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

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!

Now also at MO.

Answer 1

No, this isn't true. The requirement that $\deg u$, $\deg v \geq 2$ is annoying, so I'll first solve the problem without this and then tweak my example to add this condition. As already explained in other answers, saying that $\langle u,v \rangle$ is maximal means that the curves $\{ u = 0 \}$ and $\{ v=0 \}$ intersect at a unique point, and cross transversely there. We want to find polynomials such that this is true for $u$ and $v$, but not for $u-\lambda$ and $v-\lambda$.

An example is to take $u = xy^2 + y + x$ and $v=x$. Then $u-(y^2+1) v = y$, so it is easy to see that $\langle u,v \rangle = \langle x,y \rangle$. But $u=0$, $v=2/5$ has two roots, at $x=2/5$, $y \in \{ 2, 1/2 \}$. (Almost any value of $v$ would have worked; I chose one that made the $y$-coordinates rational.) Here is a plot of the curve $u=0$ together with $v=0$ and $v=2/5$:

Now we have to tweak this to make $\deg v$ be $\geq 2$ rather than $1$. Take $u=xy^2+y+x$ and $v_2 = xy^2 + y$, so $v_2 = u-v$. Then $\langle u,v \rangle = \langle u,v_2 \rangle$ and $\langle u,v-2/5 \rangle = \langle u, v_2+2/5 \rangle$ so $\langle u,v \rangle$ is a maximal ideal and $\langle u, v_2+2/5 \rangle$ is not.

Answer 2

Update 19/07/2020 This approach does not work and the statement is in fact, not true. See David Speyer's answer instead.

The issue is that $\overline{u - \lambda}(x,y,0) = \bar{u}(x,y,0)$ and $\overline{v - \mu}(x,y,0) = \bar{v}(x,y,0)$ only implies that $\{u = 0\} \cap \{v = 0\}$ and $\{u = \lambda \} \cap \{v = \mu\}$ have the same set of zeroes at infinity, not that the multiplicities are equal.

For example $u(x,y) = x$ and $v(x,y) = xy - 1$ versus $u'(x,y) = x - 1$. These have $[0:1:0]$ as a common zero at infinity but to get the multiplicity we need to look at $\overline{u'}(x,1,z) = x - z$ and $\overline{v}(x,1,z) = x - z^2$ rather than just setting $z = 0$.

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By the Nullstellensatz $(u, v)$ is maximal if and only if $V(u,v)$ is a single (closed) point.

Why is this? First suppose $(u,v)$ is maximal. Then you're comfortable with saying that $(u,v) = (x - a, y - b)$ by the (weak) Nullstellensatz. Thus $$V(u,v) = V(x - a, y - b) = \{(a,b)\}$$ is a single point.

Conversely, if $V(u,v) = \{(a,b)\}$ is a single point, then the (strong) Nullstellensatz tells us that $$\operatorname{rad}(u,v) = I(V(u,v)) = I(\{(a,b)\}) = (x - a, y - b).$$ The condition that $(u,v) = \operatorname{rad}(u,v)$ is the same as saying $(a,b)$ doesn't have extra multiplicity as a common zero.

Let $\bar{u}(x,y,z)$ and $\bar{v}(x,y,z)$ be the homogenizations of $u, v$. By Bézout's theorem, $V(\bar{u},\bar{v}) \subseteq \mathbb{CP}^2$ consists of $\deg(\bar{u})\deg(\bar{v}) = \deg(u)\deg(v)$ points counted with multiplicity.

By the first observation, $V(u,v)$ consists of $1$ point, meaning the other $\deg(u)\deg(v) - 1$ points occur at infinity. I.e. they are the common zeroes of $\bar{u}(x,y,0)$ and $\bar{v}(x,y,0)$

If we modify $u, v$ by adding $\lambda, \mu$ then the new homogenizations are

$$ \overline{u - \lambda} = \bar{u} + \lambda z^{\deg u}, \quad \overline{v - \mu} = \bar{u} + \mu z^{\deg u} $$

Since $\overline{u - \lambda}(x,y,0) = \bar{u}(x,y,0)$ and $\overline{v - \mu}(x,y,0) = \bar{v}(x,y,0)$ it is also true that $\overline{u - \lambda}$ and $\overline{v - \mu}$ have $\deg(u)\deg(v) - 1$ common zeroes at infinity.

Thus $u - \lambda, v - \mu$ have exactly one common zero not at infinity, so by the Nullstellensatz, $(u - \lambda, v - \mu)$ is maximal.