The following question appears in [MSE](https://math.stackexchange.com/questions/4760401/if-langle-f-1-f-2-rangle-langle-g-1-g-2-rangle-then-langle-f-1-lambd) 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](https://math.stackexchange.com/questions/4399584/generators-of-a-finite-intersection-of-maximal-ideals?noredirect=1&lq=1).


**(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!



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


