Is $\mathbb{T}$ a $1$-dim subvariety of $\mathbb{R}^3$? Well, the exercise ask me to look if $\mathbb{T} = \{ x \in \mathbb{R}^3 : 4x_1^3 = 27x_3^2 , x_2 = {0} \}$ is a $1$-dim sub variety of $\mathbb{R}^3$.
Well the thing I tried is to prove that it is by one of the definition of sub variety, using:


*

*First I create the function $F : \mathbb{R}^3\to \mathbb{R}^2$ that is $F(x) = (4x_1^3- 27x_3^2, x_2)$

*Then I prove that $F^{-1}(0) = \mathbb{T}$ and also I calculate $\mathrm{DF}(x) = \begin{equation}
\begin{pmatrix}12x_1^2 & 0 & -54x_3\\0 & 1 & 0\end{pmatrix}\end{equation}$

*Finally I look if the rank of $\mathrm{DF}(x)$ is $2,$ but I noticed that the rank is $2$ if and only if $x_1$ or $x_3$ aren't $0$, but the point $(0, 0, 0) \in \mathbb{T}$ so the rank of $\mathrm{DF}(x)$ in $\mathbb{T}$ is $1$

*Finally like the rank of $\mathrm{DF}(x)$ in $\mathbb{T}$ is $1$ we can affirm that it isn't a $1$-dim sub variety of $\mathbb{R}^3$.


I don't know if it is okay or not.
 A: I solved the problem by reducing it to a single equation $x^3 = y^2$ in $\mathbb{R}^2$ as follows.
Since $x_2 = 0$, we can reduce the problem to the curve $4x^3 = 27y^2$ in $\mathbb{R}^2$. Then, we can make a change of coordinates: $x \gets \sqrt[3]4x, y \gets \sqrt{27}y$ so what we are now looking at is $x^3 = y^2$.
This is a curve (albeit a singular curve) in $\mathbb{R}^2$ with a simple cusp at $(0,0)$. Now it depends on how you define "dimension" how you proceed. One definition is that it is the dimension of the tangent space at any non-singular point. and you can see that other than at $(0,0)$ the tangent space is $1$-dimensional.
You can also parameterize the curve as follows. Let $y = tx$ be a line of slope $t$ passing through $(0,0)$. This line intersects the curve at exactly one other point: $x^3 = t^2x^2 \implies x = t^2$. So we have $x = t^2$ and $y = tx = t^3$. So the curve is $t \mapsto (t^2, t^3)$. You can use this parameterization to compute that $\mathbb{T}$ is $1$-dimensional as well.
