Field $K (x)$ of rational functions with coefficients from $K$, if $f\in K(x)$, then $f^2 \neq x^2-1$

Question

I'm in the process of studying for an exam and I came across the following question:

Prove that if $K$ is a field and $K (x)$ is the field of rational functions with coefficients from $K$, if $f\in K(x)$, then $f^2 \neq x^2-1$.

In trying to solve this I think I came up with a contradiction.

The counterexample I came up with is the following: let $K = \Bbb Z_2$ then if $f= x+1$, $f^2=(x+1)^2$, so in $\Bbb Z_2[x], (x+1)^2= x^2 +2x+1= x^2+1= x^2-1$.

Any help with why my counterexample does not work or how to prove this would be much appreciated.

Edit: fixed a typo above, $f$ should be from the field of polynomials.

Answer 1

Set $f=\frac{Q}{P}$ where $P,Q\in\mathbb{K}[x]$ and $\gcd(Q,P)=1$. Assuming that $f^2=(x-1)(x+1)$ we get : $$Q^2=(x-1)(x+1)P^2 \qquad (*).$$ Let us work in $\mathbb{K}[x]$. You see that the prime $(x-1)$ divides $Q^2$. So $x-1$ divides $Q$ but not $P$ $(\gcd(Q,P)=1)$. Writing $Q=(x-1)Q_1$, and using $(*)$ we have : $$(x-1)Q_1^2=(x+1)P^2.$$ This leads to a condradiction if $\mathrm{Char}(\mathbb{K}) \neq 2$, since $x-1$ divides neither $P^2$ neither $x+1$. Therefore $f^2\neq (x-1)(x+1)$.

Generally, if $p$ is a prime and $b$ is a square then $\max\{k \mid p^k \text{ divides } b\}$ is even.

Answer 2

In the $char(K) \neq 2$-case, this is literally the same question as asking to show $\sqrt{6} \notin \mathbb Q$. In an UFD an element is a square in the fraction field if and only if all exponents appearing in the unique factorization are even. The critical argument is that an UFD is integrally closed in its fraction field.