\documentclass{conm-p-l} \usepackage{url,centernot} \setlength{\mathsurround}{.167em} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{question}[theorem]{Question} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{convention}[theorem]{Conventions} \newcommand{\Q}{\mathbb Q} \newcommand{\Z}{\mathbb Z} \renewcommand{\r}{\mathrm} \newcommand{\p}{^{(p)}} \newcommand{\card}{\mathrm{card}} \begin{document} \vspace{-20em} \begin{center} \texttt{ This is the final preprint version of a paper which appeared in \\[.2em] Contemporary Mathematics, {\bf 609} (2014)\\[.2em] Ring Theory and Its Applications\\[.2em] The published version is readable at\\[.2em] \url{http://dx.doi.org/10.1090/conm/609/12101} .} \end{center} \vspace{4em} \title{Thoughts on Eggert's Conjecture} \author{George M. Bergman} \address{University of California\\Berkeley, CA 94720-3840, USA} \email{gbergman@math.berkeley.edu} \urladdr{math.berkeley.edu/~gbergman} \dedicatory{To T.\,Y.\,Lam, on his 70th birthday} \thanks{Archived at \url{http://arxiv.org/abs/1206.0326}\,. After publication of this note, updates, errata, related references etc., if found, will be recorded at \url{/~gbergman/papers/} } \keywords{Eggert's Conjecture, Frobenius map on a finite-dimensional nilpotent commutative algebra, finite abelian semigroup} \subjclass[2010]{Primary: 13A35, 13E10, 16N40, % p,Frob fdim nil+np Secondary: 13H99, 16P10, 16S36, 20M14, 20M25.} % local_rgs fd' &smgp_rgs cm'smgps sgp_rgs \begin{abstract} Eggert's Conjecture says that if $R$ is a finite-dimensional \mbox{nilpotent} commutative algebra over a perfect field $F$ of characteristic $p,$ and $R\p$ is the image of the $\!p\!$-th power map on $R,$ then $\dim_F R\geq p\,\dim_F R\p.$ Whether this very elementary statement is true is not known. We examine heuristic evidence for this conjecture, versions of the \mbox{conjecture} that are not limited to positive characteristic and/or to commutative $R,$ \mbox{consequences} the conjecture would have for semigroups, and examples that give equality in the conjectured inequality. We pose several related questions, and briefly survey the literature on the subject. \end{abstract} \maketitle % - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - \section{Introduction}\label{S.intro} If $F$ is a field of characteristic $p,$ and $R$ is a commutative $\!F\!$-algebra, then the set $R\p$ of $\!p\!$-th powers of elements of $R$ is not only closed under multiplication, but also under addition, by the well-known identity % \begin{equation}\label{d.+^p} (x+y)^p\ =\ x^p+y^p\quad (x,y\in R). \end{equation} % Hence $R\p$ is a subring of $R.$ If, moreover, $F$ is a \emph{perfect} field (meaning that every element of $F$ is a $\!p\!$-th power -- as is true, in particular, if $F$ is finite, or, at the other extreme, algebraically closed), then the subring $R\p$ is also closed under multiplication by elements of $F:$ % \begin{equation}\label{d.ax^p} a\,x^p\ =\ (a^{1/p}x)^p\in R\p\quad (a\in F,\,x\in R). \end{equation} % In this situation we can ask ``how big'' the subalgebra $R\p$ is compared with the algebra $R,$ say in terms of dimension over $F.$ If we take for $R$ a polynomial algebra $F[x]$ over a perfect field $F,$ we see that $R\p=F[x^p],$ so intuitively, $R\p$ has a basis consisting of one out of every $p$ of the basis elements of $R.$ Of course, these bases are infinite, so we can't divide the cardinality of one by that of the other. But if we form finite-dimensional truncations of this algebra, letting $R=F[x]/(x^{N+1})$ for large integers $N,$ then we see that the dimension of $R\p$ is indeed about $1/p$ times the dimension of $R.