If k is a field in which 1 + 1 &#x2260;<!-- ≠ --> 0 , prove that <msqrt>

If $k$ is a field in which $1+1\ne <!--\; \ne \; -->0$, 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 $q(x)\ne <!--\; \ne \; -->0$ that are relatively prime with $\sqrt{1-<!--\; -\; -->x^2}=\frac{p(x)}{q(x)}$; clearly we may take $p(x)\ne <!--\; \ne \; -->0$. 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 –$\sqrt{}$ in the context of an arbitrary field, I am worried about not using the fact that $1+1\ne <!--\; \ne \; -->0$, 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 $2a_0=0$, and since $1+1\ne <!--\; \ne \; -->0$, $a_0=0$ 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 $1+1\ne <!--\; \ne \; -->0$, $f$ has two distinct roots in $k$ yet is only a 1-st degree polynomial, a contradiction.