The quotient field of an unique factorisation domain is never algebraically closed . $\mathbf {The \ Problem \ is}:$ Let, $R$ be a UFD with field of fractions $K(\neq R).$ Show that $K$ can't be an algebraically closed field .
$\mathbf {My \ approach}:$ Actually, I am new in learning field extension and galois theory.
A hint is given :
Define $A = \frac{R[x]}{x^2-a}$
Then I could show $A$ is a integral domain(id) iff $a$ isn't a perfect square in $R.$
Next hint is :
If $A$ is an id with quotient field $L$, show that $L$ is a field extension of $K.$ and find $[L:K].$
This part I couldn't show .
And how the answer follows from this two hints?
A small help is warmly appreciated, thanks in advance.
 A: If $a$ is not a perfect square in $R$, it's not difficult to see that the quotient field of $A = R[x]/(x^2-a)$ is going to be the field $L = K[x]/(x^2-a)$.  Since $x^2-a$ has no roots in $K$, it is irreducible, meaning $[L : K] = 2$.
A: A field $K$ is algebraically closed if and only if every polynomial of positive degree in $K[x]$ has a root. If $K$ is the field of fractions of $R$, where $R$ is a domain that's not a field, then every polynomial in $K[x]$ has the same roots as a polynomial in $R[x]$.
If $R$ is a UFD, then the rational root test works: if a polynomial $f(x)=a_0+a_1x+\dots+a_{n-1}x^{n-1}+a_nx^n$ (with $n>0$) has a root $p/q\in K$, with $p$ and $q$ having no irreducible factor in common, then $p\mid a_0$ and $q\mid a_n$. Indeed we have
$$
a_0q^n+a_1pq^{n-1}+\dots+a_{n-1}p^{n-1}q+a_np^n
$$
so we need that $q\mid a_np^n$. But $q$ and $p^n$ have no irreducible factor in common, so by uniqueness of factorization we conclude that $q\mid a_n$. Similarly $p\mid a_0$.
Thus a root in $K$ of a monic polynomial in $R[x]$ is actually in $R$. Now just take an irreducible element $a\in R$ and consider $x^2-a$. This cannot have roots in $R$, because $a$ is not invertible, so no invertible element can be a root and, on the other hand, if $b$ is associate to $a$, say $b=au$ with $u$ invertible, then $b^2-a=0$ implies $a(au^2-1)=0$, contradicting the assumption that $a$ is not invertible.
Thus we have found a polynomial of degree $2$ in $K[x]$ that has no root in $K$.

The hint you've been given works as well, but it needs some form of the rational root test to show that $x^2-a$ is irreducible, when $a\in R$ is irreducible (basically, an adaptation of the proof of irrationality of $\sqrt{2}$).
When you've proved that $x^2-a$ is irreducible in $R[x]$, then you know that $R[x]/(x^2-a)=S$ is a domain and there's an embedding $R\to S$. Thus there is an embedding of $K$ into $L$, the field of fractions of $S$ and one can prove that $L$ is a finite extension of $K$; it is a proper extension, because $x^2-a$ has a root in $L$. Therefore $K$ is not algebraically closed.