$ If we do similar constructions starting with polynomials in $d$ variables, we get $R\p$ having dimension about $1/p^d$ times that of $R.$ Is the ratio $\dim R\p/\dim R$ always small? No; a trivial counterexample is $R=F;$ a wider class of examples is given by the group algebras $R=F\,G$ of finite abelian groups $G$ of orders relatively prime to~$p.$ In $G,$ every element is a $\!p\!$-th power, hence $R\p$ contains all elements of $G,$ hence, being closed under addition and under multiplication by members of $F,$ it is all of $R;$ so again $\dim\,R\p/\dim\,R=1.$ In the above examples, the $\!p\!$-th power map eventually ``carried things back to themselves''. A way to keep this from happening is to assume the algebra $R$ is \emph{nilpotent}, i.e., that for some positive integer $n,$ $R^n=0,$ where $R^n$ denotes the space of all sums of $\!n\!$-fold products of members of $R.$ This leads us to \begin{conjecture}[Eggert's Conjecture \cite{E}]\label{Cj.E} If $R$ is a finite-dimensional nilpotent commutative algebra over a perfect field $F$ of characteristic $p>0,$ then % \begin{equation}\label{d.E} \dim_F\,R\ \geq\ p\,\dim_F\,R\p. \end{equation} % \end{conjecture} Of course, a nonzero nilpotent algebra does not have a unit. Readers who like their algebras unital may think of the $R$ occurring above and throughout this note as the maximal ideal of a finite-dimensional local unital $\!F\!$-algebra. Let us set down some conventions. \begin{convention}\label{Cv.cm_ass} Throughout this note, $F$ will be a field. The symbol ``$\dim\!$'' will always stand for ``$\dim_F\!$'', i.e., dimension as an $\!F\!$-vector-space. Except where the contrary is stated \textup{(}in a few brief remarks and two examples\textup{)}, $\!F\!$-algebras will be assumed associative, but not, in general, unital. \textup{(}Most of the time, we will be considering commutative algebras, but we will make commutativity explicit. When we simply write ``associative algebra'', this will signal ``not necessarily commutative''.\textup{)} An {\em ideal} of an $\!F\!$-algebra will mean a ring-theoretic ideal which is also an $\!F\!$-subspace. If $R$ is an $\!F\!$-algebra, $V$ an $\!F\!$-subspace of $R,$ and $n$ a positive integer, then $V^n$ will denote the \emph{\mbox{$\!F\!$-subspace}} of $R$ spanned by all $\!n\!$-fold products of elements of $V,$ while $V^{(n)}$ will denote the \emph{set} of $\!n\!$-th powers of elements of $V.$ \end{convention} Thus, if $V$ is a subspace of a commutative algebra $R$ over a perfect field $F$ of characteristic $p,$ then $V\p$ will also be a subspace of $R,$ but for a general base-field $F,$ or for noncommutative $R,$ this is not so. The map $x\mapsto x^p$ on a commutative algebra $R$ over a field of characteristic $p$ is called the \emph{Frobenius map}. We remark that the unital rings $R=F[x]/(x^{N+1})$ that we discussed before we introduced the nilpotence condition generally fail to satisfy~(\ref{d.E}). Most obvious is the case $N=0,$ where $R=F.$ More generally, writing $N=pk+r$ $(0\leq r< p),$ so that the lowest and highest powers of $x$ in the natural basis of $R$ are $x^0$ and $x^{pk+r},$ we find that $\dim\,R\p/\dim\,R=(k+1)/(pk+r+1),$ which is $>1/p$ unless $r=p-1.$ The corresponding nilpotent algebras are constructed from the ``nonunital polynomial algebra'', i.e., the algebra of polynomials with zero constant term, which we shall write % \begin{equation}\label{d.[F][x]} [F][x]\ =\ \{\,\sum_{i>0} a_i x^i\,\}\ \subseteq\ F[x]. \end{equation} % When we divide this by the ideal generated by $x^{N+1},$ again with $N=pk+r$ $(0\leq r
0,$
the $\!p\!$-th power map on
$R$ over $F$ is ``almost'' linear.
