I feel a little confused about varieties.

If I am asked to determine whether V(cos(x),sin(x)) seen in R1 is a variety, would it be true to just say it is?, because the polynomial 0=1 has no solution (since we know that cos(x) and sin(x) don't coincide at their zeros), so V(cos(x),sin(x))=R.

But if I am now being asked to determine whether V((cos(x),sin(x))) seen in R2 is a variety (if this makes sense) what can I say?. I think it should not be a variety but when trying to see how may I integrate polynomials into my thinking, I get confused. What reasoning should I be using for this case?

  • $\begingroup$ What is your definition of a variety? $\endgroup$ – Youngsu Sep 17 '18 at 2:00
  • $\begingroup$ when T is a finite set of polynomials in R=K[x1,...xn], V(T) is equal to the set of n-tuples (a1,...,an) which make all the polynomials of T equal to 0. In this case, I believe we need to see if there's a match between the values for (cosx,sinx) in R2 and the zeros of a finite set of polynomials in R=K[x]. but I guess that I feel a little confused about it as well. $\endgroup$ – Stiven G Sep 17 '18 at 2:46

I think it would be clearer if you change the variable name.

$$ (\cos t , \sin t) $$

So that you get

$$ x=\cos t\\ y=\sin t\\ x^2+y^2-1=0 $$

The t coordinate is not important for determining if your result is cut out by polynomials in the target space. It is the just the parameterization, but this question only depends on the underlying set. (Birationality is a separate question)

If it is the graph of a polynomial function $f$. You get the set of points of the form

$$ (x,f(x)) $$

$y-f(x)$ gives the polynomial in two variables you need to cut this out.


For the first question, you have the right spirit but your reasoning is wrong: what you've called $V(\cos(x),\sin(x))$ is $\varnothing$, not $\Bbb R$ (for exactly the reason you describe: it is the variety of any nonzero constant polynomial, e.g. $V(1)$).

The second question seems to be the following: Consider the set $C=\{(\cos(x),\sin(x))\in\Bbb R^2: x\in\Bbb R\}$; can this be realized as a variety. The answer is yes, which you will see immediately if you draw the graph.

Note that the polynomial(s) you are looking for will necessarily be polynomials in two variables (say, in $y$ and $z$). Clearly, if the graph satisfies the equation $f(x,y)=c$ for some polynomial $f$, then this is still a variety because the modified polynomial $\tilde f(x,y)=f(x,y)-c$ vanishes there.


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