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 $\dim\,V^i=e*2$ there exist a commutative
semigroup $S$ with zero, and a subset $X\subseteq S,$ such
that $\card(X-\{0\})=\card(X^p-\{0\}),$ but such that for
all $i$ with $1 1,$ the set $X^i-\{0\}$ has
$i+1$ elements, $x^i,\,x^{i-1}y,\dots,y^i.$
Hence $\card(X^p-\{0\})=p+1=\card(X-\{0\});$ but for $12$ in place of $p,$ including powers of $2,$ as long as
one adds to the statement corresponding to Example~\ref{E.X^(q)} the
condition that $i$ be relatively prime to $p.$\vspace{.5em}
From the construction of Example~\ref{E.X^(q)}, we can also obtain a
counterexample to a statement which, if it were true, would,
with the help of Lemma~\ref{L.dimV^i},
lead to an easy affirmative answer to Question~\ref{Q.X^i}:
\begin{example}\label{E.F0X^n}
There exists a commutative semigroup $S$ with zero, a finite
subset $X\subseteq S,$ and an integer $n>0,$
such that the $\!n\!$-th power map is one-to-one on $X$
and does not take any nonzero element of $X$ to $0,$
but such that for some field $F,$ the $\!n\!$-th power map
on the span $FX$ of $X$ in $F_0 S$
does take some nonzero element to~$0.$
\end{example}
\begin{proof}[Construction]
Let us first note that though we assumed in
Example~\ref{E.X^(q)} that $p$ was a prime to emphasize
the relationship with Corollary~\ref{C.X^i}, all we needed
was that $p$ and $i$ be relatively prime.
For the present example, let us repeat that construction
with any integer $n>2$ (possibly, but not necessarily, prime)
in place of the $p$ of that construction,
while using a prime $p*

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, {\em Sobre las unidades de \'{a}lgebras finitas,} ({\em Units of finite algebras}), Anales del Instituto de Matem\'{a}ticas, Universidad Nacional Aut\'{o}noma de M\'{e}xico, {\bf 16} (1976) 1--78. \url{http://texedores.matem.unam.mx/publicaciones/index.php?option=com_remository&Itemid=57&func=fileinfo&id=227}\,. MR~{\bf 58}\#11011. \bibitem{GMB_HL} George M. Bergman, {\em Conjugates and nth roots in Hahn-Laurent group rings,} Bull. Malaysian Math. Soc. (2) {\bf 1} (1978) 29-41; historical addendum at (2) {\bf 2} (1979) 41-42. MR~{\bf 80a}:16003, {\bf 81d}:16016. (For reasons noted in the footnote to the title, I used ``Hahn-Laurent group ring'' for what is standardly called a ``Mal'cev-Neumann group ring''. But I have subsequently followed standard usage.) \bibitem{C+P} A. H. Clifford and G. B. 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