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After asking this question and finding this relevant paper, I would like to ask the following question:

For every $a,b \in \mathbb{C}$, denote: $A_{a,b}=\mathbb{C}[(x-a)(x-b),x(x-a)(x-b),y]$ and $B_{a,b}=\mathbb{C}[(y-a)(y-b),y(y-a)(y-b),x]$.

If I am not wrong (but I may be wrong), $A_{a,b}$ and $B_{a,b}$ are maximal sub-$\mathbb{C}$-algebras of $\mathbb{C}[x,y]$, namely, there is no proper sub-$\mathbb{C}$-algebra $T_A$ of $\mathbb{C}[x,y]$ such that $A_{a,b} \subsetneq T_A$ and there is no proper sub-$\mathbb{C}$-algebra $T_B$ of $\mathbb{C}[x,y]$ such that $B_{a,b} \subsetneq T_B$.

Question: I am looking for maximal sub-$\mathbb{C}$-algebras of $\mathbb{C}[x,y]$ other than $A_{a,b},B_{a,b}$. Is it possible to describe all maximal sub-$\mathbb{C}$-algebra of $\mathbb{C}[x,y]$ in terms of generators like $A_{a,b},B_{a,b}$?

Thank you very much!

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  • $\begingroup$ Applying a change of coordinates (automorphism of $\mathbb C[x,y]$) will give you other algebras, similarly to how $A_{a,b}$ is obtained from $B_{a,b}$ by exchanging $x$ and $y$. $\endgroup$
    – red_trumpet
    Commented Oct 21 at 18:51
  • $\begingroup$ @red_trumpet, nice comment; thank you! So we can focus on $A_{a,b}$ and apply any automorphism of $\mathbb{C}[x,y]$ to obtain another (proper) maximal sub-$\mathbb{C}$-algebra. Are there additional (proper) maximal sub-$\mathbb{C}$-algebras? $\endgroup$
    – user237522
    Commented Oct 21 at 18:52

1 Answer 1

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These are not maximal. For example if you add $xy$ to $A_{a,b}$ you get something which is a larger algebra but still not proper since any polynomial in the algebra takes the same value at the points $(a,0)$ and $(b,0)$.

The following two types of subalgebras are codimension $1$, hence maximal: For any two points in $\mathbb C^2$, the algebra of polynomials whose values at those two are the same, or for any point in $\mathbb C^2$ and tangent vector at that point, the algebra of polynomials whose derivative at that point in that tangent direction vanishes.

Maybe these are the only maximal ones.

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  • $\begingroup$ What if we take the above $A_{a,b}=\mathbb{C}[(x-a)(x-b),(x-a)(x-b)x,y]$ and extend it to the larger sub-$\mathbb{C}$-algebra $\tilde{A_{a,b}}=A_{a,b}+\langle (x-a)(x-b),(x-a)(x-b)x,y \rangle$? I have added the ideal generated by $\{(x-a)(x-b),(x-a)(x-b)x,y\}$. Is the larger $\tilde{A_{a,b}}$ a proper maximal sub-$\mathbb{C}$-algebra? Notice that $xy \in \tilde{A_{a,b}}-A_{a,b}$. $x \notin \tilde{A_{a,b}}$, so it is proper. I am not sure about maximality (maybe maximality follows by the codimension $1$ argument). $\endgroup$
    – user237522
    Commented Oct 21 at 20:33
  • $\begingroup$ Sometimes $\tilde{A_{a,b}}$ is not proper: For example, $a=1,b=-1$, $\tilde{A_{a,b}}=\mathbb{C}[x^2-1,x^3-x,y]+\langle x^2-1,x^3-x,y\rangle$. If we allow to change the generators in the algebra and accordingly in the ideal, then we get $\mathbb{C}[x^2,x^3-x,y]+\langle x^2,x^3-x,y \rangle$, then $x$ is in this algebra: $x=x(x^2)-(x^3-x)$. But if we do not allow such a change of generators in the ideal, then the algebra is proper. $\endgroup$
    – user237522
    Commented Oct 21 at 20:46

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