Problem:
Solution:
First, let's look at what is $ V(x - y) $, it is the point set such that $ x - y = 0 $, so it is simply the straight line $ (t, t) $.
Next, we look at what $ \langle x - y \rangle $ is, it is the set of all polynomial with a factor $ (x - y) $, so it must vanish on the straight line, so we showed $ \langle x - y \rangle \subset I(V(x - y)) $
For the reverse inclusion, we consider the polynomials in $ I(V(x - y)) $, they must vanishes on $ (t, t) $. The challenge is to show one such polynomial must have $ (x - y) $ as a factor.
In order to solve the challenge, we need this result:
We claim that any polynomial $ f(x, y) $ can be written as $ f(x, y) = h(x, y)(x - y) + r(x) $.
The prove the claim, it suffice to show that the trick works when $ f(x, y) $ are monomials. For general polynomials we just add the terms up.
So for a general monomial, we see that we can indeed express it in the claim form as follow:
$ x^a y^b = x^a(x - (x - y))^b = x^a(x^b + c(x, y)(x-y)) = x^{a+b} + x^ac(x, y)(x - y) $.
Armed with the claim, we write any polynomial $ f(x, y) \in I(V) $, $ f(x, y) = h(x, y)(x - y) + r(x) $.
It must vanishes on $ (t, t) $, so $ h(t, t)(t - t) + r(t) = 0 \implies r(t) = 0 $.
Therefore we proved any polynomial $ f(x, y) \in I(V) \implies f(x, y) \in \langle x - y \rangle $.
Here we have the previous wrong proof - just for reference for a mistake I had.
An easy fact is that such polynomial must vanish on $ (0, 0) $, so the polynomial must not have a constant term, any such polynomial can be written as
$ f(x, y) = x a(x, y) + y b(x, y) $.
Next, we substitute $ (t, t) $ and get $ a(t, t) = -b(t, t) $ whenever $ t \ne 0 $.
So we have two polynomials that agree on infinite number of points (as the field $ k $ is infinite). The two polynomials must be equal, so we have $ a(x, y) = -b(x, y) $.
Putting back in, we have $ f(x, y) = x a(x, y) - y a(x, y) = (x - y) a(x, y) $, so we have proved that any such polynomial must have $ (x - y) $ as a factor, or $ I(V(x, y)) \subset \langle x - y \rangle $.
The key mistake is in the red line. In general, two polynomials of two or more variables, even when they agree on infinite number of points, can be different.
See:
http://math.stackexchange.com/questions/623981/check-that-two-function-fx-y-and-gx-y-are-identical
This is a good story learnt.
Solution:
First, let's look at what is $ V(x - y) $, it is the point set such that $ x - y = 0 $, so it is simply the straight line $ (t, t) $.
Next, we look at what $ \langle x - y \rangle $ is, it is the set of all polynomial with a factor $ (x - y) $, so it must vanish on the straight line, so we showed $ \langle x - y \rangle \subset I(V(x - y)) $
For the reverse inclusion, we consider the polynomials in $ I(V(x - y)) $, they must vanishes on $ (t, t) $. The challenge is to show one such polynomial must have $ (x - y) $ as a factor.
In order to solve the challenge, we need this result:
We claim that any polynomial $ f(x, y) $ can be written as $ f(x, y) = h(x, y)(x - y) + r(x) $.
The prove the claim, it suffice to show that the trick works when $ f(x, y) $ are monomials. For general polynomials we just add the terms up.
So for a general monomial, we see that we can indeed express it in the claim form as follow:
$ x^a y^b = x^a(x - (x - y))^b = x^a(x^b + c(x, y)(x-y)) = x^{a+b} + x^ac(x, y)(x - y) $.
Armed with the claim, we write any polynomial $ f(x, y) \in I(V) $, $ f(x, y) = h(x, y)(x - y) + r(x) $.
It must vanishes on $ (t, t) $, so $ h(t, t)(t - t) + r(t) = 0 \implies r(t) = 0 $.
Therefore we proved any polynomial $ f(x, y) \in I(V) \implies f(x, y) \in \langle x - y \rangle $.
Here we have the previous wrong proof - just for reference for a mistake I had.
An easy fact is that such polynomial must vanish on $ (0, 0) $, so the polynomial must not have a constant term, any such polynomial can be written as
$ f(x, y) = x a(x, y) + y b(x, y) $.
Next, we substitute $ (t, t) $ and get $ a(t, t) = -b(t, t) $ whenever $ t \ne 0 $.
So we have two polynomials that agree on infinite number of points (as the field $ k $ is infinite). The two polynomials must be equal, so we have $ a(x, y) = -b(x, y) $.
Putting back in, we have $ f(x, y) = x a(x, y) - y a(x, y) = (x - y) a(x, y) $, so we have proved that any such polynomial must have $ (x - y) $ as a factor, or $ I(V(x, y)) \subset \langle x - y \rangle $.
The key mistake is in the red line. In general, two polynomials of two or more variables, even when they agree on infinite number of points, can be different.
See:
http://math.stackexchange.com/questions/623981/check-that-two-function-fx-y-and-gx-y-are-identical
This is a good story learnt.
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