In particular, its image is a vector subspace (in fact, a subalgebra).
Pleasantly, we can even find a vector subspace $V\subseteq R$
which that map sends bijectively to $R\p.$
Namely, take any $\!F\!$-basis $B$ for $R\p,$ let $B'$
be a set consisting of exactly one $\!p\!$-th root of
each element of $B,$ and let $V$ be the $\!F\!$-subspace
of $R$ spanned by $B'.$
Since the $\!p\!$-th power map sends $\sum_{x\in B'} a_x x$
to $\sum_{x\in B'} a_x^p x^p,$ it hits each
element of $R\p$ exactly once.
This suggests the following approach to Eggert's Conjecture.
Suppose we take such a subspace $V,$ and look at the subspaces
$V,\ V^2,\ \dots,\ V^p$
(defined as in the last paragraph of Convention~\ref{Cv.cm_ass}).
Can we deduce that each of them has dimension at least
that of $V\p=R\p$ (as the first and last certainly do),
and conclude that their sum within $R$ has dimension at least $p$
times that of $R\p$?
The answer is that yes, we can show that each has
dimension at least that of $R\p,$ but no, except under
special additional hypotheses, we cannot say that
the dimension of their sum is the sum of their dimensions.
The first of these claims can be proved in a context that does
not require positive characteristic, or commutativity, or nilpotence.
We \emph{will} have to assume $F$ algebraically closed; but we will
subsequently see that for commutative algebras over a perfect field
of positive characteristic, the general case reduces to that case.
\begin{lemma}\label{L.dimV^i}
Let $F$ be an algebraically closed field, $R$ an associative
$\!F\!$-algebra, $V$ a finite-dimensional subspace of $R,$ and $n$
a positive integer such that every nonzero element of $V$ has
nonzero $\!n\!$-th power.
Then for all positive integers $i\leq n$ we have
%
\begin{equation}\label{d.dimV^i}
\dim\,V^i\ \geq\ \dim\,V.
\end{equation}
%
\end{lemma}
\begin{proof}
Let $d=\dim\,V,$ and let $x_1,\dots,x_d$ be a basis for $V$ over $F.$
Suppose, by way of contradiction, that for some $i\leq n$
we had $\dim\,V^i=e 1/(p+1).$
\end{example}
\begin{proof}[Sketch of construction]
Given $p,$ let $F$ be any field of characteristic $p$
having a proper purely inseparable extension $F'=F(u^{1/p}),$
such that every element of $F'$ has a $\!p\,{+}1\!$-st root in $F'.$
(We can get such $F$ and $F'$ starting with any
algebraically closed field $k$ of characteristic $p,$ and
any subgroup $G$ of the additive group $\Q$ of rational numbers
which is $\!p\,{+}1\!$-divisible but not $\!p\!$-divisible.
Note that $p^{-1}G\subseteq\Q$ will have the form
$G+p^{-1}h\,\Z$ for any $h\in G - p\,G.$
Take a group isomorphic to $G$ but written multiplicatively, $t^G,$ and
its overgroup $t^{p^{-1}G},$ and
let $F$ and $F'$ be the Mal'cev-Neumann power series
fields $k((t^G))$ and $k((t^{p^{-1}G}))$
\cite[\S2.4]{PMC_SF}, \cite{GMB_HL};
and let $u\in F$ be the element $t^h.$
The asserted properties are easily verified.)
Let us now form the (commutative, associative)
truncated polynomial algebra $[F'][x]/(x^{p+2}),$
graded by degree in $x,$
and let $R$ be the $\!F\!$-subspace of this algebra consisting of
those elements for which the coefficient of $x^p$ lies
in the subfield $F$ of $F'$ (all other coefficients
being unrestricted).
