Does there exist a (strictly) descending chain of translation invariant linear spaces of polynomials in $n$ variables? The answer is "no" if $n=1$.
1
-
$\begingroup$ A linear space of polynomials is translation invariant if and only if it is invariant under partial differentiation. $\endgroup$– user16456Commented Jul 16, 2011 at 6:38
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
No. Suppose that $T$ is a translation-invariant subspace of $k[x_1,\dots,x_n]$. Consider the graded ring $D=k[\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_n}]$ of constant-coefficient differential operators. Then $T$ determines a homogeneous ideal $I(T)\subset D$, by letting the purely degree $j$ part $I_j(T)$ consist of homogeneous operators that annihilate all polynomials of degree at most $j$ in $T$.
An infinite descending chain $T_\alpha$ would give an infinite ascending chain $I(T_\alpha)$, which is impossible because $D$ is noetherian.
answered Jul 16, 2011 at 10:08
-
$\begingroup$ "An infinite descending chain $T_{\alpha}$ would give an infinite ascending chain $I(T_{\alpha})$..." This is true only if $T_1\ne T_2$ implies that the closure of $T_1$ is different from the closure of $T_2$.If $T_1$ and $T_2$ have the same closure, then $I(T_1)$ and $I(T_2)$ are the same. $\endgroup$ Commented Jul 16, 2011 at 11:31
-
1$\begingroup$ If $T_1$ properly contains $T_2$ then for some $D$ there is a degree $d$ polynomial in $T_1$ but not $T_2$. Then there is an element of degree $d$ in $D$ that annihilates one but not the other. $\endgroup$ Commented Jul 16, 2011 at 13:04
-
-
$\begingroup$ I meant "homogeneous ideal". I have edited accordingly. $\endgroup$ Commented Jul 26, 2011 at 10:06
-
$\begingroup$ A more highbrow explanation: $k[x_1,\ldots,x_n]$ is, as module over the constant-coefficient DO's, the injective hull of $k$ and as such is Artinian because the ring is Noetherian as is $k$ as module (the proof of this general fact is essentially by the argument Tom gives). $\endgroup$ Commented Jul 26, 2011 at 10:12