Problem:
Solution:
a)
The hint basically give out the answer.
Basic fact: Union or Intersection of two affine varieties is an affine variety.
Induction:
Base case is trivial, Union or Intersection of 1 affine variety is an affine variety, obviously.
Suppose the Union or Intersection of $ k $ affine varieties is an affine variety, because of the basic fact, the Union or Intersection of $ k + 1$ affine varieties is again an affine variety.
Therefore, by the principle of mathematical induction, the finite union and intersection is not an affine variety.
b)
We showed that the integer grid is not an affine variety, but it is an infinite union of a single points, which are affine varieties.
c)
We showed that the punctured line is not an affine variety, but a line and a point are both affine varieties and so that we have found an example.
d)
Just put all the governing polynomials together would work, of course, we need to rename the variables.
The proof that it works is trivial, a point that is in $ V \times W $ must be a point that satisfy all the governing polynomials.
Solution:
a)
The hint basically give out the answer.
Basic fact: Union or Intersection of two affine varieties is an affine variety.
Induction:
Base case is trivial, Union or Intersection of 1 affine variety is an affine variety, obviously.
Suppose the Union or Intersection of $ k $ affine varieties is an affine variety, because of the basic fact, the Union or Intersection of $ k + 1$ affine varieties is again an affine variety.
Therefore, by the principle of mathematical induction, the finite union and intersection is not an affine variety.
b)
We showed that the integer grid is not an affine variety, but it is an infinite union of a single points, which are affine varieties.
c)
We showed that the punctured line is not an affine variety, but a line and a point are both affine varieties and so that we have found an example.
d)
Just put all the governing polynomials together would work, of course, we need to rename the variables.
The proof that it works is trivial, a point that is in $ V \times W $ must be a point that satisfy all the governing polynomials.
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