We make $R$ a graded nonassociative $\!F\!$-algebra by using the
multiplication of $[F'][x]/(x^{p+2})$ on all pairs of homogeneous
components {\em except} those having degrees summing to $p,$ while
defining the multiplication when the degrees sum to $p$ by fixing
an $\!F\!$-linear retraction $\psi: F'\to F,$ and taking the product of
$a\,x^i$ and $b\,x^{p-i}$ $(0N,$ which, in~(\ref{d.12...}), equals $\infty.$
So $h$ and all integers $\geq h$ fall together with $\infty;$
and if we follow up the consequences, we eventually find that every
integer $\geq i$ is identified with $\infty.$
Thus, we get a semigroup just like~(\ref{d.12...}), but with
$i-1$ rather than $N$ as the last finite value.
So let us instead pass to a subsemigroup of~(\ref{d.12...}).
The smallest change we can make is to drop $1,$ getting the
subsemigroup generated by $2$ and $3,$ which we shall now denote $S.$
Then $\card(S-\{\infty\})$ has gone down by $1,$ pushing the
value of~(\ref{d.nSn-S}) up by $1;$ but the integer $n$
has ceased to belong to $S^{(n)},$ decreasing~(\ref{d.nSn-S}) by $n.$
So in our attempt to find a counterexample, we have ``lost ground'',
decreasing~(\ref{d.nSn-S}) from $0$ to $-n+1.$
However, now that $1\notin S,$
we can regain some ground by imposing relations.
Suppose we impose the relation that identifies $N-1$
either (a)~with $N$ or (b)~with $\infty.$
If we add any member of $S$ (loosely speaking,
any integer $\geq 2)$ to both sides of either relation,
we get $\infty=\infty,$ so no additional identifications are implied.
Since we are assuming $N$ is divisible by $n,$ the integer $N-1$ is
not; so we have again decreased the right-hand term of~(\ref{d.nSn-S}),
this time without decreasing the left-hand term; and thus brought
the total value to $-n+2.$
In particular, if $n=2,$ we have returned to the value $0;$
but not improved on it.
I have experimented with more complicated examples of
the same sort, and gotten very similar results:
I have not found one that made the value of~(\ref{d.nSn-S}) positive;
but surprisingly often, it was possible to arrange things so that
for $n=2,$ that value was $0.$
Let me show a ``typical'' example.
We start with the additive subsemigroup of the natural
numbers generated by $4$ and $5.$
I will show it by listing an initial string of the
positive integers, with the members of our subsemigroup underlined:
%
\begin{equation}\label{d.bold}
1\ 2\ 3\ \underline{4\ 5}\ 6\ 7\ \underline{8\ 9\ 10}\ 11
\ \underline{12\ 13\ 14\ 15\ 16\ 17\ \dots\ .}
\end{equation}
Assume this to be truncated at some large integer $N$
which is a multiple of $n,$ all larger integers being
collapsed into $\infty.$
If we combine the effects on the two terms of~(\ref{d.nSn-S})
of having dropped the six integers $1,\,2,\,3,\,6,\,7,\,11$
from~(\ref{d.12...}),
we find that, assuming $N\geq 11n,$~(\ref{d.nSn-S}) is now $6(-n+1).$
Now suppose we impose the relation $i=i+1$ for some $i$ such
that $i$ and $i+1$ both lie in~(\ref{d.bold}).
Adding $4$ and $5$ to both sides of this equation, we
get $i+4=i+5=i+6;$ adding $4$ and $5$ again we get
$i+8=i+9=i+10=i+11.$
At the next two rounds, we get strings of equalities that overlap
one another; and all subsequent strings likewise overlap.
So everything from $i+12$ on falls together with $N+1$ and
hence with $\infty;$ so we may as well assume
%
\begin{equation}\label{d.n+1=i+12}
N+1\ =\ i+12.
\end{equation}
What effect has imposing the relation $i=i+1$ had on~(\ref{d.nSn-S})?
The amalgamations of the three strings of integers described
decrease $\card(S{-}\{\infty\})$ by $1,$ $2$ and $3$ respectively,
so in that way, we have gained ground, bringing~(\ref{d.nSn-S})
up from $6(-n+1)$ to possibly $6(-n+2).$
But have we decreased $\card(S^{(n)}{-}\{\infty\}),$ and so
lost ground, in the process?
