# Rational quartic curve in $\mathbb P^3$

By using similar arguments to the ones from my answer to this question, I can prove that the homogeneous coordinate ring of the rational quartic curve in $\mathbb P^3$, that is, $$R = K[x_1, x_2, x_3, x_4]/\left< x_1^2x_3-x_2^3,x_1x_3^2-x_2^2x_4,x_2x_4^2-x_3^3,x_1x_4-x_2 x_3\right>,$$ is isomorphic to $K[s^4,s^3t,st^3,t^4]$. (This is also Exercise 18.8 in Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry.)

I'm interested in geometric arguments which I expect to be simpler than the algebraic ones. References are also welcome.

Edit. There is some connection with this topic where it is proved that $R$ is an integral domain.

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I am afraid there is no simple geometric argument. – user18119 Jan 21 '13 at 23:16
The problem is at some point we have to do direct computations with the algebras. The only thing which is non-trivial is to show $R$ is integral (or enough it is reduced). By localizations or using $Proj(R)$, we can see that the only prime $\mathfrak p$ of $R$ where $R_{\mathfrak p}$ could be non-reduced is the maximal ideal generated by $x_1,\dots, x_4$. This maximal ideal is completely out of control in $Proj(R)$. So I don't have a purely geometric proof. – user18119 Jan 23 '13 at 22:52