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The following question appears here; hopefully, it is appropriate for MO.

Let $k$ be a field of characteristic zero, and let $\beta: k[x,y] \to k[x,y]$ be the following involution $\beta: (x,y) \mapsto (x,-y)$, namely, $\beta$ is an automorphism of $k[x,y]$ of order two. Denote the set of symmetric elements w.r.t $\beta$ by $S_{\beta}$ and the set of skew-symmetric elements w.r.t $\beta$ by $K_{\beta}$.

Let $s_1,s_2 \in S_{\beta}$ and $k_1,k_2 \in K_{\beta}$. Write $p=s_1+k_1$ and $q=s_2+k_2$. Denote the ideal generated by $p$ and $q$ by $I$, $I= \langle p,q \rangle$.

Assume that $s_1,s_2,k_1,k_2 \in I$; equivalently, $\beta(I) \subseteq I$.

Should $\{s_1,s_2\}$ or $\{k_1,k_2\}$ be algebraically dependent?

Remarks:

(i) The only examples I have in mind are of the form: $p=aS+bK$, $q=cS+dK$, where $a,b,c,d \in k$, $S \in S_{\beta}$, $K \in K_{\beta}$. In other words: $s_1=aS, s_2=cS, k_1=bK, k_2=dK$, so $\{s_1,s_2\}$ and $\{k_1,k_2\}$ are algebraically dependent.

(ii) Observe that, for example, we do not know if $J:=\langle p^2,q \rangle \subset I$ is invariant under $\beta$.

Any comments and hints are welcome!

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  • $\begingroup$ Take $s_1=x, s_2=y^2, k_1=y^3, k_2=0$. Then $s_1,s_2$ are algebraically independent . $\endgroup$
    – Mohan
    Commented Sep 30, 2020 at 0:27
  • $\begingroup$ @Mohan, thank you for your comment. But $k_1=y^3$ and $k_2=0$ are algebraically dependent. I apologize for not asking clearly; I meant at least one of these pairs is algebraically dependent. Anyway, there is a counterexample: $s_1=x+x^2+y^2$, $k_1=-2xy$, so $p=x+(x-y)^2$. $s_2=x^2+y^2$, $k_2=-2xy+y$, so $q=y+(x-y)^2$. Here $\langle p,q \rangle= \langle x,y \rangle$ is $\beta$-invariant. Clearly, both pairs $\{s_1,s_2\}$, $\{k_1,k_2\}$ are algebraically independent over $k$ (since the Jacobian of each pair is non-zero). $\endgroup$
    – user237522
    Commented Sep 30, 2020 at 1:26

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