Problem: Let $k$ be an algebraically closed field, $k[x,y,z,w]$ be polynomial ring, and $(wx-yz)$ be the ideal generated by $wx-yz$. Does $A(X):= k[x,y,z,w]/(wx-yz)$ contain any nonconstant invertible polynomial? i.e., Does there exist $\bar{h}, \bar{p} \in A(X)$ s.t. $\bar{h} \cdot \bar{p}=\bar{1}$ with $\bar{h}\neq \bar{a}$ for all $a\in k$?
A few observation: $wx-yz$ is irreducible in $k[x,y,z,w]$, and $k[x,y,z,w]$ is a UFD, so $(wx-yz)$ is a prime ideal. Hence $A(X)$ is an integral domain.
Motivation: Let $h\in k[x,y,z,w],X:=V(wx-yz)=\{(x,y,z,w):wx-yz=0\}$. My motivation is to determine if $h\mid_{X}$ is nonconstant ($h$ is viewed as a function) indicates $(h,wx-yz)$ is a proper ideal in $k[x,y,z,w]$.$(*)$
You see, if $A(X)$ contains any nonconstant invertible element $\bar{h}:=h+(wx-yz)$, then $h\mid_{X}$ would also be nonconstant as a function, and $\exists q\in k[x,y,z,w]$ s.t. $hq+(wx-yz)=1+(wx-yz)$, hence $1=hq+p(wx-yz)$ for some $p\in k[x,y,z,w]$. As a result $1\in (h,wx-yz)$, which means $(h,wx-yz)=k[x,y,z,w]$, then $(*)$ would be false.
No. Consider the restriction of $h$ to any line through the origin on $X$: this must be again be invertible, and therefore constant as we know the invertible functions on $\Bbb A^1$ are just the nonzero elements of $k$. As every point on $X$ lies on a line through the origin, we see that $h$ must in fact be constant.