A regular function on an open set of Affine variety that can not be extended to a regular function I have a problem with Exercise 2.3.8(iv) of Brodmann and Sharp. Here is a brief introduction to this exercise.
Let $V$ be the affine variety defined by the ideal
$$\mathfrak{p}= \left( x_1^2x_2-x_3^2, x_2^3-x_4^2, x_2x_3-x_1x_4, x_1x_2^2-x_3x_4\right) \subset \mathbb{C}[x_1,x_2,x_3,x_4]$$
and $U = V \setminus \{(0,0,0,0)\}$.
Note that $U$ is a quasi-affine variety which is isomorphic to $\mathbb{A}_\mathbb{C}^2 \setminus \{(0,0)\}$, by previous parts of this exercise. Now, take the regular function $\beta_2\colon U \to \mathbb{C}$ defined by:
$$\beta_2 \left( \left( c_1,c_2, c_3, c_4 \right) \right) = \begin{cases} c_3/c_1 & \text{if } c_1 \neq 0 \\
c_4/c_2 & \text{if } c_2 \neq 0
\end{cases}
$$
The aim is to show that $\beta_2$ can not be extended to a regular function on $V$.
I would be thankful for any possible response or hint to this problem.
Many thanks,
 A: A short calculation with Macaulay 2 shows that the ideal $\mathfrak{p}$ is indeed prime and $S=R/\mathfrak{p}$ an integral domain ($R=\mathbb{C}[x_1,\ldots,x_4]$)
First let us verify that $\beta_2$ is well defined:
$$\frac{x_3}{x_1} - \frac{x_4}{x_2} = \frac{x_3 x_2 - x_1 x_4}{x_1 x_2} = 0 \text{ in } Q(S)$$
because $x_3 x_2 - x_1 x_4 \in \mathfrak{p}$.
If $\beta_2$ would be a regular function on $\mathrm{spec}(R/\mathfrak{p})$ that is $x_4 - x_2 f \in \mathfrak{p}$ for a polynomial $f \in R$, the more it would be a regular function in $R/\mathfrak{p}'$ with
$$\mathfrak{p}' = \mathfrak{p} + (x_1,x_3) = (x_1,x_3,x_2^3-x_4^2)$$
But this is essential a semicubical parabola $T=k[x,y]/(y^2-x^3)$ where $y/x \in Q(T)$ is well known not to be in $T$, instead we have that $T[y/x] \subseteq Q(T)$ is the integral closure of $T$ in $Q(T)$. ($(y/x)^2-x = 0$ and so $y/x$ is integral over $T$, that it generates the integral closure can be verified for $k=\mathbb{Q}$ with Macaulay 2).
A: @Juergen: Many thanks for your nice response. Thanks to your note, let me restate your answer, for those who have less background in algebraic geometry.
Let $\beta'_2$ be the regular function on $V$ which extends $\beta_2$ and let
\begin{align*}
& V'=\mathcal{Z} \left( x_1, x_3, x_2^3-x_4^2 \right) \subset V \\
& U'= V' \setminus \left\{(0,0,0,0) \right\} \subset U
\end{align*}
Then, $V'$ is an affine variety and $\beta'_2\lceil_{V'}$ is a regular function on $V'$ which extends $\beta_2 \lceil_{U'}$. Note however that 
$$\beta_2 (c_1, \ldots, c_4) = \frac{c_4}{c_2}, \quad \text{for all } (c_1, \ldots, c_4) \in U'.$$
On the other hand, by the isomorhism $\mathcal{O}(V) \cong \frac{\mathbb{C}[x_1, \ldots, x_4]}{\left( x_1, x_3, x_2^3-x_4^2 \right)} \cong \frac{\mathbb{C}[x,y]}{\left( y^2-x^3 \right)}$, we conclude that there exists a regular function on $V'' = \mathcal{Z} \left(y^2-x^3 \right)$, say $\gamma'$, which extends $\gamma \colon V'' \setminus \{(0,0)\} \to \mathbb{C}$, given by $\gamma (a,b) = \frac{b}{a}$, for $(a,b) \in V'' \setminus \{(0,0)\}$. But in this case, $\gamma'$ has a converse given by $\theta \colon \mathbb{A}_{\mathbb{C}}^1 \to V''$, ($a \mapsto \left( a^2, a^3\right)$). In particular, $V''$ is a smooth variety. But an easy calculation shows that $(0,0)$ is a singular point of $V''$.