If $n>2,$ then even if there has been no such
loss, the value $6(-n+2)$ is negative; so let us assume $n=2.$
If we are to avoid bringing~(\ref{d.nSn-S}) below $0,$
we must make sure that none of the sets that were fused into
single elements,
%
\begin{equation}\label{d.i+}
\{i,\,i+1\},\quad\{i+4,\,i+5,\,i+6\},\quad\{i+8,\,i+9,\,i+10,\,i+11\},
\end{equation}
%
contained more than one member of $S^{(2)}.$
For the first of these sets, that is no problem; and for
the second, the desired conclusion can be achieved by taking
$i$ odd, so that of the three elements of that set, only $i+5$ is even.
For the last it is more difficult -- the set will contain two even
values, and if $i$ is large, these will both belong to $S^{(2)}.$
However, suppose we take $i$ not so large; say we choose it
so that the smaller of the two even values in that set is
the largest even integer that does \emph{not} belong to $S^{(2)}.$
That is $22,$ since $11$ is the largest integer not in~(\ref{d.bold}).
Then the above considerations show
that we do get a semigroup for which~(\ref{d.nSn-S}) is zero.
The above choice of $i$ makes $i+9$ (the smallest even value in
the last subset in~(\ref{d.i+})) equal to $22$ (the largest
even integer not in $S^{(2)}),$ so $i=13,$ so
by~(\ref{d.n+1=i+12}), $N+1=25.$
Let us write down formally the contracted semigroup algebras of
the two easier examples described earlier, and of the above example.
%
\begin{example}\label{E.4,5}
Let $F$ be a perfect field of characteristic~$2.$
Then the following nilpotent algebras have
equality in the inequality of Eggert's Conjecture.
%
\begin{equation}\label{d.2,3}
R\ =\ [F][x^2,\,x^3]\,/\,(x^{N-1}-x^{N},\,x^{N+1},\,x^{N+2})
\qquad \mbox{for every even $N>2,$}
\end{equation}
%
\begin{equation}\label{d.2,3'}
R\ =\ [F][x^2,\,x^3]\,/\,(x^{N-1},\,x^{N+1},\,x^{N+2})
\qquad \mbox{for every even $N>2,$}
\end{equation}
%
\begin{equation}\label{d.4,5}
R\ =\ [F][x^4,\,x^5]\,/\,(x^{13}-x^{14},\,x^{25},\dots,x^{28}).
\end{equation}
More precisely, in both~\textup{(\ref{d.2,3})}
and~\textup{(\ref{d.2,3'})}
$\dim\,R=N-2,$ and $\dim\,R^{(2)}=(N-2)/2,$
while in~\textup{(\ref{d.4,5})},
$\dim\,R=18,$ and $\dim\,R^{(2)}=9.$
\qed
\end{example}
Many examples behave like these.
A couple more are
%
\begin{equation}\label{d.more}
[F][x^2,\,x^5]/(x^{11}-x^{12},\,x^{\geq 15}),\qquad
[F][x^3,\,x^7]/(x^{13}-x^{14},\,x^{\geq 25})
\end{equation}
%
(where ``$x^{\geq n}$'' means ``$x^n$ and all higher powers''; though
in each case, only finitely many are needed).
Perhaps Eggert's Conjecture is true, and these examples
``run up against the wall'' that it asserts.
Or --~who knows -- perhaps if one pushed this sort of
exploration further, to homomorphic images of
semigroups generated by families of three or more
integers, and starting farther from $0,$
one would get counterexamples.
For values of $n$ greater than $2,$ I don't know any examples of
this flavor that even bring~(\ref{d.nSn-S}) as high as zero.
(But a class of examples of a different sort, which does, was noted in
the last paragraph of \S\ref{S.intro}.)
