Ideals: If $\langle f_1,f_2 \rangle = \langle g_1,g_2 \rangle$, then $\langle f_1-\lambda,f_2-\mu \rangle = \langle g_1-\delta,g_2-\epsilon \rangle$?

Question

The following question appears in MSE without answers.

Let $f_1,f_2,g_1,g_2 \in \mathbb{C}[x,y]-\mathbb{C}$.

Assume that $\langle f_1,f_2 \rangle = \langle g_1,g_2 \rangle \subsetneq \mathbb{C}[x,y]$, where $\langle u,v \rangle$ denotes the ideal in $\mathbb{C}[x,y]$ generated by $u$ and $v$.

Claim: Given $\lambda,\mu \in \mathbb{C}$, $(\lambda,\mu) \neq (0,0)$ (but one of $\{\lambda,\mu\}$ can be zero), there exist $\delta,\epsilon \in \mathbb{C}$ such that $\langle f_1-\lambda,f_2-\mu \rangle = \langle g_1-\delta,g_2-\epsilon \rangle$.

Notice that in the claim, necessarily $(\delta,\epsilon) \neq (0,0)$; otherwise, $\delta=\epsilon=0$, so $I:= \langle f_1-\lambda,f_2-\mu \rangle = \langle g_1,g_2\rangle= \langle f_1,f_2 \rangle$ and then $I=\mathbb{C}[x,y]$ since w.l.o.g. $\lambda \neq 0$, hence $\lambda=f_1-(f_1-\lambda) \in I$.

Question 1: Is the claim true when $\langle f_1,f_2 \rangle = \langle g_1,g_2 \rangle$ is a maximal ideal?

Question 2: Is the claim true when $\langle f_1,f_2 \rangle = \langle g_1,g_2 \rangle$ is a finite intersection of maximal ideals?

Question 3: Are the following two sets equal? $\{ \langle f_1-\lambda,f_2-\mu \rangle \}_{ \lambda,\mu \in \mathbb{C}}$ and $\{ \langle g_1-\delta,g_2-\epsilon \rangle \}_{ \delta,\epsilon \in \mathbb{C}}$.

Examples:

(i) $f_1=x+y,f_2=x-y,g_1=x,g_2=y$, namely, $\langle x+y,x-y \rangle = \langle x,y \rangle$. If we choose $\lambda=1,\mu=2$, then we can take $\delta=\frac{3}{2}, \epsilon=-\frac{1}{2}$ and get $\langle x+y-1,x-y-2 \rangle = \langle x-\frac{3}{2},y+\frac{1}{2} \rangle$.

• This example shows that it is generally not true that $\langle f_1-\lambda,f_2-\mu \rangle = \langle g_1-\lambda,g_2-\mu \rangle$.

• At least for linear $f_1,f_2,g_1,g_2$, there exist $\delta,\epsilon$ such that $\langle f_1-\lambda,f_2-\mu \rangle = \langle g_1-\delta,g_2-\epsilon \rangle$. Linear means of the form $ax+by+c$, for some $a,b,c \in \mathbb{C}$.

(ii) $f_1=x,f_2=y+x^2+5x+3,g_1=x,g_2=y+3$, namely, $\langle x,y+x^2+5x+3 \rangle = \langle x,y+3 \rangle$. If we choose $\lambda=0,\mu=3$, then we can take $\delta=0,\epsilon=3$ and get $\langle x,y+x^2+5x \rangle = \langle x,y \rangle$.

• In this particular example $(\lambda,\mu)=(\delta,\epsilon)=(0,3)$.
• It seems that at least for triangular/linear $f_1,f_2,g_1,g_2$, there exist $\delta,\epsilon$ such that $\langle f_1-\lambda,f_2-\mu \rangle = \langle g_1-\delta,g_2-\epsilon \rangle$. Triangular means of the form $ax+F(y)$ or $by+G(x)$, for some $a,b \in \mathbb{C}-\{0\}$, $F(y) \in \mathbb{C}[y]$, $G(x) \in \mathbb{C}[x]$.

Remarks:

(i) Every finite intersection of maximal ideals in $\mathbb{C}[x,y]$ is (a radical ideal) that can be generated by two elements, see this.

(ii) For the general case where $\langle f_1,f_2 \rangle = \langle g_1,g_2 \rangle \subsetneq \mathbb{C}[x,y]$ is not even radical, perhaps it is interesting to consider $f_1=x+y,f_2=xy,g_1=x+y,g_2=x^2+y^2$.

$\langle x+y,xy \rangle \subsetneq \langle x,y \rangle$ and it is not a radical ideal, since $x^2 \in \langle x+y,xy \rangle$, but $x \notin \langle x+y,xy \rangle$. (Indeed, $x^2=(x+y)x-xy$).

Thank you very much!

Answer 1

The answer to your questions is no. The ideals $\langle x, y(1-xy) \rangle$ and $\langle x, y \rangle$ are equal, and maximal; but

$$ \langle x-\lambda, y(1-xy)\rangle \neq \langle x-\delta, y-\epsilon \rangle$$

for any $\lambda,\delta,\epsilon \in \mathbb{C}$, $\lambda \neq 0$. (In the notation of your questions, $\mu=0$.)

A related example is $\langle x, 1-xy\rangle = \langle 1 \rangle$, but $\langle x-\lambda, 1-xy \rangle \neq \langle 1 \rangle$ for nonzero $\lambda$.