SIMPLIFYING SMOKTUNOWICZ’S EXTRAORDINARY EXAMPLE
PACE P. NIELSEN
Abstract. At the turn of the 21st century Agata Smoktunowicz constructed the first example of a nil
algebra over a countable field such that the polynomial ring over the algebra is not nil. This answered
an old question of Amitsur. We present a simplification of the example.
Introduction
This monograph began as the author’s earnest attempt to understand the beautiful example constructed in [1], and was originally meant only to serve as a set of personal notes. However, the author
believes the simplifications offered will draw more people to understand and appreciate Smoktunowicz’s
example. Further, as of the writing of this paper the example has been used in more than twenty
other research papers, and continues to be generalized and expanded. It is hoped that this note will
aid in this process and simplify future constructions. Almost all of the original research comes from
Smoktunowicz’s work in [1], and many of the lemmas, theorems, and corollaries of this note are direct
generalizations of results from that paper. The present paper reworks and streamlines the bounds and
numerical computations of that paper. Further, following the methods of [2], we answer the implicit
question left open at the end of the paper, showing that the main construction technique needs only
one polynomial variable (and not two) to reach the desired example.
Rather than construct each of the components of the example separately and then put them together
at the end, we take a motivational approach where we give the underlying reason for each piece before
it is constructed. It is hoped that this will aid the reader in keeping track of the important information
as it is needed. As the paper is in essence a single example, any quantities defined outside of lemmas,
theorems, or corollaries will continue to be defined throughout the paper and retain their meanings
when invoked in proofs or in the statement of propositions.
1. Nilpotence over polynomial rings
Let K be a countable field. Our goal is to construct a nil K-algebra R such that the polynomial
ring R[X] is not nil. The first half of our goal is simple. We let A = Khx1 , x2 , x3 i be the (unital) free
algebra in the three noncommuting variables x1 , x2 , and x3 . We also let A0 denote the (nonunital) Ksubalgebra of A consisting of the elements with zero constant term. We fix an indexing {f1 , f2 , f3 , . . .}
of A0 by the positive integers. This is the one and only place where we use the countability of K. For
each i ≥ 1 we let ei be a positive integer and we let I be an ideal of A containing fiei for every i ≥ 1.
We set R = A0 /I and see that R is nil.
As arbitrary ideals are difficult to work with, we take the simplifying position that I is a homogeneous
ideal. Thus, we assume:
(1)
The ideal I is generated by the homogeneous components of {fiei : i ≥ 1}.
One might ask: Why use three noncommuting variables to form A? It turns out that in fact two
variables suffice (which will be shown at the end of this section), but with three the proof is significantly
2010 Mathematics Subject Classification. Primary 16N40, Secondary 16N20, 16S36.
Key words and phrases. Jacobson radical, nil ring, polynomial ring.
1
2
PACE P. NIELSEN
simplified. In particular, condition (1) is compatible with the 3-generator construction but not with the
corresponding 2-generator construction.
The difficulty in effectuating the second half of our goal is in controlling the exponents ei enough so
that R[X] is not nil. Intuitively, if the exponents grow quickly then that should suffice. The remainder
of the paper will make formal what constitutes “quickly.” As might be expected, we demonstrate that
R[X] is not nil by exhibiting an element of the form a + bX ∈ R[X] which is not nilpotent. There is a
characterization of when a + bX is nilpotent which we can translate into information on the exponents
ei . We begin with the following easy exercise:
Lemma 1. Given a ring S and two elements a, b ∈ S, then a + bX ∈ S[X] is nilpotent if and only if
1 − (aX + bY ) ∈ S[X, Y ] is a unit in S[X, Y ].
Proof. We first introduce some notation. Given integers i, j ≥ 0, let w(i, j) denote the coefficient of
X i Y j in (aX + bY )i+j . Notice that f (X, Y ) = 1 − (aX + bY ) is always a unit in the power series ring
SJX, Y K, with inverse given by
X
X
g(X, Y ) =
(aX + bY )n =
w(i, j)X i Y j .
n≥0
i,j≥0
Thus f (X, Y ) is a unit in S[X, Y ] if and only if g(X, Y ) is a (finite) polynomial. But g(X, Y ) is a
polynomial if and only if w(i, j) = 0 whenever i + j is large enough.
