In $\mathbb{C}[x,y]$: If $\langle u,v \rangle$ is a maximal ideal, then $\langle u-\lambda,v-\mu \rangle$ is a maximal ideal?

Question

I have asked the following question at MSE and got one answer. Any further ideas are welcome:

Let $u=u(x,y), v=v(x,y) \in \mathbb{C}[x,y]$, with $\deg(u) \geq 2$ and $\deg(v) \geq 2$. Let $\lambda, \mu \in \mathbb{C}$.

Assume that the ideal generated by $u$ and $v$, $\langle u,v \rangle$, is a maximal ideal of $\mathbb{C}[x,y]$.

Is it true that $\langle u-\lambda, v-\mu \rangle$ is a maximal ideal of $\mathbb{C}[x,y]$?

My attempts to answer my question are:

(1) By Hilbert's Nullstellensatz, $\langle u,v \rangle= \langle x-a,y-b \rangle$, for some $a,b \in \mathbb{C}$, so $x-a=F_1u+G_1v$ and $y-b=F_2u+G_2v$, for some $F_1,G_1,F_2,G_2 \in \mathbb{C}[x,y]$. Then, $x=F_1u+G_1v+a$ and $y=F_2u+G_2v+b$.

(2) $\frac{\mathbb{C}[x,y]}{\langle u,v \rangle}$ is a field (since $\langle u,v \rangle$ is maximal); actually, $\frac{\mathbb{C}[x,y]}{\langle u,v \rangle}$ is isomorphic to $\mathbb{C}$. Is it true that $\frac{\mathbb{C}[x,y]}{\langle u,v \rangle}$ is isomorphic to $\frac{\mathbb{C}[x,y]}{\langle u-\lambda,v-\mu \rangle}$? In other words, is it true that $\frac{\mathbb{C}[x,y]}{\langle u-\lambda,v-\mu \rangle}$ is isomorphic to $\mathbb{C}$? See this question.

(3) If $\langle u-\lambda,v-\mu \rangle$ is not maximal, then it is contained in some maximal ideal: $\langle u-\lambda,v-\mu \rangle \subsetneq \langle x-c,y-d \rangle$, $c,d \in \mathbb{C}$. It is not difficult to see that $(u-\lambda)(c,d)=0$ and $(v-\mu)(c,d)=0$, so $u(c,d)-\lambda=0$ and $v(c,d)-\mu=0$, namely, $u(c,d)=\lambda$ and $v(c,d)=\mu$.

Remark: Is it possible that $\langle u-\lambda,v-\mu \rangle = \mathbb{C}[x,y]$? If so, then there exist $F,G \in \mathbb{C}[x,y]$ such that $F(u-\lambda)+G(v-\mu)=1$. Then at $(a,b)$ we get: $F(a,b)(-\lambda)+G(a,b)(-\mu)=1$ (since, by (1), $u(a,b)=0$ and $v(a,b)=0$).

Thank you very much!

Answer 1

While it may be, in general it is not. Consider as an example, $u=x+y+yp(x), v=x+yp(x)$ where$ \deg p(x)\geq 2$. Then $(u,v)=(x,y)$ and so maximal. Notice that $u=y+v$. So, $$(u-a, v-b)=(u-v+b-a, v-b)=(y+b-a, v-b)=(y+b-a, x-b+(a-b)p(x))$$ and so most pairs of values of $a,b$, it is not maximal, as long as $\deg (x+(a-b)p(x))\geq 2$.