Let $k$ be a field of characteristic zero, $k[x,y]$ be the polynomial ring and $\beta$ the following involution (= automorphism of order two) $\beta: (x,y) \mapsto (x,-y)$. 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 $p,q \in k[x,y]$, and write $p=s_1+k_1$ and $q=s_2+k_2$, with $s_1,s_1 \in S_\beta$ and $k_1,k_2 \in K_\beta$.
Let $I$ be the ideal in $k[x,y]$ generated by $p$ and $q$; every element of $I$ has the form $Ap+Bq$, for arbitrary $A,B \in k[x,y]$.
Assume that $s_1,s_2,k_1,k_2 \in I$; does this assumption imply something interesting about $p$, $q$ or $I$? By interesting I mean, for example, that one of $\{p,q\}$ is symmetric or skew-symmetric.
Remarks: (1) The symmetric elements with respecto to $\beta$ are of the form $a_{2n}y^{2n}+a_{2n-2}y^{2n-2}+\cdots+a_2y^2+a_0$, where $a_{2n},a_{2n-2},\ldots,a_2,a_0 \in k[x]$.
The skew-symmetric elements with respecto to $\beta$ are of the form $b_{2m+1}y^{2m+1}+b_{2m-1}y^{2m-1}+\cdots+b_3y^3+b_1y$, where $b_{2m+1},b_{2m-1},\ldots,b_3,b_1 \in k[x]$.
(2) In $k[x]$ with the involution $\beta: x \mapsto -x$, a similar situation does not tell much, for example: $p=x^2+x$ and $q=1-x$. Here $s_1=x^2$, $k_1=x$, $s_2=1$, $k_2=-x$. $s_1,k_1,s_2,k_2 \in I=\langle p,q \rangle$, but none of $p,q$ is symmetric or skew-symmetric.
(It is easy to check that $s_1,k_1,s_2,k_2 \in I$, for example: $A=\frac{1}{2}$, $B=-\frac{x}{2}$ give $Ap+Bq=\frac{1}{2} (x^2+x) -\frac{x}{2}(1-x)= \frac{x^2}{2} + \frac{x}{2} -\frac{x}{2} + \frac{x^2}{2}= x^2=s_2$).
However, this example may not hint what happens in $k[x,y]$, since this is a two-generated ideal in $k[x]$ (which is actually one generated ideal, since $k[x]$ is a PID. Actually, $I=\langle p ,q \rangle =k[x]$, since $2x=(x^2+x)+(x-x^2)=p+xq \in I$, so $x \in I$, and then $1=(1-x)+x=q+x \in I$). But if we assume that $I=\langle p \rangle$ satisfies $\beta\langle p \rangle \subseteq \langle p \rangle$, then it is not difficult to see that $p \in S_\beta \cup K_\beta$.
Thank you very much!