On the other hand, a + bX is nilpotent if and only if there exists an integer N ≥ 1 so that for all
integers n ≥ N , we have
n
X
0 = (a + bX)n =
w(n − j, j)X j .
j=0
Or equivalently, w(n − j, j) = 0 for all integers n ≥ N and all integers j ≥ 0. This is exactly the same
condition as at the end of the first paragraph.
Set xi = xi + I ∈ R and consider the element z = x1 + x2 X + x3 Y ∈ R[X, Y ]. Recall that as R is nil
this entails that x1 ∈ R is nilpotent, say of index k ≥ 1. Let y = (1−x1 )−1 = 1+x1 +x21 +· · ·+xk−1
∈ R.
1
We compute
1 − z = (1 − x1 )(1 − yx2 X + yx3 Y ).
Thus, by the previous lemma, yx2 + yx3 X ∈ R[X] is nilpotent if and only if z is a unit in R[X, Y ]. Our
job now is to:
(2)
Choose the exponents ei so that 1 − (x1 + x2 X + x3 Y ) is not a unit modulo I in R[X, Y ].
Once this is done we will know R (or even the subalgebra of R generated by the two elements a = yx2
and b = yx3 ) gives us the example we are after.
2. Homogeneous polynomials related to x1 + x2 X + x3 Y
We introduce some notation and terminology which will be central throughout the rest of the paper.
By N we will mean the set of non-negative integers. Let Mn denote the set of monomials in x1 , x2 , x3
of total degree n ∈ N, and let Hn denote the set of homogeneous polynomial of degree n. Note that
|Mn | = 3n , and these monomials form a basis for the K-vector space Hn . Let w = y1 y2 · · · yn ∈ Mn ,
where yi ∈ M1 for each i ≥ 1. Given a permutation σ ∈ Sn , we define wσ = yσ(1) yσ(2) · · · yσ(n) ∈ Mn .
We extend this action K-linearly to Hn . Given f ∈ A we let deg(f ) denote the largest total degree of
a monomial appearing with nonzero support in f , and let dxi (w) denote the degree of the monomial w
in the variable xi .
SIMPLIFYING SMOKTUNOWICZ’S EXTRAORDINARY EXAMPLE
3
For each n ∈ N we let Tn = {(n1 , n2 , n3 ) ∈ N3 : n1 + n2 + n3 = n}. Given r = (n1 , n2 , n3 ) ∈ Tn we
define
X
w(r) = w(n1 , n2 , n3 ) =
{w ∈ Mn1 +n2 +n3 : dx1 (w) = n1 , dx2 (w) = n2 , dx3 (w) = n3 }.
In other words, this is the sum of all monomials w of total degree n = n1 + n2 + n3 whose degree in xi
is ni . We call these the nice homogeneous polynomials.
There is an interesting connection between these nice polynomials and x1 + x2 X + x3 Y ∈ A[X, Y ].
Namely, given n1 , n2 , n3 ∈ N the coefficient in (x1 + x2 X + x3 Y )n of X n2 Y n3 is w(n1 , n2 , n3 ) if n1 +
n2 + n3 = n. There is another connection, which is related to what happens in the proof of Lemma 1.
Lemma 2. The element 1 − (x1 + x2 X + x3 Y ) ∈ A[X, Y ] is a unit modulo I in R[X, Y ] if and only if
there exists an integer N ≥ 1 such that for each n ≥ N and each r ∈ Tn , w(r) ∈ I.
Proof. Recall that from (1), x1 is nilpotent modulo I, say of index k ≥ 1. We see that if r = (n1 , n2 , n3 ) ∈
N3 is a triple with n1 > (k − 1)(n2 + n3 + 1) then any monomial in w(r) has a string of k successive
x1 ’s, and thus belongs to I. In particular, 1 − (x1 + x2 X + x3 Y ) has an inverse modulo I in the power
series ring RJX, Y K, given by


X
X
X

w(n1 , n2 , n3 ) X n2 Y n3 + IJX, Y K.
