Let $k$ be a field of characteristic zero.
The two-dimensional case: Let $p,q \in k[x,y]$, $I=\langle p,q \rangle$ a proper ideal of $k[x,y]$ and $\delta$ an involution on $k[x,y]$, namely, a $k$-algebra automorphism of order two. Denote the set of symmetric elements with respect to $\delta$ by $S_{\delta}$ and the set of skew-symmetric elements with respect to $\delta$ by $K_{\delta}$.
Assume that $\delta(I) \subseteq I$. Further assume that: (i) $\delta: (x,y) \mapsto (x,-y)$. (ii) $(p,q)$ is a Jacobian pair, namely, there exists a $k$-algebra endomorphism $f: (x,y) \mapsto (p,q)$ such that $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in k-\{0\}$.
Is the following claim true: There exist $k$-algebra automorphisms $g,h$ of $k[x,y]$, such that one of $\{(gfh)(x),(gfh)(y)\}$ belongs to $S_\epsilon \cup K_\epsilon$, where $\epsilon$ is an involution on $k[x,y]$.
Remarks:
(a) It can be shown that there exist two conjugacy classes of involutions on $k[x,y]$: That of $(x,y) \mapsto (x,-y)$ (Jacobian $-1$) and that of $(x,y) \mapsto (-x,-y)$ (Jacobian $1$).
(b) An easy observation: If $p$ is symmetric w.r.t. an involution $\beta$, then $g^{-1}(p)$ is symmetric w.r.t. $\gamma= g^{-1}\beta g$; indeed, $p=\beta(p)$, so $p=(g \gamma g^{-1})(p)$, then $g^{-1}(p)=\gamma (g^{-1}(p))$. (Same argument with skew-symmetric instead of symmetric).
(c) Notice that Arthur's comment is included in the above claim; indeed, $k-\{0\} \ni \operatorname{Jac}(aS+bK,cS+dK)=(ad-bc)\operatorname{Jac}(S,K)$, so $ad-bc \in k-\{0\}$ (which means that $aS+bK, cS+dK$ are linearly independent over $k$). If one of $\{a,b,c,d\}$ equals to zero, then we are done. Otherwise, it is not difficult to find an automorphism g of $k[x,y]$ such that $g(p)=S$ or $g(p)=K$, and we are also done.
(d) Am I missing a known theorem that can be applied here? Perhaps a theorem concerning group actions? Fixed points theorems? Algebraic geometry? C-star algebras?
Please see this question.
The one-dimensional case: Notice that in $k[x]$ an analog result holds, at least for $\delta: x \mapsto -x$, (which is actually the 'unique' involution on $k[x]$) namely:
If $I$ is an ideal of $k[x]$ which is invariant under $\delta$, then $I=\langle h \rangle$, where $h$ is either an odd polynomial (= a skew-symmetric element with respect to $\delta$) or an even polynomial (= a symmetric element with respect to $\delta$).
Indeed: $k[x]$ is a PID, so $I=\langle h \rangle$, for some $h \in k[x]$.
We assumed that $\delta(I) \subseteq I$, hence, in particular, $\delta(h) \in I$, so there exists $g \in k[x]$ such that $\delta(h)=gh$.
Of course, $\deg(\delta(h))=\deg(h)$ and $\deg(gh)=\deg(g)+\deg(h)$. Therefore, $\deg(g)=0$, so $g=\lambda \in k$.
Write $h=s+k$ with $s$ even (= symmetric w.r.t. $\delta$ and $k$ odd (= skew-symmetric w.r.t. $\delta$). We have $s-k=\delta(h)=\lambda h= \lambda (s+k)$. Then, $(1-\lambda)s=(1+\lambda)k$.
Denote $w:=(1-\lambda)s=(1+\lambda)k$.
From $S_{\delta} \cap K_{\delta} =\{0\}$, we obtain that $w=0$, so $(1-\lambda)s=0$ and $(1+\lambda)k=0$.
(i) $(1-\lambda)s=0$ implies that $\lambda=1$ or $s=0$. If $\lambda=1$, then $\delta(h)=h$, so $h$ is symmetric. If $s=0$, then $h$ is skew-symmetric.
