If $k$ is a field in which $1 + 1 \neq 0$, 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 $q(x) \neq 0$ that are relatively prime with $\sqrt{1-x^2} = \frac{p(x)}{q(x)}$; clearly we may take $p(x) \neq 0$. Then $1-x^2 = \frac{p(x)^2}{q(x)^2}$ or $q(x)^2(1-x^2) = p(x)^2$, which says $q^2 | p^2$, 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 $1+1 \neq 0$, 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 $x=0$, $1 = a^2_0$ 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 $1+1 \neq 0$, $a_0 = 0$ which contradicts the fact that $a_0$ is a unit.
How does this sound?
Another attemtpt:
If $f(x)^2 = 1-x^2$, then $2 = \deg(f^2) = 2 \deg (f)$ implies $\deg(f) = 1$. 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 $1+1 \neq 0$, $f$ has two distinct roots in $k$ yet is only a $1$-st degree polynomial, a contradiction.
Your proof is not quite complete, because $q|p$ is not a complete contradiction to $(p,q)=1$: we could have $q=1$. We must rule out that case, i.e., we must show that $1-x^2$ is not the square of a polynomial. That is where we will use the assumption that $1 + 1 \ne 0$.