(x1 + x2 X + x3 Y )n + IJX, Y K =
n∈N
n2 ,n3 ∈N
0≤n1 ≤(k−1)(n2 +n3 +1)
Thus 1 − (x1 + x2 X + x3 Y ) is a unit modulo I in R[X, Y ] if and only if the power series above is a
finite polynomial. As I is homogeneous, the power series is a polynomial if and only if w(r) ∈ I for all
r ∈ Tn , for all sufficiently large integers n.
This lemma lets us rephrase our goal (2) in the following form:
(3)
Choose the ei and an increasing sequence of integers 1 ≤ m1 < m2 < m3 < . . . so that
for each j ≥ 1 there is some triple r ∈ Tmj with w(r) ∈
/ I.
This will guarantee the noninvertibility of 1 − (x1 + x2 X + x3 Y ) modulo I in R[X, Y ].
It is difficult turn information about the exponents ei and the sequence of integers mi into a proof
that w(r) ∈
/ I for some r ∈ Tmi . Our way around this difficulty is to construct a map (one for each
i ≥ 1) which is zero on I ∩ Hmi but nonzero on at least one nice polynomial in Hmi . We begin this
process by first investigating how nice polynomials behave under K-linear mappings in §3. Then in §4
we study the behavior of the homogeneous components of fiei . Finally, we put our knowledge together
to actually construct the needed maps in §5.
3. Mappings on the nice polynomials
We start with two easy lemmas. The proofs are left to the reader.
Lemma 3. The following hold:
(a) For each n ∈ N, the set S = {w(r) : r ∈ Tn } is right linearly independent over A.
(b) Given a, b ∈ N, if t ∈ Ta+b then
X
w(t) =
w(r)w(s).
r∈Ta , s∈Tb , r+s=t
Let ≺ denote the lexicographic ordering on Z3 . So (m1 , m2 , m3 ) ≺ (n1 , n2 , n3 ) holds if and only if
(i) m1 < n1 , or (ii) m1 = n1 and m2 < n2 , or (iii) m1 = n1 , m2 = n2 , and m3 < n3 .
Lemma 4. Let r, r0 , s ∈ Z3 . If r ≺ r0 then r + s ≺ r0 + s.
4
PACE P. NIELSEN
Assume a, b ∈ N, and let r0 ∈ Ta and s0 ∈ Tb . Putting Lemma 3(b) and Lemma 4 together we have
X
w(r)w(s).
(4)
w(r0 + s0 ) = w(r0 )w(s0 ) +
r∈Ta , s∈Tb , r+s=r0 +s0 ,
r≺r0 or s≺s0
The next lemma tells us that a multiplicative K-linear function cannot be zero on nice polynomials,
unless one of the component functions is zero on nice polynomials.
Lemma 5. Let a, b ∈ N. Suppose we have three K-linear maps ϕ : Ha+b → A, γ : Ha → A, and
δ : Hb → A satisfying ϕ(uv) = γ(u)δ(v) for each u ∈ Ha and each v ∈ Hb . If γ(w(r)) 6= 0 for some
r ∈ Ta and δ(w(s)) 6= 0 for some s ∈ Tb then ϕ(w(t)) 6= 0 for some t ∈ Ta+b .
Proof. Let r0 = (a1 , a2 , a3 ) ∈ Ta and s0 = (b1 , b2 , b3 ) ∈ Tb be minimal with respect to the lexicographical
ordering such that γ(w(r0 )) 6= 0 and δ(w(s0 )) 6= 0. By equation (4), and the multiplicative nature of
ϕ, we have
X
ϕ(w(r0 + s0 )) = ϕ(w(r0 )w(s0 )) +
ϕ(w(r)w(s))
r∈Ta , s∈Tb , r+s=r0 +s0 ,
r≺r0 or s≺s0
X
= γ(w(r0 ))δ(w(s0 )) +
γ(w(r))δ(w(s)) = γ(w(r0 ))δ(w(s0 )) 6= 0.
r∈Ta , s∈Tb , r+s=r0 +s0 ,
r≺r0 or s≺s0
It may be of interest to note that the previous lemma is true more generally if each of ϕ, γ, and δ
map to a common K-algebra without zero-divisors.
The previous lemma told us that if ϕ is multiplicative and the components γ and δ are not zero on
all nice polynomials, then ϕ is not zero on all nice polynomials. If one of the component functions is
the identity then we have even more information available.
