$u^{2}-1$ is a unit in a ring $A$ Hello I have the next question that I want to prove (or one counterexample if there is one)
Let $A$ be a ring with maximal ideal $M$ and quotient field $k=A/M$ of size at least $4$ such that the natural induced map $p:A^{\ast}\rightarrow k^{\ast}$ is surjective, then there exist $u\in A^{\ast}$ such that $u^{2}-1\in A^{\ast}$
I tried this
Since $|k|\geq 4$ then there exist $x\in k^{\ast}$ with $x^{2}\not=1$. Now since $p$ is surjective, we get that $x=p(u)=\overline{u}$ with $u\in A^{\ast}$, then $\overline{u}^{2}\not=1$ in $k$. Thus $\overline{u}^{2}-1\in k^{\ast}$. From this I have that $p(a)=\overline{u}^{2}-1$ with $a\in  A^{\ast}$.
It follows that $u^{2}-1=a+m$ with $a\in  A^{\ast}$, $m\in  M$.
From this I am not sure how to continue in order to prove that $u^{2}-1\in A^{\ast}$.
I was thinking that since $a$ is unit then $(u^{2}-1)a^{-1}=1+m'$ with $m'=ma^{-1}\in M$.
Thus  $1+m'$ is a unit if and only if $u^{2}-1$ is a unit. So I need to prove that $1+m'$ is unit (That is true if $A$ is a local ring)
That's my idea if there is a suggestion or hint  I would appreciate.
Thanks!
 A: Edit: To answer your question in the comments, the result is not true in general either even if $A$ is assumed to be finite, and in fact a counterexample here is perhaps even easier to find. Let $A=\mathbb{Z}\big/10\mathbb{Z}$, and let $M=5\mathbb{Z}\big/10\mathbb{Z}$. Then $A^\times=\{\overline{1},\overline{3},\overline{7},\overline{9}\}$, and $A\big/M\cong\mathbb{Z}\big/5\mathbb{Z}$, so it is easy to verify by hand that (i) the map $A^\times\to k^\times$ is surjective, and that (ii) $u^2-1$ is not a unit for any $u\in A^\times$.
On the other hand, suppose that $A$ is any ring such that $n:=\operatorname{char} A$ is positive and not divisible by $2$ or $3$. There is a (unique) injective ring morphism $\mathbb{Z}\big/n\mathbb{Z}\hookrightarrow A$, and so to find a unit $u\in A^\times$ with $u^2-1\in A^\times$ it suffices to find one in $\left(\mathbb{Z}\big/n\mathbb{Z}\right)^\times$. Now, recall that every non-zero divisor of a finite commutative ring is a unit. In particular, the elements $\overline{2},\overline{3},\overline{4}$ are all units in $\mathbb{Z}\big/n\mathbb{Z}$, since they are coprime with $n$, and thus so is $\overline{3}^2-1=\overline{8}=\overline{2}\cdot\overline{4}$, as desired. So, put differently, by Lagrange's theorem you will never be able to find a finite counterexample where $|A|$ is not divisible by $2$ or $3$.

Here's a counterexample where both $A$ and $k$ have characteristic $0$. Let $B=\mathbb{Z}[t_n:n\in\mathbb{N}]$ be the polynomial ring in $\mathbb{N}$-many variables, where $\mathbb{N}$ denotes the non-zero natural numbers, let $S\subset B$ be the multiplicatively closed subset generated by the set $\{t_n:n\in\mathbb{N}\}$, and take  $A=B_S$ to be the localization of $B$ at $S$. If you prefer, $A$ can be considered as the subring $\mathbb{Z}[t_n,t_n^{-1}:n\in\mathbb{N}]$ of the field of fractions of $B$. Now, define the ideal $$I=\langle t_n-n:n\in\mathbb{N}\rangle<B,$$ and denote by $M=I_S$ the ideal of $A$ generated by the images of each $t_n-n\in B$ under the localization map $B\hookrightarrow A$. Since localization commutes with quotients, we have $$A\big/M=B_S\big/ I_S\cong \left(B\big/I\right)_S\cong\mathbb{Z}_\mathbb{N}\cong\mathbb{Q},$$ so indeed $M$ is a maximal ideal of $A$. Also, letting $\varphi:A\big/M\cong\mathbb{Q}$ denote this isomorphism, note that for any $m/n\in\mathbb{Q}$ with $m\neq 0$ we have $m/n=\varphi(t_mt_n^{-1})$. Since $t_mt_n^{-1}\in A^\times$, this shows that the surjectivity condition on units that you desire holds. However, by construction, every unit of $A$ is of the form $\pm t^{\varepsilon_1}_{n_1}\dots t^{\varepsilon _k}_{n_k}$ for some $\varepsilon_i\in\{\pm 1\}$ and some $n_i\in\mathbb{N}$, and so there is no unit $u\in A^\times$ with $u^2-1\in A^\times$.
