# Checking the ring k[x,y,u,v]/(xy-uv) is a normal domain

Let $A= k[x,y,u,v]/(xy=uv)$, is this ring normal (integral closed in its field of fraction)?

Edit (see comments below): Because $p$ is of height 1, then there exists one among $x,y,u,v,$ say $x$ such that $f(x)\notin p$ for any $f(x)\in k[x]$. Otherwise, if there are polynomials in one variable for each of the variables $x,y,u,v,$ contained in $p$, then $p$ is of height $>1$. Thus $f(x)$ becomes invertible in $A_p$ for any $f(x)\in k[x]$.