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Results 1 - 2 of 2 |
1. CJM 2007 (vol 59 pp. 186)
| Endomorphism Algebras of Kronecker Modules Regulated by Quadratic Function Fields Purely simple Kronecker modules ${\mathcal M}$, built from an algebraically closed field $K$,
arise from a triplet $(m,h,\alpha)$ where $m$ is a positive integer,
$h\colon\ktil\ar \{\infty,0,1,2,3,\dots\}$ is a height function, and
$\alpha$ is a $K$-linear functional on the space $\krx$ of rational
functions in one variable $X$. Every pair $(h,\alpha)$ comes with a
polynomial $f$ in $K(X)[Y]$ called the regulator. When the module
${\mathcal M}$ admits non-trivial endomorphisms, $f$ must be linear or
quadratic in $Y$. In that case ${\mathcal M}$ is purely simple if and
only if $f$ is an irreducible quadratic. Then the $K$-algebra
$\edm\cm$ embeds in the quadratic function field $\krx[Y]/(f)$. For
some height functions $h$ of infinite support $I$, the search for a
functional $\alpha$ for which $(h,\alpha)$ has regulator $0$ comes
down to having functions $\eta\colon I\ar K$ such that no planar curve
intersects the graph of $\eta$ on a cofinite subset. If $K$ has
characterictic not $2$, and the triplet $(m,h,\alpha)$ gives a
purely-simple Kronecker module ${\mathcal M}$ having non-trivial
endomorphisms, then $h$ attains the value $\infty$ at least once on
$\ktil$ and $h$ is finite-valued at least twice on
$\ktil$. Conversely all these $h$ form part of such triplets. The
proof of this result hinges on the fact that a rational function $r$
is a perfect square in $\krx$ if and only if $r$ is a perfect square
in the completions of $\krx$ with respect to all of its valuations.
Keywords:Purely simple Kronecker module, regulating polynomial, Laurent expansions, endomorphism algebra Categories:16S50, 15A27 |
2. CJM 1998 (vol 50 pp. 719)
| Indecomposable almost free modules---the local case Let $R$ be a countable, principal ideal domain which is not a field and
$A$ be a countable $R$-algebra which is free as an $R$-module. Then we
will construct an $\aleph_1$-free $R$-module $G$ of rank $\aleph_1$
with endomorphism algebra End$_RG = A$. Clearly the result does not
hold for fields. Recall that an $R$-module is $\aleph_1$-free if all
its countable submodules are free, a condition closely related to
Pontryagin's theorem. This result has many consequences, depending on
the algebra $A$ in use. For instance, if we choose $A = R$, then
clearly $G$ is an indecomposable `almost free' module. The existence of
such modules was unknown for rings with only finitely many primes like
$R = \hbox{\Bbbvii Z}_{(p)}$, the integers localized at some prime $p$. The result
complements a classical realization theorem of Corner's showing that
any such algebra is an endomorphism algebra of some torsion-free,
reduced $R$-module $G$ of countable rank. Its proof is based on new
combinatorial-algebraic techniques related with what we call {\it rigid
tree-elements\/} coming from a module generated over a forest of trees.
Keywords:indecomposable modules of local rings, $\aleph_1$-free modules of rank $\aleph_1$, realizing rings as endomorphism rings Categories:20K20, 20K26, 20K30, 13C10 |

