1
THE ACKERMANN FUNCTION IN ELEMENTARY ALGEBRAIC GEOMETRY
by
Harvey M. Friedman
[email protected]
http://www.math.ohiostate.edu/~friedman/
October 21, 2000
1. Exact subideals.
2. Algebraic approximations.
3. Upper bounds for algebraic approximations.
4. Ascending chains of ideals 
historical notes.
5. Ascending chains of ideals.
6. Decreasing chains of algebraic sets.
7. Degrees in ideal bases.
8. Degrees in polynomial presentations of algebraic sets.
9. Additional formulations.
1. THE ACKERMANN HIERARCHY.
Let f:Z
+
into Z
+
be strictly increasing.
Define f':Z
+
into Z
+
by f'(n) = f.
..f(1), where there are n f's.
Define A
1
(n) = 2n. For each k
≥
1, define A
k+1
= A
k
'. Finally,
define A(k,n) = A
k
(n), and A(k) = A(k,k).
We can equivalently present this by the recursion equations
f
1
(n) = 2n, f
k+1
(1) = f
k
(1), f
k+1
(n+1) = f
k
(f
k+1
(n)), where k,n
≥
1. We define A(k,n) = f
k
(n).
A(3,5) = 2
65,536
. A(4,3) = 65,536. A(4,4) = E*(65,536). And
A(4,5) is E*(E*(65,536)).
It seems safe to assert, e.g., that A(5,5) is incomprehen
sibly large. We propose this number as a sort of benchmark.
The following facts about A are useful, and are easily proved
in the order stated.
THEOREM 1.1. For all k,n
≥
1, n < A
k
(n) < A
k
(n+1). For all k
≥
1 and n
≥
3, A
k
(n) < A
k+1
(n). For all k,n
≥
1, A
k
(n)
£
A
k+1
(n).
For all k
≥
1, A
k
(1) = 2, A
k
(2) = 4, and A
k
(3)
≥
2
k+1
. For all
k
≥
3, A
k
(3)
≥
A
k2
(2
k
) > A
k2
(k2). If k
≥
n+5 then A
k
(3) >
A
n
(k).
Ackerman's original definition is similar but is ternary. The
growth rates are the same.
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1. FULL SUBIDEALS.
Here we take the degree of an ideal in a polynomial ring over
a field is the least d such that the ideal has a set of
generators all of which have degree at most d. When we
discuss algebraic sets, we will consider the usual notion of
degree in algebraic geometry.
Let k
≥
0 and I be an ideal in the polynomial ring
F[x
1
,...,x
k
], F a field. I has finite degree by the Hilbert
basis theorem. We write d(I) for the degree of I.
For each n
≥
0, we let I*n be the ideal generated by the
elements of I of degree n. We call these the full subideals
of I.
We let I*
£
n be the ideal generated by the elements of I of
degree
£
n.
THEOREM 1.1. Let F be a field, k
≥
1, and I be an ideal in
F[x1,.
..,xk]. For all n
≥
0, I*n = I*
£
n. I*0 .
.. I*d(I) =
I = I*d(I)+1 = .
.. . For all n
≥
0, I*n = I*d(I*n). Any two
full subideals with the same degree are equal. If I
≠
F[x
1
,...,x
n
] then I*0 = {0}.
Proof: For the first claim, let P in I have degree < n. By
multiplication, let Q in I have degree n. Then Q+P in I has
degree n. Hence P lies in I*n.
Now suppose I*n has degree d. Let K be a set of generators of
I each of degree
£
d. Then I*n I*d. Also d
£
n since I*n is
generated by the elements of I of degree
£
n. Hence I*d
I*
£
n = I*n.
From the first claim, clearly the full subideals of I are the
ideals I*
£
n, which form a chain under inclusion of length at
most d(I)+1.
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 Fall '08
 JOSHUA
 Math, Algebra, Geometry, Approximation, Algebraic geometry, A*n

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