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Let $F=F(x,y),G=G(x,y) \in \mathbb{C}[x,y]$, $u=u(x),v=v(x) \in \mathbb{C}[x]$, $c \in \mathbb{C}$.

Denote:

$A=F(x,y)y+u(x)(x-c)=Fy+u(x-c)$,

$B=G(x,y)y+v(x)(x-c)=Gy+v(x-c)$.

When $I_{A,B}:=\langle A,B \rangle$ is a maximal ideal of $\mathbb{C}[x,y]$?

It is well-known that maximal ideals of $\mathbb{C}[x,y]$ are of the form: $\langle x-a,y-b \rangle$, $a,b \in \mathbb{C}$, see this question. Observe that, for example, $\langle x-a, y-b+(x-a)^5 \rangle$ is maximal, since $\langle x-a, y-b+(x-a)^5 \rangle = \langle x-a,y-b \rangle$.

Remarks: 1. Notice that $I_{A,B} \subseteq \langle x-c, y \rangle$.

  1. What if we further assume that $\gcd(u,v)=\gcd(u(x),v(x))=1$?

Any hints and comments are welcome!

user237522
  • 6,467

1 Answers1

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This is true if and only if $Fv-Gu\in \Bbb{C}$.

To see why, let $$ M = \newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}}\bmat F & u\\G & v\emat.$$ Then $$\bmat A \\ B \emat = M \bmat y \\ x-c \emat.$$

Then if $(A,B)=(y,x-c)$, we have that $y=e_{11}A + e_{12}B$ and $(x-c)=e_{21}A + e_{22}B$. Then the matrix $$ E = \bmat e_{11} & e_{12} \\ e_{21} & e_{22} \emat $$ satisfies $$ E\bmat A \\ B \emat = \bmat y \\ x-c \emat.$$ Thus $$ EM\bmat y \\ x-c \emat = \bmat y \\ x-c \emat.$$

Expanding this gives $$ \bmat e_{11} F +e_{12} G & e_{11}u+e_{12}v \\ e_{21}F+e_{22}G & e_{21}u+e_{22}v\emat\bmat y \\ x-c\emat = \bmat y \\ x-c \emat.$$

Then $$(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y,$$ which tells us immediately that $e_{11}u+e_{12}v=0$, and thus that $e_{11}F+e_{12}G=1$. Similarly, the fact that $$(e_{21}F+e_{22}G)y + (e_{21}u+e_{22}v)(x-c) = (x-c)$$ tells us that $e_{21}F+e_{22}G=0$ and $e_{21}u+e_{22}v=1$.

Hence $F$ and $G$ are relatively prime (and not just relatively prime, they generate the unit ideal), and $u$ and $v$ also generate the unit ideal.

Moreover, since $e_{11}u+e_{12}v=0$, we have $e_{11}u=-e_{12}v$, so since $u$ and $v$ are relatively prime, $e_{11} = h_1v$ and $e_{12}=-h_1u$. Then since $e_{11}F+e_{12}G=1$, we have $h_1vF-h_1uG=1$. Therefore $Fv-Gu=1/h_1\in \Bbb{C}$.

Conversely if $Fv-Gu\in\Bbb{C}$, then $M$ is invertible, so $A$ and $B$ generate $(y,x-c)$.

jgon
  • 28,469
  • Thank you very much! Your answer is very nice. – user237522 Dec 12 '18 at 21:44
  • Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n \in I$ implies that $a \in I$. – user237522 Dec 13 '18 at 14:38
  • Please, I do not understand something in your proof: What I see is that $(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y$ implies that $y$ divides $e_{11}u+e_{12}v$, namely $e_{11}u+e_{12}v=yw$ for some $w$. Equivalently, I think that you meant to take $y=0$ in $(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y$ and get that $(e_{11}u+e_{12}v)(y=0)=0$ which is not as saying that $e_{11}u+e_{12}v=0$ itself. Or maybe I am missing something? (Same with the second equation). – user237522 Jun 30 '20 at 14:52