Invertibility of a Polynomial map.

Question

Given following polynomial map $f:\mathbb{R}^2\to V\subset \mathbb{R}^3 $: $$ (z_1,z_2)\mapsto (2z_1-z_2, 2z_1^2-z_2^2, 2z_1^3-z_2^3) $$

Is this map a bijection? If so, how?

Answer 1

Since your map is polynomial, it cannot be bijective from $\mathbb{R}^2$ to $\mathbb{R}^3$ as it goes from dimension $2$ to dimension $3$.

In fact the image needs to be in a variety of dimension $\le 2$, which corresponds to say that the points of the image satisfy some polynomial equation.

In your case, you can easily check that $(0,0,1)$ is not in the image.

However, if you denote by $V$ the image, then yes it is bijective.

Indeed, write $(a,b,c)$ a point of the image, and working with the explicit description you cand eliminate $z_1,z_2$ in terms of the others. You find

$$z_1=\frac{a^3-c}{2(a^2-b)}, z_2=\frac{a^3-3ab+2c}{3b-3a^2}.$$ Moreover, the equation of the image is $a^6-3a^4b-8a^3c+9a^2b^2+12abc-9b^3-2c^2$.