Incidentally, observe that in the semigroup-theoretic context that
led to~(\ref{d.2,3}) and~(\ref{d.2,3'}), we had the choice of
imposing either the relation $N-1=N$ or the relation $N-1=\infty.$
However, in the development that
gave~(\ref{d.4,5}), setting a semigroup element equal to $\infty$
would not have done the same job as setting two such elements equal.
If we set $i=\infty,$ then, for example, $i+4$ and $i+5$ would each
become $\infty,$ so looking at the latter two elements,
we would lose one from $S^{(2)}$ as well as one not in $S^{(2)}.$
Above, we instead set $i=i+1,$ and the resulting
pair of equalities $i+4=i+5=i+6$ turned a family consisting of
two elements not in $S^{(2)}$ and one in $S^{(2)}$
into a single element of~$S^{(2)}.$
Turning back to Eggert's ring-theoretic conjecture,
it might be worthwhile to experiment with
imposing on subalgebras of $[F][x]$ relations ``close to'' those
of the sort used above, but not expressible in purely
semigroup-theoretic terms; for instance, $x^i+x^{i+1}+x^{i+2}=0,$ or
$x^i-2x^{i+1}+x^{i+2}=0.$
\section{Sketch of the literature}\label{S.literature}
The main positive results in the
literature on Eggert's Conjecture concern two kinds of
cases: where $\dim(R\p)$ (or some related
invariant) is quite small, and where $R$ is graded.
N.\,H.\,Eggert \cite{E}, after
making the conjecture, in connection with
the study of groups that can appear as the group of units of a finite
unital ring $A$ (the nonunital
ring $R$ to which the conjecture would be applied being
the Jacobson radical of $A),$ proved it for $\dim(R\p)\leq 2.$
That result was extended to $\dim(R\p)\leq 3$ by R.\,Bautista \cite{RB},
both results were re-proved more simply by
C.\,Stack \cite{CS2}, \cite{CS3}, and most recently pushed up to
$\dim(R\p)\leq 4$ by B.\,Amberg and L.\,Kazarin \cite{A+K2}.
Amberg and Kazarin also prove in \cite{A+K1} some similar
results over an arbitrary field,
in the spirit of our Questions~\ref{Q.V->} and~\ref{Q.->W},
and they show in \cite{A+K4} that, at least when the values
$\dim(R^i/R^{i+1})$ are small, these give a nonincreasing function
of $i.$
In \cite{A+K4} they give an extensive survey of results on this
subject and related group-theoretic questions.
K.\,R.\,McLean \cite{McL1}, \cite{McL2} has obtained strong positive
results in the case where $R$ is graded and generated by its
homogeneous component of degree~$1.$
In particular, in \cite{McL1} he proves Eggert's Conjecture in
that case if $(R_3)\p=0$ (recall that in Corollary~\ref{C.graded}
we could not get beyond the case $(R_2)\p=0),$
or if $R\p$ is generated by two elements.
Moreover, without either assumption (but still assuming
$R$ graded and generated in degree~$1),$ he proves
that $\dim\,R\p/\dim\,R\leq 1/(p\,{-}1).$
His technique involves taking a subspace $V\subseteq R_1$ as at the
start of \S\ref{S.V^n} above, and constructing recursively
a family of direct-sum decompositions of $V,$ each new summand arising
as a vector-space complement of the kernel of multiplication by an
element obtained using the previous steps of the construction.
He also shows in \cite{McL1} that Eggert's Conjecture
holds for the radicals of group algebras of finite
abelian groups over perfect fields $F$ of nonzero characteristic.
S.\,Kim and J.\,Park \cite{K+P} prove Eggert's Conjecture
when $R$ is a commutative nilpotent \emph{monomial algebra},
i.e., an algebra with a presentation in which all relators
are monomials in the given generators.
M.\,Korbel\'{a}\v{r} \cite{MK} has recently shown that Eggert's
Conjecture holds whenever $R\p$ can be generated as
an $\!F\!$-algebra by two elements.
