Dimension of polynomially parameterized set of points

Question

I'm seeking a general method for determining the dimension of a set of points that have polynomial parameterizations. Any information about what this type of problem would be called would be helpful, any resources, or a clear method for the problem below. I have very little background in this area of mathematics.

---

Example Problem:

Suppose we have the parameterized functions \begin{equation} x(a,b,c,d) = ac\end{equation} \begin{equation} y(a,b,c,d) = bc + ad \end{equation} \begin{equation} z(a,b,c,d) = bd \end{equation}

What is the intrinsic dimension of the set \begin{equation} M = \{ (x,y,z) \in \mathbb{R}^3 : (a,b,c,d) \in \mathbb{R}^4\}? \end{equation}

Clearly it is not all of $\mathbb{R}^3$ since points are missing from $M$, such as $(1,0,1)$. But could it be a volume, surface or just a curve?

---

Possible method of solution: It was suggested be a friend that maximizing the rank of the Jacobian matrix where the parameterization is smooth gives the dimension of $M$.

The Jacobian is \begin{equation} J = \begin{pmatrix} c&0&a&0 \\ d&c&b&a \\ 0&d&0&b \end{pmatrix}. \end{equation} Since the three rows here are linearly independent in general, one can conclude that $\dim(M) = 3$. Is this a valid method to determine the dimension of $M$? If so, I could imagine that this becomes increasingly difficult for larger and more complicated systems. References on general methods would be greatly appreciated.

Answer 1

The algebraic-geometry perspective on this is that you have a morphism of varieties $X\to Y$ and you're trying to find the equations of (the closure of) the image. The general way to do this is to write $x=ac$, $y=bc+ad$, and $z=bd$ and then eliminate $a,b,c,d$ from these equations using elimination theory.

If you're instead focusing on more of the real-valued aspect of the problem, you can analyze this via semialgebraic geometry. A good starting point to become familiar with this (depending on your level) is Coste's notes.

In this case, there's a no-tech way to approach the problem: the equations are simple enough that we can just solve for $a,b,c,d$ on a suitable open set. When $x\neq 0$, we can set $a=1$, $b=\frac{y\pm\sqrt{y^2-4xz}}{2x}$, $c=x$, and $d=\frac{y\mp\sqrt{y^2-4xz}}{2}$ which shows that image of our map contains the set $\{(x,y,z)\in \Bbb R^3\mid x\neq 0, y^2-4xz>0\}$, which is open in the usual topology. So the dimension of the image is three (we're glossing over a bunch of different ways to define definition and their possible equivalence here, hope you don't mind).

---

I wrote the above answer before your edit discussing the Jacobian loaded. If you want to use the Jacobian to answer this problem, you may refer to the answers here which discuss how locally in the standard topology, a map which has Jacobian of rank $k$ in a neighborhood of a point locally looks like a coordinate projection, and the image has dimension $k$. So this is indeed a valid way to solve your problem, as we again speed past the correct notion of dimension to apply here.