Theorem 6. Let a, b ∈ N. Suppose ϕ : Ha+b → A and δ : Hb → A are K-linear maps, with
ϕ(uv) = uδ(v) for each u ∈ Ha and each v ∈ Hb . If δ(w(s)) 6= 0 for some s ∈ Tb then the dimension of
the K-subspace of A spanned by {ϕ(w(t)) : t ∈ Ta+b } is ≥ (a + 1)(a + 2)/2.
Proof. Let s0 ∈ Tb be the unique element of Tb which is minimal with respect to the lexicographical
ordering and subject to the condition δ(w(s0 )) 6= 0. Fix r0 = (a1 , a2 , a3 ) ∈ Ta . Notice that there are
exactly (a + 1)(a + 2)/2 = |Ta | choices for such an r0 .
By Lemma 3(a) we know that
X
(5)
w(r)A
r∈Ta
is a direct sum. Applying ϕ to (4), we have
ϕ(w(r0 + s0 )) = ϕ(w(r0 )w(s0 )) +
X
ϕ(w(r)w(s))
r∈Ta , s∈Tb , r+s=r0 +s0 ,
r≺r0 or s≺s0
= w(r0 )δ(w(s0 )) +
X
w(r)δ(w(s))
r∈Ta , s∈Tb , r+s=r0 +s0 ,
r≺r0 or s≺s0
= w(r0 )δ(w(s0 )) +
X
w(r)δ(w(s)).
r∈Ta , s∈Tb , r+s=r0 +s0 ,
r≺r0
Thus, we see that the largest r ∈ Ta for which ϕ(w(r0 + s0 )) has a nonzero summand occurring in
the decomposition (5) is r0 . As r0 ranges over Ta , the elements ϕ(w(r0 + s0 )) have their highest
SIMPLIFYING SMOKTUNOWICZ’S EXTRAORDINARY EXAMPLE
5
nonzero components in different summands in (5), hence the set {ϕ(w(r0 + s0 )) : r0 ∈ Ta } is K-linearly
independent and has cardinality (a + 1)(a + 2)/2.
4. The homogeneous components of powers of elements
We now understand how the nice polynomials behave under K-linear mappings. What we need next
is an understanding of how the homogeneous components of fiei behave. In particular, as our goal is to
construct maps which are zero on I ∩ Hmi we need to understand how “large” this set is. The following
lemma and theorem address what happens for a single element f ∈ A0 .
Lemma 7. Let f ∈ A0 have degree t ≥ 1. For each n ≥ t there exists a set G(f, n) ⊆ Hn with
2(t − 1)(3t − 1)n − 3t (t − 3)(2t − 1) − 5t + 3
4t
such that, for every integer m ≥ n, f m belongs to the right ideal of A generated by polynomials of the
form hgf m−n , where h ∈ G(f, n) and deg(g) < t.
Pt
Proof. Write f = i=1 fi where fi ∈ Hi . Thus f m is a sum of terms of the form fi1 fi2 · · · fik fik+1 · · · fim ,
where k denotes the largest subscript such that i1 + i2 + · · · + ik ≤ n. We then have the decomposition:



t−1
X
X
X
X


fm =
fi1 · · · fik f m−k +
(fi1 · · · fik )fik+1 f m−k−1 
|G(f, n)| ≤
(6)
k≤n
i1 +···+ik =n
j=1

=
X
v0,k f
m−k
k≤n
X
+
t−1
X

k≤n−1
i1 +···+ik =n−j, ik+1 >j

vj,k (fj+1 + fj+2 + · · · + ft )f
m−k−1 
.
j=1
where vj,k is the homogeneous
component of degree n − j in f k .
S
S
Taking G(f, n) = 0≤j≤t−1 k vj,k Mj , the representation above has the desired properties. To see
this, note that in the second summation each term vj,k fj+i f m−k−1 lies in the span of all members of
vj,k Mj+i f m−n f n−k−1 ⊆ vj,k Mj Mi f m−n A. As elements of vj,k Mj lie in G(f, n) and elements of Mi
have degree < t, the product lives in the desired right ideal. The same is still more easily seen for terms
of the first summation.