(So a counterexample in the spirit of the
preceding section would require at least $3$ generators.)
\cite{MK} ends with a generalization of Eggert's conjecture,
which is equivalent to the case of Question~\ref{Q.V->} above
in which $F$ is a field of positive characteristic $p$
and $n=p,$ but $F$ is not assumed perfect.
In \cite{LH}, a full proof of Eggert's
Conjecture was claimed, but the argument was flawed.
(The claim in the erratum to that paper, that the proof is at
least valid for the graded case, is also incorrect.)
There is considerable variation in notation and language
in these papers.
E.g., what I have written $R\p$ is denoted $R^{(1)}$ in
Amberg and Kazarin's papers, $R\p$ in Stack's and Korbel\'{a}\v{r}'s,
and $R[p]$ in McLean's (modulo differences in the letter
used for the algebra $R).$
McLean, nonstandardly, takes the statement that $R$ is graded to
include the condition that it is generated by its
degree~$\!1\!$ component.
Though I do not discuss this above, I have,
also examined the behavior of the
sequence of dimensions of quotients $R^i/R^{i+1}$ for a
commutative algebra $R.$
Most of my results seem to be subsumed by those of Amberg and Kazarin,
but I will record here a question which that line of thought suggested,
which seems of independent interest for its simplicity.
Given two subspaces $V$ and $W$ of a commutative algebra, let
$\r{Ann}_V\,W$ denote the subspace $\{x\in V\mid xW=\{0\}\}\subseteq V.$
\begin{question}\label{Q.AnnV^n}
If $R$ is a commutative algebra over a
field $F,$ $V$ a finite-dimensional
subspace of $R,$ and $n$ a positive integer, must
%
\begin{equation}\label{d.AnnV^n}
\dim(V/\r{Ann}_V\,V^n)\ \leq\ \dim\,V^n\ \mbox{?}
\end{equation}
%
\end{question}
I believe I have proved~(\ref{d.AnnV^n}) for $\dim\,V^n\leq 4.$
The arguments become more intricate with each succeeding value
$1,$~$2,$~$3,$~$4.$
I am indebted to Cora Stack for bringing Eggert's Conjecture
to my attention and providing a packet of relevant literature,
to Martin Olsson for pointing me to the result in \cite{Mumford}
used in the proof of Lemma~\ref{L.dimV^i}, and to the referee for
making me justify an assertion that was not as straightforward as I
had thought.
\begin{thebibliography}{00}
\bibitem{A+K1} Bernhard Amberg and Lev Kazarin,
{\em On the dimension of a nilpotent algebra,}
Math. Notes {\bf 70} (2001) 439--446.
MR~{\bf 2002m}:13006.
\bibitem{A+K2} Bernhard Amberg and Lev Kazarin,
{\em Commutative nilpotent $p$-algebras with small dimension,}
pp.\,1--19 in {\em Topics in infinite groups,}
Quad. Mat., 8, Dept. Math., Seconda Univ. Napoli, Caserta,
2001.
MR~{\bf 2003k}:13002.
\bibitem{A+K4} Bernhard Amberg and Lev Kazarin,
{\em On the powers of a commutative nilpotent algebra,} pp.1--12 in
{\em Advances in Algebra,} World Sci. Publ., River Edge, NJ, 2003.
MR~{\bf 2005f}:16031.
\bibitem{A+K5} Bernhard Amberg and Lev Kazarin,
{\em Nilpotent $p$-algebras and factorized $p$-groups,}
pp.\,130--147 in {\em Groups St. Andrews 2005,} Vol. 1,
London Math. Soc. Lecture Note Ser., 339.
MR~{\bf 2008f}:16045.
% http://books.google.com/books?hl=en&lr=&id=EUkaQBFMOtAC
\bibitem{RB} R.\,Bautista,
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(For reasons noted in the footnote to the title, I
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\bibitem{E} N.\,H.\,Eggert,
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\bibitem{Mumford} David Mumford, % QA3.L35 v.1358
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\end{thebibliography}
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