FUNDAMENTALNAYA
I PRIKLADNAYA MATEMATIKA

(FUNDAMENTAL AND APPLIED MATHEMATICS)

2013, VOLUME 18, NUMBER 3, PAGES 69-76

**Symmetric polynomials and nonfinitely generated
$Sym(N)$-invariant ideals**

E. A. da Costa

A. N. Krasilnikov

Abstract

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Let $K$ be
a field and let $N\; =\; \{1,2,$¼}
be the set of all positive integers.
Let $R$_{n} =
K[x_{ij} | 1 £ i
£ n,
j Î N] be the
ring of polynomials in
$x$_{ij}
($1$£ i
£ n,
$j$Î N)
over $K$.
Let $S$_{n} =
Sym({1,2,¼,n})
and $Sym(N)$
be the groups of permutations of the sets
$\{1,2,$¼,n}
and $N$,
respectively.
Then $S$_{n}
and $Sym(N)$ act
on $R$_{n} in
a natural way: $$t(x_{ij}) =
x_{t(i)j} and
$$s(x_{ij}) =
x_{is(j)},
for all $i$Î
{1,2,¼,n}
and $j$Î
N,
$$t
Î
S_{n} and
$$s
Î Sym(N).
Let $\$\backslash bar\; R\_n\$$
be the subalgebra of ($S$_{n}-)symmetric
polynomials in $R$_{n}, i.e.,

$R$_{n} =
{f Î R_{n} |
t(f) = f for each
t Î
S_{n}}.
An ideal $I$
in $\$\backslash bar\; R\_n\$$
is called $Sym(N)$-invariant if
$$s(I) = I for each
$$s
Î Sym(N).
In 1992, the second author proved that if
$char(K)\; =\; 0$
or $char(K)\; =\; p\; >\; n$,
then every $Sym(N)$-invariant ideal
in $\$\backslash bar\; \{R\}\_n\$$
is finitely generated (as such).
In this note, we prove that this is not the case if $char(K)\; =\; p$£ n.
We also survey some results about $Sym(N)$-invariant ideals in
polynomial algebras and some related topics.

Location: http://mech.math.msu.su/~fpm/eng/k13/k133/k13306h.htm

Last modified: March 4, 2014