Ideal of $(u^3,u^2v,uv^2,v^3)$ Let $k$ be a algebraically closed field, consider $f: \mathbb A_k^2\rightarrow \mathbb A_k^4$ given by $f(u,v)=(u^3,u^2v,uv^2,v^3)$. Let $X=f(\mathbb A_k^2)$. Then how to determine $I(X)$?
If I let $s=u,t=v/u$, then it becomes $(s^3,s^3t,s^3t^2,s^3t^3)$, looking at this form, I am quite confident $I(X)$ must be $(x_2^2-x_1x_3, x^3_2-x_1^2x_4)$, but I don't know how to prove it, help please.
 A: Let $X,Y,Z,W$ be the four coordinates on $A^4$. Then I claim that the ideal of your variety, call it $C$, is given by $J = (XW - YZ, Y^2 - XZ, Z^2 - YW)$. 
Honestly I'm not sure how to check this without a bit of geometric machinery - I'm sure it is possible to do it just on the level of equation manipulation though.
(It makes more sense to me to think of this projectively, but I guess it doesn't matter too much.)
Outline:


*

*You can check that these polynomials vanish on the image of your map.

*The ideal $J$ is prime. (You can choose judicious patches, meaning inverting some of the elements $X$, $Y$, $Z$, $W$ to check this. After some simplification you should get just a localization of the polynomial ring in two variables. Really this tells us that the corresponding projective variety is integral, and a little work will show that this implies its affine cone is as well.) (Looking at some of the other answers - there are easier ways to prove that this ideal is prime. For example, what Mariano did in one of the links.)

*The vanishing of $J$ is 2 dimensional. (You can use the previous computation.)

*So $V(J)$ is a closed, irreducible algebraic set that contains $C$, and is of the same dimension. So in fact $V(J) = C$. (Why is C Zariski closed? This is where it is more natural for me to think of it projectively. But it is still true.) 


BTW The projectivization of this variety, thought of as a curve in $\mathbb{P}^3$ is called the twisted cubic.
A: It is not hard to check that the image of $f$ is contained in the variety $Z(x_1x_3 - x_2^2, x_2x_4 - x_3^2, x_1x_4 - x_2x_3)$, so the real difficulty lies in showing that any point on this variety is contained in the image of $f$.
Let's take a point $a = (a_1, a_2, a_3, a_4)\in \Bbb A^4_k$ on the variety. This means that the relations above hold for the coordinates of $a$. We want to show that there are $u, v$ such that $a = f(u, v)$. If $a$ is the origin, then $u = v = 0$ works. For the following, assume that $a$ is not the origin.
If we write up the matrix
$$
M = \begin{pmatrix}a_1&a_2&a_3\\a_2&a_3&a_4\end{pmatrix}
$$
then we see that this matrix has rank $1$: it cannot have rank $0$ since we assumed that $a$ had components that were non-zero, and the relations show exactly that every $2\times 2$ minor vanishes, so the rank cannot be $2$.
(Note that this matrix is tailored to the given relations. This is not a "magic matrix" that solves all such problems, and there are certainly problems that cannot be solved using this approach. However, whenever you have several second-degree relations, I would suggest looking for a matrix which happens to have minors that coincide with the relations.)
From linear algebra, this tells us that there are two non-zero vectors $b = (b_1, b_2)\in k^2$ and $c = (c_1, c_2, c_3)\in k^3$ such that $M = b^Tc$, in other words
$$
M = \begin{pmatrix}b_1c_1&b_1c_2&b_1c_3\\b_2c_1&b_2c_2&b_2c_3\end{pmatrix}
$$If either $b_1$ or $b_2$ is zero, then that means that an entire row of $M$ is zero, which means that $a$ has only one of $a_1$ or $a_4$ as non-zero component. In either case let $u$ or $v$ be a cube root of the respective $a_i$, and we're done.
From now on, assume that neither $b_1$ nor $b_2$ is zero. The rest of the proof consists of showing that we may take $b = (u, v)$ and $c = (u^2, uv, v^2)$.
By comparing the two forms of $M$ above, we see that if one $c_i$ is zero, then all the components of $c$ are zero, and that cannot be the case (again since we assumed that $a$ is not the origin). Therefore $c$ cannot have any zero components.
Since $k$ is a field, we may scale $b$ by any non-zero factor we want, as long as we scale $c$ by the inverse of that factor. The trick is to find the right factor.
$k$ is algebraically closed, so $b_1$ and $c_1$ have cube roots. Pick one for each, and name the chosen cube root of $b_1$ $\beta$, and $\gamma$ for $c_1$. Now scale $b$ by $\frac{\gamma}{\beta^2}$ and $c$ by $\frac{\beta^2}{\gamma}$. $M$ remains unchanged, while the first element of this new $b$ is $\beta\gamma$ and the first element of $c$ is $\beta^2\gamma^2$. We take $\beta\gamma$ as our $u$.
Finally, we know that $b_1c_2 = b_2c_1$ and $b_1c_3 = b_2c_2$ from the two expressions for $M$ above, which means that $c$ truly is of the form $(b_1^2, b_1b_2, b_2^2)$, and we are finished.
