Let $D$ be a UFD, let $F$ be the field of fractions of $D$, let $a \in D$ be such that $x^2 \ne a, \forall x \in D$. Then is it true that $x^2\ne a ,\forall x \in F$ ?
(This problem is motivated from the fact that if square root of an integer is not an integer, then the square root is not rational. I know that in the problem, if we drop the UFD condition on $D$ then it is not valid, for example $D=\{f(x) \in \mathbb Z[x] : x^2 | f(x)-f(0) \}$ and $a=x^2$.)
I think it follows from Gauss Lemma. $x^2-a$ is irreducible in $D$ and therefore it must be irreducible in $F$. In particular, it can't have any root.
Yes, it's true. To see this, let $p$ an irreducible element in $D$. Consider the valuation $v_p$ on $F$. If we have $x=y^2$ in $F$, then $$v_p(x)=2v_p(y),$$ so every irreducible factor appears with an even exponent. As units can be incorporated into irreducible elements, this proves that, if $x$ is a square in $F$, it is a square in $D$.
The standard proof of the Rational Root Test works in any UFD, so any root of $\,x^2-a\,$ in the fraction field of $D$ must be in $D,\,$ exactly as in the classical case $\,D = \Bbb Z$.
