p. We give lower bounds for L(F,k,p) that
work for all infinite fields F simultaneously.
By the upper bound results for ideals, we can, for each k,
find primitive recursive upper bounds for L(F,k,p) that work
for all fields F simultaneously.
Fix dimension k, an infinite field F, and
that L(F,k,p) is large by showing that it
large as a number arising out of a purely
problem about finite trees. We do this by
long sequence of sets algebraic in Fk of
p,p+1,...,p+n. p ≥ 1. We will show
is at least as
combinatorial
constructing a very
presentation degrees Our long sequence of algebraic sets will actually be finite
unions of finite intersections of equations xi = ci, where
only some of the 1 £ i £ n are mentioned.
The first order of business is to show that if we have a
decreasing sequence of algebraic sets A1,A2,...,An Œ Fk of
degrees £p,£p+1,...,£p+n, where each Ai+1\Ai is infinite, then
we can adjust them to have degrees p,p+1,...,p+n.
LEMMA 2.3. Let A be algebraic in Fk of degree d and b in Fk.
Then A » {b} has degree £ d+1.
Proof: Let b
intersection
common zeros
common zeros = (b1,...,bk), and A’ = A » {b}. Then A’ is the
of the sets Bi = A » {x: xi = bi}. Let A be the
of a set of polynomials X. Now Bi consists of the
of the set of polynomials obtained by multi 7
plying the elements of X by xi  bi. Hence A’ has degree £
d+1. QED
LEMMA 2.4. Let A be algebraic in Fk of degree d and d’ ≥ d.
Then finitely many elements can be added to A so that the
degree is exactly d’.
Proof: By Lemma 2.3, we can succeively add points without
increasing the degree by more than one. Thus it suffices to
show that we cannot successively add points forever and keep
the degrees bounded. If we could, we would get arbitrarily
long strictly descending sequences of algebraic sets, all
with a bound on the degrees. This violates Theorem 2.1, or
even linear algebra. QED
LEMMA 2.5. Let A1,A2,...,An be strictly decreasing sets
algebraic in Fk with degrees £p,£p+1,...,£p+n, where each
Ai+1\Ai is infinite. Then we can find such A1’,A2’,...,An’ with
degress p,p+1,...,p+n.
Now we construct long such A1,A2,...,An. Let T be a finite
tree with at least two vertices, where every path excluding
the root is of length £ k. Assume that every vertex except
the root is labeled by an element of F, where no label is
repeated. The algebraic meaning of a vertex at the ith level
above the root with label c is that xi = c. The algebraic
meaning of a path is the conjunction of the algebraic meaning
of each vertex along that path other than the root. The
algebraic meaning of the tree T is the set [T] of all
elements of Fk that obey the algebraic meaning of at least one
of the paths.
These T are called the good trees.
Thus [T] is a certain union of intersections. The length of
the union is #T = the number of terminal vertices of T. When
we write [T] as an intersection of unions, note that the
unions are still of length #T, although there may be
redundancies. Write each union as a monomial by multiplying
the relevant factors xc. Hence the degree of [T] is at most
#T.
We need to develop a sufficient criterion for [T] to properly
contain [T’]. 8
LEMMA 2.6. Let T,T’ be good trees. Suppose T’ is obtained
from T by adding one or more child to a terminal vertex. Or
suppose T’ is obtained from T by deleting one of the children of a vertex that has at least two children (and of course
all vertices above the one deleted). Then [T] properly
contains [T’]. Also [T]\[T’] is infinite.
Now all we have to do is to deal with the combinatorics of
these two tree operations. Obviously any good tree with only
a root and children of the root has degree equaled to the
number of children.
Let G(k,p) be the longest sequence of good trees of height k
starting with a root with p children, where each tree is
obtained from the preceding tree by one of these two
operations, and has 1 more terminal vertex.
Let’s look at G(k+1,p). Make p children to a root. Off of the
first child create two children, and imitate the sequence
used there for G(k,1). When finished there, we will have
G(k,1)+p1 terminal vertices. Remove this left child and all
vertices above it, leaving p1 terminal vertices, and then
create G(k,1)+2 children above the second original child of
the root. Continue this process p times. This is at least the
iteration of G(k,_) p times starting at 1. So we have at
least primitive recursion, provided we can get going. We
don’t get going with k = 1 since G(1,p) = p. But G(2,p) =
(p)(p+1). So we get G(k,p) ≥ A(k1,p), k ≥ 2. In particular,
G(3,p) > 22^p, G(4,p) > E*(2p). Detailed calculations give
G(4,1) = 22^72 and G(4,1) = 2^2^2^2^72.
3. SEMILINEAR DYNAMICS
Semilinear dynamics appears not only of interest in its own
right, but also can serve as an elegant interpretation of
recursion theory.
The basic idea is that rational semilinear functions simulate Turing degrees. In particular, the structure of orbits
has the requisite power.
A semilinear set in Rn is a finite Boolean combination of
halfplanes in Rn. Halfplanes are given by T(x) > 0 or T(x) ≥
0, where T is an affine tranformation (the shift of a linear
transformation). 9
A rational semilinear set in Rn is a finite Boolean combination of rational halfplanes in Rn. Rational halfplanes are
given by T(x) > 0 or T(x) ≥ 0, where T is an affine transformation whose matrix (including coordinates of the constant
term) has only rational entries.
