Fulton 1.6

Posted on November 15, 2017

1.30. Let k=\mathbb{R}.

(a) Show that I(V(X^2+Y^2+1))=(1).

Observe a^2+b^2+1\ge 1 for all a,b\in\mathbb{R}. Thus I(V(X^2+Y^2+1))=I(\emptyset) =(1).

(b) Show every algebraic subset of \mathbb{A}^2(\mathbb{R}) is equal to V(F) for some F\in\mathbb{R}[X,Y].

By Corollary 2, it suffices to show points are V(F) for some F\in\mathbb{R}[X,Y]. Let (a_1,a_2)\in\mathbb{A}^2(\mathbb{R}). Define F=(X-a_1)^2+(Y-a_2)^2. Observe that V(F)=\{(a_1,a_2)\}.

1.31. (a) Find the irreducible components of V(Y^2-XY-X^2Y+X^3) in \mathbb{A}^2(\mathbb{R}) and \mathbb{A}^2(\mathbb{C}).

Observe that Y^2-XY-X^2Y+X^3=(Y-X)(Y-X^2). So V(Y^2-XY-X^2Y+X^3)=V(Y-X)\cup V(Y-X^2). Since Y-X and Y-X^2 are irreducible over both \mathbb{R} and \mathbb{C}, these are irreducible components

(b) Do the same for V(Y^2-X(X^2-1)), and for V(X^3+X-X^2Y-Y).

In K[X] observe that X(X^2-1)\in (X) and X(X^2-1)\not\in (X^2). Therefore Y^2-X(X^2-1) is irreducible in K[X][Y] by Eisenstein’s criterion. Since Y^2-X(X^2-1) has infinitely many solutions over both the real and complex numbers, V(Y^2-X(X^2-1)) is itself irreducible in \mathbb{A}^2(\mathbb{R}) and \mathbb{A}^2(\mathbb{C}).

Observe that X^3+X-X^2Y-Y = (X^2+1)(X-Y). Note that X^2+1 has no solutions in \mathbb{R}. Thus in \mathbb{A}^2(\mathbb{R}) we have V(X^3+X-X^2Y-Y)=V(X-Y), which is irreducible. In \mathbb{C} we have X^3+X-X^2Y-Y=(X-i)(X+i)(X-Y) and V(X^3+X-X^2Y-Y)=V(X-i)\cup V(X+i)\cup V(X-Y).