Let $R$ be a local ring (commutative and with $1$) with maximal ideal $M$, with an involution $\theta$. Let $h$ be a Hermitian form on $R^n$, i.e. $h:R^n\times R^n\rightarrow R$ such that $h$ is $R$-linear in its first argument and $h(v,w)=h(w,v)^\theta$. Denote $$H(R) = \{v\in R^n\mid h(v,v)=0\}\;,$$ and look at a map $\pi:H(R/M^i)\rightarrow H(R/M)$ for some $i\geq1$.

Question 1 Is $\pi$ always surjective (for any local ring and any $i$)?

There seems to be no reason for this to be the case, but I can't seem to find a counterexample. So, if the answer to question 1 turns out to be no:

Question 2 Find a local ring $R$ such that $\pi$ is not surjective.

My motivation for asking this, is that I am looking at groups acting on these solutions. If $\pi$ is surjective, I can limit myself to looking at the solutions over $R$, whereas otherwise, I'd also have to consider the solutions over all projections $R/M^i$.

Possible assumptions: If one can prove question 1 with (some of these) extra assumptions, that would be sufficient. If one could find solution to question 1, it would be nice if they satisfy these conditions:

  • $\bigcap_{i=0}^\infty M^i = 0$ (or even $M^N=0$ for some $N$);
  • $n=3$ and $h\big((x_1,y_1,z_1), (x_2,y_2,z_2)\big) = x_1z_2^\theta+z_1x_2^\theta+y_1y_2^\theta$.

I will answer the question in a slightly more specific context, where

  • We restrict ourselves to the projective plane $\mathbb{P}_2(R)$ over $R$, i.e. triples $(x,y,z)\in R^3$ such that $xR+yR+zR=R$, up to multiples in $R^\ast$.
  • We use the given Hermitian form $h\big((x_1,y_1,z_1),(x_2,y_2,z_2)\big) = x_1z_2^\theta+z_1x_2^\theta+y_1y_2^\theta$

If we have a triple $v=(x,y,z)$ in the projective plane satisfying $h(v,v)=0$, that implies that either $x$ or $z$ is invertible, so we may assume $z$ is invertible, and hence we can assume $z=1$.

In this formulation, the question reduces to the following:

If $x+x^\theta+yy^\theta\in M$, can we find $x',y'$ such that $x'+x'^\theta+y'y'^\theta=0$ and $x-x',y-y'\in M^i$.

Define for any ideal $I$ in $R$ for which $I^\theta=I$ $$S_I = \{x+x^\theta+yy^\theta\mid x,y\in I\}.$$ Property: The following two statements are equivalent for $I$ an ideal in $R$ (with $I^\theta=I$):

  1. $S_I = S_R\cap I$
  2. If we have a solution $x+x^\theta+y^{1+\theta}\in I$, then we can find $x',y'$ such that $x'+x'^\theta+y'^{1+\theta}=0$ and $x-x',y-y'\in I$.

Proof: Assume that $S_I = S_R\cap I$, and that we have $x+x^\theta+y^{1+\theta}= i\in I$, then by the assumption there are $s,t\in I$ such that $s+s^\theta+t^{1+\theta}=i$. We can take $x' = x-s+y^\theta t-t^{1+\theta}$ and $y' = y-t$ and get $$x'+x'^\theta+y'^{1+\theta} = (x+x^\theta+y^{1+\theta})-(s+s^\theta+t^{1+\theta}) = 0\;.$$ Clearly, $x-x'$ and $y-y'$ are in $I$.

Next, assume 2. holds. The inclusion $S_I\subseteq S_R\cap I$ is immediate, so we show the other inclusion. Assume we have $i\in S_R\cap I$, i.e. we have $x,y\in R$ such that $x+x^\theta+y^{1+\theta}= i$. By the assumption, we can find $x',y'$ such that $x'+x'^\theta+y'^{1+\theta}=0$ and $x-x',y-y'\in I$. We can now set $s = x-x'+y'^\theta(y-y')$ and $t = y-y'$. Then $$s+s^\theta+t^{1+\theta} = x+x^\theta+y^{1+\theta}=i$$ and $s,t\in I$. Hence $i\in S_I$.

This seemingly easy reformulation does give an easier check, as one can usually compute $S\cap I$ and $S_I$ if a ring and involution are given.

Corollary: If the residue field $R/M$ has characteristic different from $2$, the answer to question 1 is yes.

Proof: We apply the previous with as the ring $R/M^i$ and as ideal $M$. If $x+x^\theta+y^{1+\theta}= m \in M$, then $m=m^\theta$, so we can take $s=m/2$ and $t=0$ to get $s+s^\theta+t^{1+\theta}=m$ with $s,t\in M$. Hence $S_{R/M^i}\cap M \subseteq S_M$, so we get an affirmative answer to question 1.

Lastly, we now know that to get a counterexample, we need to look at local rings for which the residue field has characteristic $2$. One can check that if we take $R = \mathbb{Z}_2$ with trivial involution $\theta$, that $(1,0,1)$, a solution in $\mathbb{Z}_2/2\mathbb{Z}_2$, is never reached as a projection from a solution in $\mathbb{Z}_2/2^k\mathbb{Z}_2$ for $k\geq2$.

Credit for the equivalency goes to one of my colleagues, who is not active on MO.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.