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 $i\geq1$.



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



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> **Question 2** Find a local ring $R$ such that $\pi$ is not surjective.
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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 $h\big((x\_1,y\_1,z\_1), (x\_2,y\_2,z\_2)\big) = x\_1z\_2^\theta+z\_1x\_2^\theta+y\_1y\_2^\theta$.



Projecting solutions of Hermitian forms over local rings
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:



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




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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.




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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.


