Problem:
Solution:
According to my proof for exercise 6 in section 1, we have proved the following fact.
If a polynomial vanish on the whole integer grid, it must be the zero polynomial.
Now it is easy to prove this with that fact. Suppose $ \mathbf{Z}^n $ is an affine variety, it must be written as a finite collection of polynomials. All of its polynomials must vanish on the integer grid, and therefore all of them must be zero polynomials, but then $ \mathbf{Z}^n $ does not contain non-integer points yet its polynomial must also vanish there as well, so we have reached a contradiction and that $ \mathbf{Z}^n $ is not an affine variety.
This problem also give us a nice corollary, we know any single point is an affine variety. Now we show that the infinite union of affine variety need not be an affine variety. (Although finite unions are)
Solution:
According to my proof for exercise 6 in section 1, we have proved the following fact.
If a polynomial vanish on the whole integer grid, it must be the zero polynomial.
Now it is easy to prove this with that fact. Suppose $ \mathbf{Z}^n $ is an affine variety, it must be written as a finite collection of polynomials. All of its polynomials must vanish on the integer grid, and therefore all of them must be zero polynomials, but then $ \mathbf{Z}^n $ does not contain non-integer points yet its polynomial must also vanish there as well, so we have reached a contradiction and that $ \mathbf{Z}^n $ is not an affine variety.
This problem also give us a nice corollary, we know any single point is an affine variety. Now we show that the infinite union of affine variety need not be an affine variety. (Although finite unions are)
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