(a) Show that $V_f$ is a Zariski open set in $V$ (called a principal open set in $V$).
(b) Let $J$ be the ideal in $k[x_1,...,x_n,x_{n+1}]$ generated by $\mathcal{I}(V)$ and $x_{n+1}f-1$, and let $W = \mathcal{Z}(J) ⊆ \mathbb{A}^{n+1}$. Show that $J = \mathcal{I}(W)$ and that the map $π : \mathbb{A}^{n+1}→\mathbb{A}^n$ by projection onto the first $n$ coordinates is a Zariski continuous bijection from $W$ to $V_f$.
(c) If $U$ is any open set in $V$ show that $U=V_{f_1}∪...∪V_{f_m}$ for some $f_1,...,f_m∈k[V]$ (so that the principal open sets form a basis for the Zariski topology).
25. Suppose $f(x)=x^3+ax^2+bx+c$ is an irreducible cubic over a field $k$ of characteristic $≠2$, and let $D$ be its discriminant. Let $I=(x+y+z+a,xy+xz+yz-b,xyz+c)⊆k[x,y,z]$.
(a) Show that $I$ is a prime ideal if and only if $D$ is not a square in $k$, and that in this case, $I$ is actually a maximal ideal and $k[x,y,z]/I$ is the splitting field over $k$ for $f(x)$.
(b) If $D=r^2$ is a square over $k$, then show that if $Q_{\pm}=I+((x-y)(x-z)(y-z) \pm r)$ then $I=Q_- ∩ Q_+$ is a primary decomposition for $I$ and both $k[x,y,z]$ modulo $Q_-$ or $Q_+$ is the splitting field over $k$ for $f(x)$.
48. Show that $\mathcal{Z}(x^3-xyz+z^2)$ is the smallest algebraic set in $ℝ^3$ containing the points $S=\{(st,s+t,s^2t)~|~s,t∈ℝ\}$.
Proof: (21)(a) We see that $V_f$ is the complement in $V$ of $\mathcal{Z}(f)$, rendering $V_f$ an open set in the Zariski topology on $V$.
(b) It is clear that $W=\{(a,\dfrac{1}{f(a)})~|~a∈V_f\}$ by construction. Now to show $\mathcal{I}(W)=J$ we must show that if $$g = g_0(x_1,...,x_n) + x_{n+1} g_1(x_1,...,x_n) + ... + x_{n+1}^m g_m(x_1,...,x_n)$$ is zero on all of $W$ then $g≡0 \mod J$. Note that multiplication by a unit biconditionally preserves the property of being non-zero, and since $fx_n≡1 \mod J$, this is equivalent to showing $$gf^m ≡ f^m g_0(x_1,...,x_n) + f^{m-1} g_1(x_1,...,x_n) + ... + g_m(x_1,...,x_n)$$ is zero modulo $J$. Since this latter is a polynomial in variables $x_1,...,x_n$ that is zero on the nonempty open set $V_f⊆V$ hence on all of $V$ (cf. exercise 11), we see it can be written as a $k[x_1,...,x_n]$-linear combination of the generators of $\mathcal{I}(V)$, hence particularly is zero modulo $J$.
(c) If $U=V-\mathcal{Z}(f_1,...,f_m)$ for some $f_1,...,f_m∈k[V]$, then $$U=V-(\mathcal{Z}(f_1)∩...∩\mathcal{Z}(f_m))=$$$$(V-\mathcal{Z}(f_1))∪...∪(V-\mathcal{Z}(f_m))=V_{f_1}∪...∪V_{f_m}$$ as desired.
(25)(a+b) With some preliminary canceling of leading terms, we rewrite $I$ in the following form: $$I=(z+y+x+a,y^2+xy+ay+x^2+ax+b,x^3+ax^2+by+c)$$ At this point, we may observe $k[x,y,z]/I$ is generated as a vector space over $k$ by $1,x,x^2,y,yx,yx^2$, hence is of rank at most $6$. As well, if $φ : k[x,y,z]→K$ is the (surjective) ring homomorphism to the splitting field $K$ of $f(x)$ over $k$ by sending variables to roots of $f(x)$, it is clear $I⊆\text{ker } φ$ so that there is a factored (surjective) morphism $\overline{φ} : k[x,y,z]/I→K$. Since $K$ is a degree $6$ extension of $k$ when $D$ is not a square, it follows that in this case $\overline{φ}$ is an isomorphism so that $I$ is maximal and $k[x,y,z]/I$ is a splitting field.
Suppose $D=r^2$. Notice that one of $Q_{\pm}$, say $Q_+$, is contained in $\text{ker }φ$, and that $I⊆Q_-∩Q_+⊂Q_+$ so that we have a series of factoring surjective morphisms $k[x,y,z]→k[x,y,z]/I→k[x,y,z]/Q_+→K$. As $D$ is square, $K$ is of degree $3$ over $k$, so that necessarily $k[x,y,z]/Q_+$ is of rank $3$, implying $Q_+=\text{ker }φ$ and is thus a maximal ideal with $k[x,y,z]/Q_+≅K$. Since $Q_-$ is the image of $Q_+$ under the automorphism on $k[x,y,z]$ swapping $x$ and $y$, it follows $Q_-$ is also maximal and $k[x,y,z]/Q_-≅K$ as well. As for $I$, it is an easy fact to verify that for $k$-algebras satisfying finite generation as a vector space over $k$, being an integral domain is equivalent to being a field (consider linear dependence over powers of an arbitrary element in order to obtain an inverse), so that $I$ not being maximal ensures $I$ is not prime in this case.
(48) The most routine way to prove this would be to show $(x^3-xyz+z^2)$ is the kernel of the morphism $\overline{φ} : ℝ[x,y,z] → ℝ[s,t]$ given by $\overline{φ}(x)=st$, $\overline{φ}(y)=s+t$, $\overline{φ}(z)=s^2t$ by the method of Proposition 8. But instead, we shall show it more directly without the use of Grobner bases. As can be easily checked, $(x^3-xyz+z^2)⊆\text{ker }\overline{φ}$. Now, suppose $g∈\text{ker }\overline{φ}$ may be written in the form $g=f(x,y)+z·g(x,y)$ for at least one of $f(x,y), g(x,y)$ nonzero. That is to say, $f(xy,x+y)+x^2y·g(xy,x+y)=0$. Taken modulo $(x)$, this means $f(xy,x+y)≡f(0,y)≡f(x,y)≡0 \mod (x)$, implying $x$ is a factor of $f(x,y)$. This implies the existence of a nontrivial solution to the equation $f'(xy,x+y)+y·g(xy,x+y)=0$. Taken modulo $y$ this shows $y$ is a factor of $f'(y,x)$, i.e. again $x$ is a factor of $f'(x,y)$. Continuing in this fashion, we may perform infinite descent on the $x$-factor multiplicity of $f(x,y)$ and $g(x,y)$ which is impossible unless both of them are zero.$~\square$
