I have asked the following question at MSE and got one answer. It would be nice to have more elaboration on it, please:

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!