In this special case the quotient polynomial ring is a UFD? I would like to know if the following is true. 

Let $F$ be a field and let $p\in F[x]$ be a square-free polynomial. Then, the quotient ring $F[x,y]/\langle y^2-p\rangle$ is a UFD. 

I am not sure how to approach this question.
Thanks!  
 A: No, it's not necessarily a UFD.
Here's an example, taken from Gauss's Lemma for Number Fields.
Let $F=\mathbb{C}$, and $p(x) = x^3-3x+49$. Then let $A=\mathbb{C}[x,y]/(y^2-p)$. $A$ is a Dedekind domain, so it is a UFD if and only if it is a PID.
(I'll be a bit sloppy below and write things like $y+7$ for elements of $A$, when in fact I mean $y+7+(y^2-p)$, the class of $y+7$.)
The ring $A$ is closely related to the elliptic curve $C: y^2=x^3-3x+49$; the nonzero prime ideals of $A$ are associated with points lying on the curve $C$ by the correspondence that sends $(a,b)$ to $(x-a,y-b)$. 
The ideal $(x,7+y)$ is not principal in this ring (in fact, no power of $(x,7+y)$ is principal in the integral closure of $\mathbb{C}[x]$ in any finite extension of the ring of fractions of $A$). Therefore, $A$ is not a PID, hence not a UFD. 
(The ideal $(x,7+y)$ corresponds to the point $(0,-7)$ on the curve; the ideal $(x,7+y)^k$ is principal if and only if the order of $(0,-7)$ in the group of points of $C$ divides $k$; since only countably many points on $C$ have finite order, "most" ideals will not be principal. In fact, the point $(0,-7)$ has infinite order, so the ideal $(x,7+y)^k$ is not principal for any $k$).
You can also verify that $A$ is not a UFD directly by noting that even though $x$ divides $(y+7)(y-7)$, and $y+7$ and $y-7$ are coprime, there is no way to factor $x$ in $A$ into a product $ab$ such that $a|y+7$ and $b|y-7$; if $A$ were a UFD, then this would be possible. 