A (rational) semilinear function from Rn into Rm is a function whose graph is (rational) semilinear in Rn+m.
Quantifier elimination for the field of real numbers is more
than enough to establish various nice properties of semilinear functions. Quantifier elimination for the ordered
group of real numbers is enough to establish various nice
properties of rational semilinear functions.
An important special case is that of rational continuous
semilinear functions from R2 into R2, and also from [0,1]2
into [0,1]2.
The basic idea behind the results is that there is a usable
simulation of Turing machines in the rational semilinear
functions on R and on I2, and also in the continuous ones.
A Turing machine computation can be viewed as proceeding on a
finite tape of squares, each holding a symbol from a finite
alphabet of “symbols”, with the reading head hovering above
one of the squares, and where the machine as a whole is
viewed as being in one of another finite alphabet of
“states”. We assume that these two alphabets are disjoint.
The machine instructions are a list of quadruples
qaa’q’
qaRq’
qaLq’
The first means “if you are in state q and reading symbol a
then replace it with symbol a’ and go into state q’.” The
second means “if you are in state q and reading symbol a then
move to the right and go in state q’.” The third means “if
you are in state q and reading symbol a then move to the left
and go in state q’.”
It is required that there are no conflicting quadruples;
i.e., no two quadruples have the same first two terms. 10
There is a special element among the states called the
initial state, and there is a special symbol called the
blank.
Certain of the symbols are designated as input symbols, and
certain are designated as output symbols. Some symbols are
neither input nor output symbols. The blank cannot be an
input or output symbol.
Inputs are strings of input symbols, and outputs are strings
of output symbols. Computation is initiated by creating
squares of tape to hold the input string preceded by the
first square holding the blank, and on which the reading head
rests. The machine is put into the initial state.
Output occurs if and when no instruction applies. The output
consists of the string formed by the output symbols on the
tape from left to right.
No squares of tape are ever destroyed. However, a new square
of tape is created whenever a left or right instruction goes
over the left or right edge. The new square is considered to
have the blank placed on it.
Suppose the combined set of symbols and states is {v0,...,vp},
where v0 is the blank. At any stage of computation, the global
state of the machine is represented by
x1x2...xn:y1y2...ym
where n ≥ 0 and m ≥ 2. The colon indicates that the reading
head is hovering over y1, where y1 is the state, and is
reading the symbol y2. The x’s indicate the symbols to the
left of the reading head, and the remaining y’s indicate
those to the right of the reading head.
We can represent this as an element of [0,1]2 as follows.
(.xnxn1...x1, .y1y2...ym)
where these are decimals in base 2p, with the blank v0
recorded as 0, and v1,...,vp recorded as 2,4,...,2p.
Then the quadruples can be executed by simple rational
semilinear functions on [0,1]. We only care about the action
on the set of finite decimals in base 2p with even digits. In 11
fact, we can make the provisional domain the set of all
elements of [0,1]2 that can be written in base 2p, whose first
coordinate has an even first digit, and whose second
coordinate has its first two coordinates even. This amounts
to having a provisional domain consisting of the union of a
finite set of pairwise disjoint closed rectangles. The
function is rational affine on each rectangle. The mapping
sends elements of [0,1]2 that can be written as an infinite
pair of decimals with even digits into themselves.
Now any rational continuous semilinear function defined on
this provisional domain can be extended to a rational continuous semilinear function on I2, or even on R2.
In some constructions, we need to venture into decimals with
no restrictions on the digits, rather than the more safe
situation above. This has to be done carefully, especially if
we are to stay continuous. So some of the results below are
stated without continuity and some with. I won’t be sure that
I can stay within the continuous functions until a detailed
writeup.
Let f:X Æ X and x Æ X. The orbit of f at x is
{x,fx,ffx,...}. The diameter of a nonempty bounded subset of
the plane is the sup of the Euclidean distance between pairs
of points in the subset.
THEOREM 3.1. The diameters of the bounded orbits of rational
semilinear functions on R2 at rational points are exactly the
r.e. positive real numbers. These are the positive real
numbers whose left cut is an r.e. set of rationals.
The r.e. positive reals seem interesting. E.g., what can you
say about their base 2 expansions? They are closed under
addition and multiplication, but not subtraction or
division. Any recursive increasing homeomorphism of R+ is an
automorphism of the r.e. positive reals, but what other kinds
of automorphisms should be sought?
THEOREM 3.2. The diameters of the bounded orbits of rational
semilinear functions on R2 at x are exactly the positive real
numbers that are r.e. in x. As a consequence, let x,y Œ R2.
Then x is recursive in y if and only if the diamaters
obtained in this way from x are obtained in this way from y. 12
THEOREM 3.3. The r.e. subsets of Z2 are exactly the integer
points in orbits of rational continuous semilinear functions
on R2. ...
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This note was uploaded on 08/05/2011 for the course MATH 366 taught by Professor Joshua during the Fall '08 term at Ohio State.
 Fall '08
 JOSHUA
 Math, Algebra, Sets

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