Notice that if we want vj,k to be nonzero then we necessarily must have (n − j)/t ≤ k ≤ n − j. We
compute
|G(f, n)| ≤
t−1
X
n−j
X
j=0 k=d(n−j)/te
3j ≤
2(t − 1)(3t − 1)n − 3t (t − 3)(2t − 1) − 5t + 3
.
4t
Given an integer n ≥ 1 and a subset F ⊆ Hm we let
∞
X
(7)
Bn (F ) =
Mni F A,
i=0
which is a right ideal. Intuitively, Bn (F ) is nearly the two-sided ideal in A generated by the subset F
except that on the left we only assume closure under multiplication by monomials of degree divisible
by n.
Theorem 8. Let f ∈ A0 have degree t ≥ 1. For each r ≥ 2t − 1 there exists a set F ⊆ Hr such that
(t − 1)(3t − 1)r − 3t (2t2 − 6t + 3) − 4t + 3
< 32t−1 r
4t
and, for every integer s ≥ r, the ideal hf 3s i is contained in Bs (F ).
(8)
|F | ≤ (3t − 1)
6
PACE P. NIELSEN
S
Proof. Let F = 0≤i≤t−1 Mi G(f, r − i). The first inequality in (8) now follows from the previous
lemma. One checks the second inequality for t = 1, 2 directly, and for t ≥ 3 it is straightforward.
To prove the remainder of the lemma, it suffices to show that uf 3s ∈ Bs (F ), for each monomial
u of degree d with 0 < d < s. Clearly t < 2s − d < 3s so we apply the previous lemma to f with
n = 2s − d and m = 3s. Thus f 3s belongs to the right ideal of A generated by elements of the form
hgf 3s−2s+d = hgf s+d , where h ∈ G(f, 2s − d) and deg(g) < t. As uh ∈ H2s it suffices to show that
gf s+d ∈ Bs (F ), where we may assume g is a monomial of degree ` < t. As t ≤ r − ` < s + d we again
apply the previous lemma to f , but this time with n = r − ` and m = s + d. It follows that f s+d belongs
to the right ideal of A generated by expressions of the form h0 g 0 f s+d−r+` , where h0 ∈ G(f, r − `) and
deg(g 0 ) < t. By definition, gh0 ∈ F , and thus gf s+d ∈ F A ⊆ Bs (F ) as wanted.
Remarks: (a) In the case that t = 1 we have F = {f r } and one can prove that s + r − 1 works in
place of 3s.
(b) As Bs (F ) is a homogeneous right ideal, it contains not only the ideal hf 3s i but also the homogeneous components of this ideal.
We wish to rephrase Theorem 8 in terms of all elements of A0 simultaneously. To do this we make a
simplifying assumption about the indexing {f1 , f2 , . . .} = A0 . We choose this indexing so that
(9)
f1 = x1 and deg(fi ) ≤ i/2 when i > 1.
Corollary 9. Let 1 ≤ m1 < m2 < m3 < . . . be an increasing sequence of positive integers. There exist
subsets Fi ⊆ Hmi with
(10)
|Fi | ≤ 3i−1 mi
3mi+1 ∞
}i=1
such that the two-sided ideal of A generated
P∞ by the homogeneous components of {fi
in the K-subspace (and right ideal) i=1 Bmi+1 (Fi ).
is contained
Given a sequence 1 ≤ m1 < m2 < m3 < . . . fix subsets Fi (once and for all) satisfying this corollary.
The corollary tells us how to choose our exponents:
(11)
For each integer i ≥ 1, fix ei = 3mi+1 .
The P
sequence {mi }i≥1 will be determined later. Note that with this choice of exponents we have
∞
I ⊆ i=1 Bmi+1 (Fi ). We thus restate our ultimate goal (3) as:
(12)
Choose an increasing sequence of integers 1 ≤ m1 < m2P
< m3 < . . . so that
∞
for each i ≥ 1 there is some triple r ∈ Tmi with w(r) ∈
/ i=1 Bmi+1 (Fi ).
5. Linear maps
There is another simplifying assumption we enforce on the sequence 1 ≤ m1 < m2 < m3 < . . .,
namely:
(13)
For each i ≥ 1, mi |mi+1 .
This divisibility property will be essential for what happens next.
