# Is the restriction of a graded automorphism linearizable in characteristic zero?

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.

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.

• In your example, $A$ is not integral over $B$. Any hope of getting a counterexample when you throw in this condition? – David E Speyer Mar 19 at 20:21
• Yeah, I just noticed that and edited the answer. At this point I'm 50-50 on if it's possible :) – benblumsmith Mar 19 at 20:22
• 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. – benblumsmith Mar 19 at 20:25