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Let $f(t)\in\mathbb{C}[t]$, and let $I_f$ be the ideal in $\mathbb{C}[t]$ generated by $f(t)$.

The ideal $I_f$ has a natural $\mathbb{C}$-algebra structure. My question is the following:

For which polynomials $f\in \mathbb{C}[t]$ is the group of ($\mathbb{C}$-algebra) automorphisms of $I_f$ trivial?

Some examples:

$1)$ If $f(t)=t$, there are (of course) infinitely many automorphims, defined by $t\mapsto \alpha t$, for any $\alpha\in \mathbb{C}[t]-\{0\}$. This case is very simple to understand, since the ideal generated by $t$ is a free $\mathbb{C}$-algebra (without unit) and any homomorphism can be defined just by defining the image of $t$.

$2)$ If $f(t)=t(t-1)$ we were able to find only one automorphism, given by $f\mapsto f$, and $tf\mapsto f-tf$. It is an automorphims of order $2$.

One can observe that it is enough to define the above in $f$ and $tf$, since these two elements generate $I_f$, as a $\mathbb{C}$-algebra (although the algebra generated by these two polynomials are not free, so, one needs to be careful in defining homomorphisms)

$3)$ As a last example, we were not able to find any non-trivial automorphism of the ideal generated by $f(t)=t(t-1)(t-2)$.

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    $\begingroup$ 0. An "algebra" has an unit in my book, but whatever... 1. Are you sure that $f$ and $tf$ generate $I_f$ ? Isn't $\{t^i f \mid 0\leq i<deg(f)\}$ a minimal generating system (consider$f$-adic expansions)? Or was this just for that particular example? 2. It seems to me that $Aut(I_f)$ embeds into $Aut(\mathbb{C}[t])$ via $\alpha \mapsto (t \mapsto \alpha(f)^{-1}\alpha(tf))$. This could be useful. $\endgroup$
    – Johannes Hahn
    Commented Jan 29, 2015 at 16:45
  • $\begingroup$ 0. For us an algebra does not need to be unitary. 1. It is just for that particular example. In the general case, they are generated by $\{t^if | 0\leq i< \deg(f)\}$, as you mentioned. 2. You mean $\alpha^{-1}(f)$? I did not understand. $\endgroup$
    – Thiago
    Commented Jan 29, 2015 at 16:56
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    $\begingroup$ No, I really mean $\alpha(f)^{-1} = \frac{1}{\alpha(f)}$. $\endgroup$
    – Johannes Hahn
    Commented Jan 29, 2015 at 17:08
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    $\begingroup$ In case 3), I believe that $t\mapsto 2-t$ induces an automorphism of $I_f$. $\endgroup$
    – ACL
    Commented Jan 29, 2015 at 21:17

1 Answer 1

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We may assume that $f$ is not a unit. Let $A_f=I_f+\mathbb{C}\cdot 1$, the free unital algebra on $I_f$; this embeds naturally in $\mathbb{C}[t]$. Furthermore, $A_f$ and $\mathbb{C}[t]$ have the same fraction field, and $\mathbb{C}[t]$ is the integral closure of $A_f$ in its fraction field. Thus every automorphism of $A_f$ extends uniquely to an automorphism of $\mathbb{C}[t]$. It is now easy to see that automorphisms of $I_f$ are in bijection with automorphisms of $\mathbb{A}^1$ that map the closed subscheme defined by $f$ to itself. Concretely, such an automorphism is a linear polynomial that permutes the roots of $f$ (counting multiplicity). I don't know of a concise characterization of when any such automorphism must be trivial, but I expect this description of the automorphisms is good enough for most purposes. In particular, for instance, if $f$ has more than one distinct root it is easy to show from this description that every automorphism of $I_f$ is given by multiplication by a root of unity conjugated by translation by the average of the roots.

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