For each subset S ⊆ Hn , fix (once and for all) a K-vector space decomposition span(S) ⊕ S ⊥ = Hn .
We let ψS : Hn → Hn be the projection map which is the identity on S ⊥ and zero
P∞ on S. We are now
ready to define K-linear maps which will be zero on homogeneous components of i=1 Bmi+1 (Fi ). First,
define ϕ1 : Hm1 → Hm1 to be the identity map. Now we use condition (13); given a monomial w ∈ Mm2
we can write w = w1 w2 · · · wm2 /m1 where wj ∈ Mm1 for each j ≥ 1. We define ϕ2 : Mm2 → Mm2 by
the rule
m2 /m1
Y
ϕ2 (w) = ψϕ1 (F1 ) (ϕ1 (w1 ))
ϕ1 (wj ).
j=2
SIMPLIFYING SMOKTUNOWICZ’S EXTRAORDINARY EXAMPLE
7
In other words, we repeat ϕ1 a total of m2 /m1 times, except that on the first instance we post-compose
with a projection map. Thus ϕ2 is simply the identity, except that on the first m1 variables we kill
anything that lived in the span of ϕ1 (F1 ), and project to a complement. This map is defined on all of
Hm2 by extending K-linearly.
We repeat this process and define ϕi+1 inductively by the rule
mi+1 /mi
(14)
ϕi+1 (w) = ψϕi (Fi ) (ϕi (w1 ))
Y
ϕi (wj )
j=2
where w = w1 w2 · · · wmi+1 /mi ∈ Mmi+1 and wj ∈ Mmi for each j ≥ 1.
Theorem 10. For each ` ≥ 1 we have
(15)
Hm`+1 ∩ I ⊆ Hm`+1
`
\ X
!
Bmi+1 (Fi )
⊆ ker(ϕ`+1 ).
i=1
Proof. As ϕ`+1 is K-linear it suffices to show that ϕ`+1 (x) = 0 for each x ∈ Hm`+1 ∩ Bmi+1 (Fi ), for each
i in the range 1 ≤ i ≤ `. From (7) we know x is a sum of terms of the form gf h where f ∈ Fi ⊆ Hmi ,
g ∈ Mtmi+1 for some t ∈ N, and h ∈ Mm`+1 −tmi+1 −mi . Without loss of generality we may assume
x = gf h.
We want to break x = gf h into a product of m`+1 /m` homogeneous pieces each of degree m` . Write
g = x1 x2 · · · xk−1 g 0 where xj ∈ Mm` for each j ≥ 1 and deg(g 0 ) < m` . As mi |mi+1 and mi |m` |m`+1
we have deg(g 0 f ) ≤ m` , and so we can also write h = h0 h00 with deg(g 0 f h0 ) = m` . We fix xk = g 0 f h0 .
As deg(x) = m`+1 = (m`+1 /m` )m` we can further decompose h00 as h00 = xk+1 xk+2 · · · xm`+1 /m` with
xj ∈ Mm` for each j ≥ k + 1, and in particular x = x1 x2 · · · xm`+1 /m` . The inductive definition of ϕ`+1
given by (14) yields
m`+1 /m`
(16)
ϕ`+1 (x) = ψϕ` (F` ) (ϕ` (x1 ))
Y
ϕ` (xj ).
j=2
We establish ϕ`+1 (x) = 0 by induction on ` ≥ 1.
Case 1: Assume i = `. Notice that this includes the base case when ` = 1. We have g = 1, k = 1
and x1 = f ∈ F` . Thus the term ψϕ` (F` ) (ϕ` (x1 )) in equation (16) is zero.
Case 2: Assume ` > 1 and i < `. We have mi+1 |m` and so mi+1 | deg(g 0 ). Thus xk = g 0 f h0 ∈
Hm` ∩ Bmi+1 (Fi ), and so our inductive hypothesis implies ϕ` (xk ) = 0 and hence (16) is still zero. We have now accomplished the first half of what we set out to do. We have defined maps ϕi which
are zero on Hmi ∩ I, for each i ≥ 2. All that remains is to show that they do not annihilate all nice
polynomials in Hmi . To do this we will need an auxiliary sequence of integers which measures how
much ϕi acts as the identity map (in a sense to be made formal shortly). Define
mi+1
− 1 for i ≥ 1.
