If $k$ is a field in which $1+1\ne 0$, prove that $\sqrt{1-{x}^{2}}$ is not a rational function. Hint. Mimic the classical proof that $\sqrt{2}$ is irrational

It seems rather odd to discuss the square root function in the context of an arbitrary field, but here goes nothing!

Proof:

Suppose that $\sqrt{1-{x}^{2}}$ is a rational function in $k(x)$. Then there exist polynomials $p(x)$ and $q(x)\ne 0$ that are relatively prime with $\sqrt{1-{x}^{2}}=\frac{p(x)}{q(x)}$; clearly we may take $p(x)\ne 0$. Then $1-{x}^{2}=\frac{p(x{)}^{2}}{q(x{)}^{2}}$ or $q(x{)}^{2}(1-{x}^{2})=p(x{)}^{2}$, which says $q|{p}^{2}$, and since $q|{q}^{2}$, we can infer $q|{p}^{2}$. Moreover, since $p|{p}^{2}$ and $(p,q)=1$, then $pq|{p}^{2}$ or $q|p$, which contradicts the fact that the polynomials are relatively prime. Hence $\sqrt{1-{x}^{2}}$ cannot be a rational function.

How does this sound? Aside from the problem of discussing –$\sqrt{\text{}\text{}}$ in the context of an arbitrary field, I am worried about not using the fact that $1+1\ne 0$, at least not explicitly. Where exactly is this assumed used, if at all?

EDIT:

Suppose that $q(x)=1$, and let $f(x)={a}_{n}{x}^{n}+...{a}_{1}x+{a}_{0}$. Then $1-{x}^{2}=f(x{)}^{2}$. If $x=0$, $1={a}_{0}^{2}$ and therefore ${a}_{0}$ must be a unit. Letting $x=1$ we get $0=({a}_{n}+...+{a}_{1}+{a}_{0}{)}^{2}$; and letting $x=-1$ we get $0=(-{a}_{n}-...-{a}_{1}+{a}_{0}{)}^{2}$. Since we are working in a field, there can be no nonzero nilpotent which means that ${a}_{n}+...+{a}_{1}+{a}_{0}$ and $-{a}_{n}-...-{a}_{1}+{a}_{0}$ are both zero. Adding the two equations together yields $2{a}_{0}=0$, and since $1+1\ne 0$, ${a}_{0}=0$ which contradicts the fact that ${a}_{0}$ is a unit.

How does this sound?

Another attemtpt:

If $f(x{)}^{2}=1-{x}^{2}$, then $2=\mathrm{deg}({f}^{2})=2\mathrm{deg}(f)$ implies $\mathrm{deg}(f)=1$. Yet if $x=-1$, we get $f(-1{)}^{2}=1-1=0$ and therefore $f(-1)=0$ since there are no nonzero nilpotent elements in a field; similarly, $f(1)=0$. Since $1+1\ne 0$, $f$ has two distinct roots in $k$ yet is only a 1-st degree polynomial, a contradiction.