8
$\begingroup$

This question follows up a previous one which was answered by Todd Leason. I want to impose two new requirements on the setup.

Let $k$ be a characteristic zero field. Let $A=k[x_1,\dots,x_n]$ be the polynomial algebra with the usual grading. Let $g$ be a graded automorphism of $A$ and let $B$ be a graded subring of $A$ such that:

  • $A$ is integral over $B$.
  • $B$ is fixed setwise by $g$.
  • $B$ is itself a polynomial algebra.

Is $g$'s restriction to $B$ linearizable?

By linearizable I mean that there exists a set of algebra generators $f_1,\dots,f_n$ of $B$ such that the $k$-vector space $V = \langle f_1,\dots,f_n\rangle_k$ is invariant under $g$. (So that $B$ can be seen as the symmetric algebra over $V$ and $g|_B$ the automorphism induced on the symmetric algebra by $g|_V$.)

The two new requirements are integrality and characteristic zero.

Todd Leason's response to the previous question shows that without the characteristic zero assumption the answer is no. I think it's probably no in general, since if $B$ is generated in distinct degrees then $g$ must act on its generators diagonally in order to be linearizable, and that seems a lot to ask. But Todd's example used the characteristic $p$-ness in an essential way, so I remain curious.

$\endgroup$
2
$\begingroup$

Gregor Kemper answered a related question with a technique that can be used to answer this one affirmatively in the case that $g$ has finite order. If $g$ does not have finite order and we drop the integrality assumption, the answer is negative.

If $g$'s restriction to $B$ has finite order, then its action on $B$ is linearizable.

Proof: Note that $B$ is connected (i.e., its degree 0 component is just $k$) since $A$ is. Let $I$ be $B$'s positively-graded ideal. By the graded Nakayama lemma, any set of homogeneous elements of $I$ that generate the $k$-vector space $I/I^2$ will also generate $I$ as an ideal in $B$. By a standard induction argument, any homogeneous ideal generators for $I\triangleleft B$ are actually algebra generators for $B$. (To express an arbitrary homogeneous element $f$ in $B$, which without loss of generality can be taken to have positive degree since $B$ is connected, as a polynomial in these generators, express it first as a linear combination of them with coefficients in $B$. This is possible since $f$ has positive degree, and is therefore in $I$, and $I$ is an ideal. Since $B$ is graded, this linear relation can be taken to be homogeneous just by discarding all terms of degree different than $f$. But then the coefficients are all homogeneous of lower degree, so apply the induction hypothesis.) Since $B$ is presumed to be a polynomial algebra, $\dim_k I/I^2 = \dim_{\text{Krull}}B$. Thus any homogeneous lifts to $I$ of a $k$-basis for $I/I^2$ will automatically be polynomial-algebra generators for $B$.

Since $g$ acts as a graded automorphism on $B$, it preserves $I$. Therefore, it also preserves $I^2$. Since $k$ is of characteristic zero and $g$ has finite order, we have access to Maschke's theorem, which asserts that $I^2$ has a $g$-invariant complement $V$ in $I$. Because the action of $g$ on $B$ respects the grading, and $I$ and therefore $I^2$ are graded ideals, $V$ is even a graded vector space, i.e., it is the direct sum of its intersections with the graded components of $B$. (Indeed, for any $d\in\mathbb{N}$, we can apply Maschke's theorem in $I\cap B_d$ to find a $g$-invariant complement $V_d$ for $I^2\cap B_d$, and then take $V=\bigoplus V_d$.) Therefore, it has a homogeneous basis $\mathcal{B}$. Because $I=I^2\oplus V$, this basis descends to a basis of $I/I^2$. By the work in the previous paragraph, $\mathcal{B}$ is a set of polynomial algebra generators for $B$, so $V$ is the desired subspace.

Remark: This argument actually also works if $k$ has positive characteristic, as long as the order of $g$'s action on $B$ is not divisible by the characteristic. The argument also makes no use of the assumption that $A$ is integral over $B$, although if we know integrality, then we know that $\dim_{\text{Krull}}B = n$, so we know what to expect for the dimension of $I/I^2$.

If $g$'s restriction to $B$ has infinite order, and we drop the hypothesis that $A$ is integral over $B$, then $g$'s action on $B$ may not be linearizable.

Let $A=\mathbb{C}[x,y]$ and let $B=\mathbb{C}[x,xy]$. Let $g$ act on $A$ by $x\mapsto x$, $y\mapsto x+y$. Then $gB\subset B$ since $xy\mapsto x^2+xy\in B$, and $B\subset gB$ since $xy\in \mathbb{C}[x,x^2+xy]=gB$. Thus $B$ is fixed setwise by $G$. But $B$ is algebra-generated in distinct degrees and $g|_B$ does not act diagonally on the generators, so it is not linearizable.

The previous argument fails in this situation because the conclusion of Maschke's theorem fails: while $I^2$ is still a $g$-invariant subspace of $I$, it does not have a $g$-invariant complement.

But note that in this situation, $A$ is not integral over $B$, as $y$ is not integral over $B$.

I do not know if there is an example of infinite order $g$ and $A/B$ integral in which $g|_B$ is not linearizable.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ In your example, $A$ is not integral over $B$. Any hope of getting a counterexample when you throw in this condition? $\endgroup$ – David E Speyer Mar 19 at 20:21
  • $\begingroup$ Yeah, I just noticed that and edited the answer. At this point I'm 50-50 on if it's possible :) $\endgroup$ – benblumsmith Mar 19 at 20:22
  • $\begingroup$ I'm actually kind of gratified that integrality turns out to be relevant at all. When I asked the question, I threw it in because it was relevant to the motivating context ($B$ is an invariant ring for a finite group action on $A$) but not because I had a more-than-vague reason to believe it mattered. $\endgroup$ – benblumsmith Mar 19 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.