square-root of a square-free element (Matsumura)

Question

Example 4, page 65 in Matsumura's Commutative Ring Theory reads as follows:

Let $A$ be a UFD in which $2$ is a unit. Let $f \in A$ be square-free (that is, not divisible by the square of any prime of $A$). Then $A[\sqrt{f}]$ is an integrally closed domain. Proof: Let $a$ be a square root of $f$..."

Question: Since $f$ is square-free, how can it can have a square root, even in the field of fractions of $A$?

Answer 1

Just look at $A[x]/(x^2-f)$. Clearly $f$ has a root in this ring -- the element $x$.

Answer 2

$-1$ is square free in $\Bbb Z$ and yet it has a square root in $\Bbb Z[i]$. The point is that the square root doesn't lie in $\Bbb Z$. The bigger the ring containing $f$ is, the more likely it will be to pick up a square root of $f$. It will be guaranteed to happen in the algebraic closure of the field of fractions.