Let $k[x,y]$ be the $k$-algebra of polynomials in two commuting variables $x$ and $y$ over $k$, where $k$ is a field of characteristic zero.
Given an arbitrary polynomial of degree three, is it possible to find an involution such that the given polynomial is a symmetric (or skew-symmetric) element with respect to it? Can one find an answer to my question in the language of algebraic geometry?
Remark: For an arbitrary element of degree two, namely: $ax^2+bxy+cy^2+dx+ey+f$, I have managed to show, by some lengthy calculations, that such element is symmetric (or skew-symmetric) with respect to at least one of the involutions $\{ \alpha, \beta, \epsilon\}$ defined below.
More elaborately: An involution is an automorphism of $k[x,y]$ of order two; for example: $\alpha: (x,y) \mapsto (y,x)$, $\beta:(x,y) \mapsto (x,-y)$ and $\epsilon:(x,y) \mapsto (-x,-y)$.
It is not difficult to show that a $\alpha$ and $\beta$ are conjugate, while $\epsilon$ is in a different conjugacy class (the Jacobian of $\alpha$ and $\beta$ is $-1$, while the Jacobian of $\epsilon$ is $1$, and conjugates have the same Jacobian).
For an arbitrary involution $\iota$, there is the set of symmetric elements with respect to it, namely, the elements $w$ of $k[x,y]$ such that $\iota(w)=w$.
Now, let $p$ be an arbitrary polynomial of degree three: $p= a_{30}x^3+a_{21}x^2y+a_{12}xy^2+a_{03}y^3+a_{20}x^2+a_{11}xy+a_{02}y^2+a_{10}x+a_{01}y+a_{00}$, where $a_{ij} \in k$.
By coordinate change I only managed to obtain that $a_{21}=0$ (and $a_{00}=0$), but I do not see why the new $p$ should be a symmetric element for some involution.
Thank you very much!
There exists cubics which are not invariant under any non-trivial involutions. I do not know an easy proof, but I will explain.
Let $E$ be any elliptic curve and let $P,Q,R$ be three general points on $E$, so that they are linearly independent in the Picard group. Then, we can embed $E$ in $\mathbb{P}^2$ using the divisor $P+Q+R$ as a smooth cubic and $P,Q,R$ are collinear. Then consider the $\mathbb{A}^2$, which is the complement of the line joining these point and $X\subset\mathbb{A}^2$ be $E-\{P,Q,R\}$. Then $X$ is defined by a cubic polynomial say $f$.
If $\sigma$ is a non-trivial involution of the affine plane with $\sigma(f)=\pm f$, then $\sigma$ descends to an involution of $X$. If this involution is trivial, this means $X$ is contained in the fixed locus of $\sigma$. But, this is known to be not possible since it is known that such involution can have fixed locus either a single point or an affine line (not necessarily linear) and $X$ has genus one. (This is a consequence of the structure of the automorphism group of the plane as a suitable amalgamated product.)
So, $\sigma$ is a non-trivial involution of $X$ and thus gives a non-trivial involution of $E$. But, since $\sigma$ takes the three points to the three points, at least one of them must be a fixed point of $\sigma$. Let us say $\sigma(P)=P$, then either $\sigma$ fixes all three points or $\sigma(Q)=R$. Since $\sigma$ has a fixed point, $E/\sigma=\mathbb{P}^1$. Fiber of one point of this map is $2P$ and fiber of another point is $2Q$ or $Q+R$, depending on whether $\sigma$ fixes $Q$ or not. In the first case, we have $2(P-Q)=0$ and in the second case we have $2P-Q-R=0$ in the Picard group, contrary to our assumption of linear independence.