Problem:
Solution:
Recall the ideal of a variety is the set of all polynomials vanishes on $ V $, but what is $ V $ as a point set?
We claim $ V(x^n, y^m) = \{(0,0)\} $, this is actually pretty obvious, for this is the only solution for $ x^n = y^m = 0 $.
So the problem becomes, what is the set of all polynomials that vanish on $ \{(0,0)\} $?
For one thing, it must not have a constant term, otherwise $ \{(0,0)\} $ is not a solution. With that, we pick all the terms with $ x $ and factor out $ x $, and then all the term without $ x $ must have $ y $ and therefore factor out $ y $, we get
$ f(x, y) \in I(V(x^n, y^m)) \implies f(x, y) = a(x, y) x + b(x, y) y \in \langle x, y \rangle $, this show $ I(V(x^n, y^m)) \subset \langle x, y \rangle $.
The other inclusion is just as easy, notice if $ f(x, y) = c_1 x + c_2 y $, then $ f(0, 0) = 0 \in I(V(x^n, y^m)) $.
Therefore we proved $ I(V(x^n, y^m)) = \langle x, y \rangle $.
Solution:
Recall the ideal of a variety is the set of all polynomials vanishes on $ V $, but what is $ V $ as a point set?
We claim $ V(x^n, y^m) = \{(0,0)\} $, this is actually pretty obvious, for this is the only solution for $ x^n = y^m = 0 $.
So the problem becomes, what is the set of all polynomials that vanish on $ \{(0,0)\} $?
For one thing, it must not have a constant term, otherwise $ \{(0,0)\} $ is not a solution. With that, we pick all the terms with $ x $ and factor out $ x $, and then all the term without $ x $ must have $ y $ and therefore factor out $ y $, we get
$ f(x, y) \in I(V(x^n, y^m)) \implies f(x, y) = a(x, y) x + b(x, y) y \in \langle x, y \rangle $, this show $ I(V(x^n, y^m)) \subset \langle x, y \rangle $.
The other inclusion is just as easy, notice if $ f(x, y) = c_1 x + c_2 y $, then $ f(0, 0) = 0 \in I(V(x^n, y^m)) $.
Therefore we proved $ I(V(x^n, y^m)) = \langle x, y \rangle $.
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