Let $R$ be a commutative ring with identity and $A$ and $B$ be two proper ideals of $R$ such that $A+B=R$ and for each $r^2=r\in R$ we have either $r\not\in A$ or $r-1\not\in B$. How can we prove the existance of two maximal ideals $m_1$ and $m_2$ of $R$ such that $A\subseteq m_1$, $B\subseteq m_2$ and $\{r\in m_1\mid r^2=r\}=\{r\in m_2\mid r^2=r\}$?
Commutative Algebra – Existence of Two Maximal Ideals with the Same Set of Idempotents
ac.commutative-algebraag.algebraic-geometryra.rings-and-algebras
Related Solutions
Your ring $L$ is a localization of the power series rings $R= k[[x_1,\cdots,x_n]]$ at the multiplicative set $M$ of monomials in $R$.
So the prime ideals of $L$ correspond to prime ideals in $R$ which do not meet $M$. Clearly, the maximal primes are the set of biggest primes which do not contain any variable. Since $R$ is local and the maximal ideal contains all the variables, these ideals are of dimension one (this is the key difference to the Laurent polynomials case, since in that case these ideals would be maximal in the polynomial ring, so Nullstellensatz applies)
For example, when $n=2$, you get all ideals of height one which does not contains $x_1$ or $x_2$. So they are all principal ideal $(f)$ with $f$ irreducible and $f\neq x_1$, $f\neq x_2$.
It shows that the answer to your refined question is NO. The ideals $(x-y)$ and $(x-y^2)$ can not be in the same orbit.
In general one can only describe the set of maximal ideals in $L$ as follows: they are prime ideals of height $n-1$ in $S$, minus the set $(x_i, P)$, where $P$ is a prime of height $n-2$ in $k[[x_1,...x_{i-1}, x_{i+1},...,x_n]]$.
I doubt one can say more in general, since there are many open questions about affine curves.
This reminds me of what someone once said to me:
The geometers can always take a hyperplane section. We can't!
The purpose of this post is to analyze the question to see how close it is to Bertini's theorem (it is not obvious to me). The statement is immediately equivalent to: one can always find a nonzero prime ideal $P \subseteq m_1\cap m_2$ (since if we can, then induction on dimension proves the original).
Now, how can such $P$ exist? Let $U_1 = R-m_1$ and $U_2=R-m_2$. Let $U=\{xy \| x\in U_1, y\in U_2\}$. $U$ is multiplicative and our prime $P$ obviously just has to avoid $U$. So $P$ exists unless the localization $U^{-1}R$ has dimension $0$. But it is a domain, so we have to make sure $U^{-1}R$ is not a field. That statement is equivalent to the existence of some element $f\in R$ such that $f$ does not become an unit in $U^{-1}R$. In other words:
there are no $a,b \in U$ such that $af=b$.
Since $b$ is itself a product of elements in $U_1,U_2$, our condition is obvious if $fR$ is a prime ideal and $f\in m_1\cap m_2$. But it is technically weaker, although not clear to me by how much. Note that we do not require $f$ to be linear, which is common for Bertini's type statement.
This analysis would seem to rule out certain clever arguments. But may be one can find some!
For "algebraic" version of Bertini's theorem, I will look at the reference given here. Also, how to rescue Bertini over finite fields using hypersurface instead of hyperplane (which suits your purpose), looks here.
Best Answer
Sketch: First, if $e$ is an idempotent in $A$, show that $B$ can be replaced with $B+Re$, and the hypotheses still holds. Use this to reduce to the case that $A$ and $B$ contain the same idempotents.
Second, if $e$ is an idempotent of $R$ with $e,1-e\notin A$ (and hence also not in $B$) show that we can replace $A$ and $B$ with the new pair $A+Re$ and $B+Re$, still satisfying the same conditions. Use this to reduce to the case that $A$ and $B$ contain the same idempotents, and either an idempotent or its complement belongs to $A$.
Now, extend to any maximal ideals containing $A$ and $B$, and note that a maximal ideal cannot contain an idempotent and its complement at the same time.