(17)
p1 = m1 and pi+1 = pi
mi
Note that (13) implies that this new sequence also consists of integers satisfying pi |pi+1 .
Lemma 11. Given an integer c ≥ 2, if for every i ≥ 1 we have mi = ci−1 then pi = (c − 1)i−1 .
The sequence of numbers pi is defined so that (in retrospect) the proof of the following theorem
flows smoothly. Before we begin the proof, we introduce one piece of notation. Given a K-linear map
ϕ : Hn → Hn and a permutation σ ∈ Sn we define a new linear map ϕσ : Hn → Hn by
ϕσ (w) = (ϕ(wσ ))σ
−1
.
8
PACE P. NIELSEN
Theorem 12. For each integer i ≥ 1 there exists σi ∈ Smi and a K-linear map δi : Hmi −pi → Hmi −pi
satisfying, for each u ∈ Hpi and each v ∈ Hmi −pi , the formula
ϕσi i (uv) = uδi (v).
(18)
Proof. To prove the theorem we induct on i ≥ 1. The case i = 1 is trivial.
Given σ ∈ Sk and any integer r ≥ 1, define σ ∈ Srk by taking a monomial w1 w2 · · · wr (where
wj ∈ Mk ) to w1σ w2σ · · · wrσ . In other words, we just repeat σ on the appropriate length pieces.
Each monomial u ∈ Mpi is uniquely written as u = u1 u2 · · · u(mi /mi−1 )−1 with each uj ∈ Mpi−1 .
Similarly, v ∈ Mmi −pi can be written uniquely as v = wv1 v2 · · · v(mi /mi−1 )−1 where vj ∈ Mmi−1 −pi−1
and w ∈ Mmi−1 . Define a permutation ρi ∈ Smi by
(uv)ρi = wu1 v1 u2 v2 · · · u(mi /mi−1 )−1 v(mi /mi−1 )−1 .
Recursively, we set σi = σ i−1 ◦ ρi . Working out what this does to the left-hand side of (18), we
compute
σ−1
−1
ϕσi i (uv) = (ϕi ((uv)σi ))σi = ϕi (wσi−1 (u1 v1 )σi−1 (u2 v2 )σi−1 · · · (u(mi /mi−1 )−1 v(mi /mi−1 )−1 )σi−1 ) i
ρ−1
σi−1
σi−1
σi−1
= (ψϕi−1 (Fi−1 ) ◦ ϕi−1 )σi−1 (w)ϕi−1
(u1 v1 )ϕi−1
(u2 v2 ) · · · ϕi−1
(u(mi /mi−1 )−1 v(mi /mi−1 )−1 ) i
ρ−1
= (ψϕi−1 (Fi−1 ) ◦ ϕi−1 )σi−1 (w)u1 δi−1 (v1 )u2 δi−1 (v2 ) · · · u(mi /mi−1 )−1 δi−1 (v(mi /mi−1 )−1 ) i
= u (ψϕi−1 (Fi−1 ) ◦ ϕi−1 )σi−1 (w)δi−1 (v1 )δi−1 (v2 ) · · · δi−1 (v(mi /mi−1 )−1 ) .
Thus, we may define


(mi /mi−1 )−1
δi (v) = (ψϕi−1 (Fi−1 ) ◦ ϕi−1 )σi−1 (w) 
Y
δi−1 (vj ) .
j=1
The map δi is extended K-linearly to all of Hmi −pi .
Hereafter, we suppose that σi and δi have been chosen satisfying the previous theorem. Notice that
the theorem says that ϕσi i behaves as a multiplicative map like the map ϕ in Theorem 6. In point of
fact, we now combine Theorem 12 with Lemma 5 and Theorem 6 to finish our goal (12). In particular,
we are ready to set the mi to be specific integers, and hence by (11) also fix the exponents ei .
Theorem 13. For every i ≥ 1 set mi = 6i−1 . For every i ≥ 1 there exists a triple ri ∈ Tmi with
ϕi (w(ri )) 6= 0.
Proof. We work by induction on i ≥ 1, the case when i = 1 being trivial. So assume that ri ∈ Tmi has
been chosen so that ϕi (w(ri )) 6= 0.
