0

What is a good way to determine whether a subset of $ \mathbb{C}^2$ is an algebraic set?

For example, I want to determine this for the following cases:
- $\{(t^2,t^3)\}$
- $\{(t,\sin t)\}$
- $\{(\cos t,\sin t)\}$
- $\{(e^t, \sin t)\}$
- $\{(e^t+e^{-t}, e^t-e^{-t})\}$
- $\{(e^{2t},e^{3t})\}$
for all $t\in\mathbb{C}$.

I thought a way to do this is to look for functions $\mathbb{C}^2\rightarrow\mathbb{C}$ that send all elements of a subset to zero (if they exist of course). Is this a good way and if so, what would these functions be in the cases above?

Edit: I don't want to use topological dimensions.

1 Answers1

0

I believe that looking for annihilating polynomials would be the best option, at least for simple cases. For example, the parametrized curve $(x,y)=\left(t^2,t^3\right)$ is the zero set of $x^3-y^2$. For the curve $(x,y)=\big(\cos(t),\sin(t)\big)$, you can take the polynomial $x^2+y^2-1$. For the curve $(x,y)=\big(\exp(t)+\exp(-t),\exp(t)-\exp(-t)\big)$, you can take $x^2-y^2-4$.

On the other hand, $(x,y)=\big(\exp(2t),\exp(3t)\big)$ is not an affine variety unless the point $(x,y)=(0,0)$ is added. This is because the annihilating ideal of $(x,y)=\big(\exp(2t),\exp(3t)\big)$ is generated by $x^3-y^2$.

For $(x,y)=\big(t,\sin(t)\big)$, you can show that, if $f(x,y)$ is an annihilating polynomial of this curve, then $y$ divides $f(x,y)$, whence the curve must contain all points $(x,y)=(t,0)$, which is false. Thus, this curve is not an affine variety.

For $(x,y)=\big(\exp(t),\sin(t)\big)$, you may employ the same argument as that of $(x,y)=\big(t,\sin(t)\big)$. This curve is not an affine variety either.

Batominovski
  • 49,629