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Let $k$ be a field and let $A=k[x_1,\dots,x_n]$ be a polynomial algebra over $k$, and let $B\subset A$ be a graded subalgebra that is itself a polynomial ring, i.e. $B=k[f_1,\dots,f_m]$ for some algebraically independent homogeneous polynomials $f_1,\dots,f_m\in A$. Let $g$ be a linear (i.e. preserving the $k$-vector space spanned by the $x_i$'s) automorphism of $A$ that fixes $B$ as a set (and so acts on $B$).

(For example, $B$ might be the invariant ring of a finite linear reflection group acting on $A$, and $g$ might be a linear automorphism of $A$ that normalizes this group. This was my motivating setup.)

An automorphism of a polynomial ring is called "linearizeable" if it is conjugate to a linear automorphism in the automorphism group of the ring. Equivalently, if there exists a set of algebraically independent $k$-algebra generators for the ring such that the $k$-vector space they span is fixed by the automorphism.

My question is, is $g$'s restriction to $B$ linearizeable?

On the one hand, I would be sort of upset if the answer is "no" since $g$ begins life as a completely linear object. On the other hand, because $g$ is degree-preserving with respect to the standard grading on $A$, it actually seems like a strong requirement in general for $g|_B$ to be linearizeable. Suppose for example that all the generators of $B$ have distinct degrees. (Because $B$ is a graded subring, these degrees are invariants of $B$, obtainable from $B$'s Hilbert series, even though the individual generators are not.) Then a degree-preserving automorphism of $B$ can't preserve the linear space spanned by these generators unless it is actually acting diagonally on them. Thus if $g|_B$ is linearizeable, then $B$ has a set of algebra generators $f_1,\dots,f_m$ on which $g$ acts diagonally. Intuitively to me this feels like a stronger requirement than the setup can force by itself. Choose a subalgebra $B$ of $A$ generated in distinct degrees, and then you have the freedom to pick any $g\in GL(n,k) = \operatorname{Aut}_{graded}A$ that fixes $B$ setwise; can you really guarantee from this that $g$ acts diagonally on some set of algebra generators for $B$?

If the general answer is "no" but the answer becomes "yes" under some condition relating to my motivating setup, for example if $A$ is integral over $B$, then I'd love to know that too.

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No. Counterexample: Let $\deg x = \deg y = 1$ and take $$A=\mathbb{F}_2[x,y],\,\,\,B=\mathbb{F}_2[x,y^2],\,\,\,\,\,g(x)=x,\,\,g(y)=x+y$$ As already observed by the OP, since the generators of $B$ have distinct degrees, any linear automorphism of $B$ must act diagonally on the generators. Thus, over $\mathbb{F}_2$, the only linear automorphism of $B$ is the identity. But $g|B$ isn't the identity since $g(y^2)=x^2 + y^2$.

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  • $\begingroup$ Beautiful. In your opinion do you think there is likely to be a characteristic zero counterexample as well? $\endgroup$ – benblumsmith May 2 '16 at 17:50
  • $\begingroup$ Sorry, I missed your comment. But I've seen the new question mathoverflow.net/questions/238103/… and will think about it. $\endgroup$ – Todd Leason May 5 '16 at 20:16

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