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Look at the example 2.1.7 at page 19 of these notes (which is the same example present in Mumford's red book at page 21). The author shows that a regular function isn't a ratio of two polynomial, and in particular in the above example inside the open set $U$ we have the following regular function: $\phi(p)=\left\{\begin{array} {lll} \frac{X_1}{X_2} & \textrm{if $p\in X_2\neq 0$}\\\\ \frac{X_3}{X_4} & \textrm{if $p\in X_4\neq 0$} \end{array}\right.$

I agree with the fact that $\frac{X_1}{X_2}=\frac{X_3}{X_4}$ in $K(X)$ when $X_2\neq0$ and $X_4\neq0$, but the point is that such two functions shoud be the same in the whole $V$ to make sensible the example. Infact in in the intersection $\{X_2\neq0\}\cap\{X_4\neq0\}$ we have the regular function described by $\frac{X_1}{X_2}$ or $\frac{X_3}{X_4}$, but what about the rest of $U$?

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    But the point is that a single formula cannot work everywhere, so you shouldn't expect them to agree everywhere because they are not even defined everywhere on the open set!2012-05-25

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The situation is that we have a cone $Y\subset \mathbb A^4_k$ of equation $X_1X_4=X_2X_3$, which contains the closed plane $P\subset Y$ given by $X_2=X_4=0$.
We are interested in the complementary open subset $U=Y\setminus P\subset Y$ and its ring of regular functions $\mathcal O_Y(U)$.

The function $\phi\in \mathcal O_Y(U)$ that you describe, which has $P$ as its polar set , cannot be written $\phi=\frac{f\mid U}{g\mid U}$ with $f,g\in \mathcal O_Y(Y)$ (where $\mathcal O_Y(Y)$ consists of restrictions to $Y$ of genuine polynomials of $k[X_1,X_2,X_3,X_4]$).
And why is that?

The reason is that if we had such a description of $\phi$, then the plane $P$ would be the zero locus of the single regular function $g\in \mathcal O_Y(Y)$ i.e. would be a Cartier divisor, and this is actually not the case.