I'm trying to solve the following problem:

Find a generating set for the algebra of invariant polynomials $\mathbb C[x_1, x_2]^\Gamma$, where $\Gamma$ is a dihedral group $D_n$, generated by matrices $\begin{pmatrix}\zeta && 0 \\ 0 && \zeta^{-1}\end{pmatrix}$ and $\begin{pmatrix} 0 && 1 \\ 1 && 0 \end{pmatrix}$. $\zeta$ is a primitive $n$th root of unity.

Here is my solution:

Polynomial is invariant under the action of $\Gamma$ iff it's invariant under the action of generators. Second generator of $\Gamma$ just permutes $x_1$ and $x_2$. So any invariant polynomial is in fact a symmetric polynomial of two variables. Any such polynomial is of the form $\sum a_i(x_1 + x_2)^{k_i} + \sum b_i (x_1x_2)^{k'_i} + \sum c_i (x_1 + x_2)^{k''_i}(x_1 x_2)^{k'''_i} + d$.

It is clear that $x_1 x_2$ is invariant under the action of the first generator of $\Gamma$. So we only want to check for which $k$ it is true that $(\zeta x_1 + \zeta^{-1} x_2)^k = (x_1 + x_2)^k$. If we expand both expressions, we can see that it means $\zeta^{k - 2 l} = 1$ for all $l \in [0, k]$. Since $\zeta$ is a primitive $n$th root of unity, it follows that $k - 2 l$ is a multiple of $n$. If $n = 2$, $k$ can be any even number. In this case, generator set for $\mathbb C[x_1, x_2]^\Gamma$ is $((x_1 + x_2)^2, x_1 x_2, 1)$.

If $n > 2$, such $k$ does not exist. Indeed, if $k = r_1 n$ and $k - 2 = r_2 n$ then $(r_1 - r_2)n = 2$ which is impossible. Thus generating set for $\mathbb C[x_1, x_2]^\Gamma$ in this case is $(x_1 x_2, 1)$.

Am I right? Am I missing something?


  • 2
    $\begingroup$ Looks OK to me. $\endgroup$ Apr 4, 2020 at 18:32

1 Answer 1


I don't think you are fully right because there is no need for $(x_1+x_2)^k$ to be an invariant polynomial.

Since $D_n$ is a finite Coxeter Group, By Chevalley's theorem, $\mathbb{C}[x_1,x_2]^\Gamma$ is a polynomial algbera over $\mathbb{C}$ with two homogeneous generators say $f_1$ and $f_2$. If $f_1$ and $f_2$ are with degree $d_1$ and $d_2$, then $d_1d_2=2n$ ( the order of the Dihedral Group). Which is met in the first case but not for the others.

Or course $f_1=x_1x_2$ is invariant under $D_n$.

Let $f_2=x_1^n+x_2^n$. Then it's easy to check that $f_2$ is also $D_n$ invariant.

Now, since the Jacobian of $f_1$ and $f_2$, $J(f_1,f_2)=n(x_2^n-x_1^n)\neq 0$, it follows that $f_1$ and $f_2$ are algebraically independent over $\mathbb{C}$ satisfying $d_1d_2=2n$.

Now by Proposition of 3.12 ( Page no-67) of 'Reflection Groups and Coxeter Groups' by James E Humphreys, it follows that $\mathbb{C}[x_1,x_2]^\Gamma=\mathbb{C}[f_1,f_2]$ for any $n$.


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