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!


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?


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.

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    $\begingroup$ One doesn't "discuss $\sqrt{\phantom{0} }$", what one does is show that $f(x)^2=1-x$ is insoluble in the rational function field $k(x)$. In characteristic two, one has $1-x^2=1+x^2=(1+x)^2$. In your argument you say $g^2\mid f^2$, but you have not defined $f$ or $g$. $\endgroup$ – Lord Shark the Unknown Oct 1 '17 at 15:48
  • $\begingroup$ @LordSharktheUnknown Sorry about mixing up the symbols. I used $f$ and $g$ when I worked out a solution by hand. Let me fix that now. Aside from that, is my proof correct? $\endgroup$ – user193319 Oct 1 '17 at 16:46

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 $1 + 1 \ne 0$.

  • $\begingroup$ I edited my post; perhaps you could take a look when you get a chance. $\endgroup$ – user193319 Oct 3 '17 at 16:25
  • $\begingroup$ Your argument is not quite right because when you plug in -1, the signs alternate rather than all become minus. An easier argument is to use unique factorization in $k[x]$. $\endgroup$ – Ted Oct 4 '17 at 4:07
  • $\begingroup$ I gave it another shot, although I wasn't able to see how to use the unique factorization in $k[x]$, and edited my post. Perhaps you could spell this factorization idea, if you have a moment. $\endgroup$ – user193319 Oct 4 '17 at 19:06
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    $\begingroup$ Now your argument is correct. The argument using unique factorization is to note that if $1-x^2 = p(x)^2$ then that's the factorization of $1-x^2$ into irreducibles, but in fact $1-x^2 = (1-x)(1+x)$ and the right side consists of 2 distinct factors if $1+1 \ne 0$. $\endgroup$ – Ted Oct 5 '17 at 5:57

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