We first recall some information. By Lemma 11, pi = 5i−1 , so the sets Fi of Corollary 9 satisfy
(19)
|Fi | ≤ 3i−1 mi ≤ 18i−1 < (pi + 1)(pi + 2)/2.
We also have w(ri )σi = w(ri ). Thus ϕσi i (w(ri )) 6= 0, and in particular δi (w(ti )) 6= 0 for some ti ∈ Tmi −pi
(by Lemma 3(b)). By Theorem 6, with ϕ = ϕσi i , δ = δi , a = pi and b = mi −pi , we conclude that the set
{ϕi (w(r)) : r ∈ Tmi } spans a K-vector space of dimension at least (pi + 1)(pi + 2)/2. In particular, by
(19) the span cannot belong to ker(ψϕi (Fi ) ). This proves that ψϕi (Fi ) (ϕi (w(ri0 ))) 6= 0 for some ri0 ∈ Tmi .
From the inductive definition of ϕi+1 given by (14) and by Lemma 5, we must have ϕi+1 (w(ri+1 )) 6= 0
for some ri+1 ∈ Tmi+1 .
This finishes our construction.
SIMPLIFYING SMOKTUNOWICZ’S EXTRAORDINARY EXAMPLE
9
6. Summary
Theorem 14. Let K be a countable field, let A = Khx1 , x2 , x3 i, and let A0 ⊂ A be the K-subalgebra
of polynomials with zero constant term. Let {f1 , f2 , . . .} = A0 be an indexing by the positive integers,
with f1 = x1 and deg(fi ) ≤ i/2 for each i ≥ 2. Let I be the ideal of A generated by the homogeneous
i
components of the set {fi3·6 }. The ring R = A0 /I is nil but R[X] is not nil. In fact, the subring of R
generated by yx2 + I and yx3 + I (where y = (1 − x1 + I)−1 ) has the same properties.
We end with three interesting observations.
1. Let I be an ideal in the 2-generated algebra B = Kha, bi and let B 0 be the subalgebra of
polynomials with zero constant term. It is impossible to choose I with B 0 /I nil and yet guarantee
w(n1 , n2 ) ∈
/ I (where w(n1 , n2 ) is as defined in Lemma 1), if I is homogeneous in terms of a and b.
This is because 1 − (a + bX) = (1 − a)(1 − (1 − a)−1 bX) always has a polynomial inverse when a and
(1 − a)−1 b are nilpotent. However, taking a = yx2 + I and b = yx3 + I as above, our construction does
guarantee that for each n ∈ N there is some pair (n1 , n2 ) ∈ N2 with n1 + n2 = n and w(n1 , n2 ) ∈
/ I,
even when I is homogeneous. This is possible since I is not homogeneous in the letters a and b.
2. If we are a little more careful, we can take m1 = 1 and m2 = 2 in the construction above, although
we must then make the rest of the mi increase very quickly. Using the first remark after Theorem 8,
this means we can take x21 ∈ I and still achieve our example.
3. In Theorem 13, if we take mi = 5i−1 then the inequalities work out for i ≥ 12. We can make
the inequalities work when i < 12 if we choose the ordering of A0 = {f1 , f2 , . . .} a little more carefully.
Or even more trivially, we may pad our indexing of the set A0 with extra copies of 0 and make the
growth rate of the index of nilpotence as slow as we like. The essential point to take away is that if
we make only the mild assumptions on the indexing of our polynomials given in (9), all growth rate
considerations are subsumed in Corollary 9 and in the choice of the mi so that (19) holds.
Acknowledgements
I wish to thank Agata Smoktunowicz for encouraging me to make this work public, Victor Camillo
for the countless hours of reading manuscripts and talking about simplifications, and George Bergman
for some extensive and much appreciated suggestions.
References
1. Agata Smoktunowicz, Polynomial rings over nil rings need not be nil, J. Algebra 233 (2000), no. 2, 427–436.
2. Agata Smoktunowicz and E. R. Puczylowski, A polynomial ring that is Jacobson radical and not nil, Israel J. Math.
124 (2001), 317–325.
Department of Mathematics, Brigham Young University, Provo, UT 84602, United States
E-mail address: pace@math.byu.edu