(ii) $(1+\lambda)k=0$ implies that $\lambda=-1$ or $k=0$. If $\lambda=-1$, then $\delta(h)=-h$, so $h$ is skew-symmetric. If $k=0$, then $h$ is symmetric.
Concluding that $h$ is symmetric or skew-symmetric, as claimed.
Remark: Concerning the 'uniqueness' of the involution $x \mapsto -x$; there exists other involutions, for example, $f_c: x \mapsto -x+c$, for arbitrary $c \in k$. ($f_c^2(x)=f_c(f_c(x))=f_c(-x+c)=-f_c(x)+c=-(-x+c)+c=x-c+c=x$).
Luckily, my above argument (for $\delta=f_0$) still shows that if $f_c(I) \subseteq I$, $c \in k$, then $I=\langle h \rangle$ with $h$ symmetric or skew-symmetric with respect to $f_c$; indeed, the only place where we used the specific form of $\delta=f_0$ was "Of course, $\deg(\delta(h))=\deg(h)$", and this trivial observation still holds if we replace $\delta=f_0$ by $f_c$, $c \in k-\{0\}$.
Concerning the two-dimensional case: Write $p=s_1+k_1$ and $q=s_2+k_2$, where $s_1,s_2 \in S_{\delta}$ and $k_1,k_2 \in K_{\delta}$, $\delta: (x,y) \mapsto (x,-y)$.
Several attempts/ideas to answer my question:
(1) By assumption, $\delta(p) \in I=\langle p,q \rangle$, so $s_1-k_1 = A(s_1+k_1)+B(s_2+k_2)$, for some $A,B \in k[x,y]$. Then $(1-A)s_1-Bs_2=(1+A)k_1+Bk_2$; however, this does not tell much, since $A,B \in k[x,y]$, not in $k$.
(If $A,B \in k$, then we obtain that $w=0$, where $w:=(1-A)s_1-Bs_2=(1+A)k_1+Bk_2$. Therefore, $s_1=\mu s_2$ and $k_1= \nu k_2$, where $\mu,\nu \in k$, and we have obtained the special form that Arthur has suggested).
We can further write: $A=A_s+A_k$ and $B=B_s+B_k$, where $A_s,B_s \in S_\delta$ and $A_k,B_k \in K_\delta$.
(2) I have written $s_1,s_2,k_1,k_2$ as elements of $I$, and also applied $\delta$ on both sides, but the equations I have reached did not help much. Perhaps I will add my calculations later.
(3) We have assumed that $p \neq q$, since otherwise $I=\langle p \rangle$, so the one-dimensional proof can be adjusted here, in short:
We assumed that $\delta(I) \subseteq I$, hence, in particular, $\delta(p) \in I=\langle p \rangle$, so there exists $g \in k[x,y]$ such that $\delta(p)=gp$.
Of course, $\deg(\delta(p))=\deg(p)$ and $\deg(gp)=\deg(g)+\deg(p)$. Therefore, $\deg(g)=0$, so $g=\lambda \in k$.
From $p=s_1+k_1$ with $s_1$ symmetric w.r.t. $\delta$ and $k_1$ skew-symmetric w.r.t. $\delta$, we have $s_1-k_1=\delta(p)=\lambda p= \lambda (s_1+k_1)$. Then, $(1-\lambda)s_1=(1+\lambda)k_1$.
Denote $w:=(1-\lambda)s_1=(1+\lambda)k_1$.
From $S_{\delta} \cap K_{\delta} =\{0\}$, we obtain that $w=0$, so $(1-\lambda)s_1=0$ and $(1+\lambda)k_1=0$, and the same arguments as above show that $p$ is symmetric or skew-symmetric.
Same question for the first Weyl algebra $A_1(k)$ over $k$, $k\langle x,y | yx-xy=1 \rangle$, instead of $k[x,y]$. Replace the Jacobian $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x$ by the commutator $[p,q]:=pq-qp$. Notice that in the non-commutative case there are involutions and anti-involutions.
I have now noticed this slightly similar question.
I have asked my above question also in MO.
Thank you very much!
