Fulton 1.6

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 $V(X^3+X-X^2Y-Y)=V(X-i)\cup V(X+i)\cup V(X-Y).$