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This question follows up a previous onea 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.

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.

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.

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benblumsmith
benblumsmith
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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.