Problem:
Solution:
Again, the hint is holding my hand. To be frank, here is what I thought without the hints
First, the variety must be written as a set of polynomials, so if I substitute $ (0, 0) $ to the polynomials, then all the constant terms must be 0, then if I substitute $ (1, 1) $ to the polynomials, then get least one of the polynomial has a non-zero coefficients. But then I haven't go any further than that.
With the hint, it is easy.
$ g_k(x) = f_k(x, x) $ vanishes on infinitely many points, so it is the zero polynomial, so $f_k(1, 1) = g_k(1) = 0 $, so we get the desired contradiction.
The hint goes too far. It could have just hint me to use the proposition ...
Solution:
Again, the hint is holding my hand. To be frank, here is what I thought without the hints
First, the variety must be written as a set of polynomials, so if I substitute $ (0, 0) $ to the polynomials, then all the constant terms must be 0, then if I substitute $ (1, 1) $ to the polynomials, then get least one of the polynomial has a non-zero coefficients. But then I haven't go any further than that.
With the hint, it is easy.
$ g_k(x) = f_k(x, x) $ vanishes on infinitely many points, so it is the zero polynomial, so $f_k(1, 1) = g_k(1) = 0 $, so we get the desired contradiction.
The hint goes too far. It could have just hint me to use the proposition ...
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