# Projecting solutions of Hermitian forms over local rings

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.