Fulton 1.7

1.32. Show that both the weak and strong Nullstellensatz and all of the corollaries are false if $k$ is not algebraically closed.

For the Weak Nullstellensatz, observe $(X^2+1)\subset\mathbb{R}[X]$ is a proper ideal but $V(X^2+1)=\emptyset.$

For the Strong Nullstellensatz, consider Problem 1.26. Let $F=Y^2+X^2(X-1)^2\in\mathbb{R}[X,Y].$ Note $V(F)=\{(0,0),(1,0)\}.$ Thus $I(V(F))=(X(X-1), Y)\ne\text{Rad}{(F)}=F.$ This is also a counterexample to Corollary 1.

For Corollaries 2 and 3 we again consider Problem 1.26. Observe $F=Y^2+X^2(X-1)^2$ is irreducible, so $(F)$ is prime. But $V(F)=V(X, Y)\cup V(X-1,Y).$ Let $G=X^2+1.$ Additionally if $G=X^2+1,$ then $(G)$ is a maximal ideal but $V(G)=\emptyset.$

For Corollary 4 let $I=(Y^2+X(X-1)^2).$ Above we saw $V(I)=\{(0,0),(1,0)\}.$ However, $k[X,Y]/I$ has the infinite linearly independent subset $\{\overline{1},\overline{Y},\overline{Y}^2,\ldots\}.$

1.33 (a) Decompose $V(X^2+Y^2-1, X^2-Z^2-1)\subset\mathbb{A}^3(\mathbb{C})$ into irreducible components.

Observe that $(X^2+Y^2-1) - (X^2-Z^2+1) = Y^2+Z^2.$ Therefore $(Y+iZ)(Y-iZ)\in (X^2+Y^2+1, X^2-Z^2-1).$ If $Y+iZ=X^2+Y^2-1=0$ we have $Y=-iZ.$ So $X^2-Z^2-1=0.$ The same is true if $Y-iZ=X^2+Y^2-1=0.$ Therefore $V(X^2+Y^2-1, X^2-Z^2-1)=V(X^2+Y^2-1, Y+iZ)\cup V(X^2+Y^2, Y-iZ).$ Observe $\begin{aligned}[t] \mathbb{C}[X,Y,Z]/(X^2+Y^2-1, X+iZ) &\cong (\mathbb{C}[X,Z]/(X+iZ))[Y]/(X^2+Y^2-1)\\ &\cong \mathbb{C}[X,Y]/(X^2+Y^2-1). \end{aligned}$ We may observe that $X^2+Y^2-1$ is irreducible over $\mathbb{C}[X][Y].$ Therefore $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ is an integral domain and $(X^2+Y^2-1, X+iZ)$ is a prime ideal. So $V(X^2+Y^2-1, X+iZ)$ is irreducible. A similar argument follows for $V(X^2+Y^2-1, X-iZ).$

(b) Let $V=\{(t,t^2,t^3)\in\mathbb{A}^3(\mathbb{C}) \ | \ t\in\mathbb{C}\}.$ Find $I(V),$ and show that $V$ is irreducible.

Note that $V=V(Y-X^2, Z-X^3).$ We will show that $(Y-X^2, Z-X^3)$ is a prime ideal by showing that $\mathbb{C}[X,Y,Z]/(Y-X^2, Z-X^3)$ is an integral domain. Observe $\begin{aligned}[t] \mathbb{C}[X,Y,Z]/(Y-X^2, Z-X^3) &\cong (\mathbb{C}[X,Y]/(Y-X^2))[Z]/(Z-X^3)\\ &\cong \mathbb{C}[X,Z]/(Z-X^3)\\ &\cong\mathbb{C}[X]. \end{aligned}$ The first equality follows from the third isomorphism theorem (numbering from Dummit and Foote). The second equality follows from the homomorphism $Y\mapsto X^2.$ The final equality follows from the homomorphism $Z\mapsto X^3.$

1.34. Let $R$ be a UFD.

(a) Show a monic polynomial of degree two or three in $R[X]$ is irreducible if and only if it has no roots in $R.$

Let $F\in R[X]$ with $2\le\deg(F)\le 3.$

Suppose $G,H\in R[X]$ are nonconstant such that $F=GH.$ Note that $G$ and $H$ must be monic. Since $\deg(F)=\deg(G)+\deg(H),$ one of $\deg(G)$ or $\deg(H)$ must be $1.$ Therefore one is of the form $X-a,$ $a\in R.$ So $a$ is a root of $F.$

Suppose $a\in R$ is a root of $F.$ Then by the division algorithm there exists $G\in R[X]$ such that $F(X)=(X-a)G(X).$

(b) The polynomial $X^2-a$ is irreducible if and only if $a$ is not a square in $R.$

Note that $X^2-a$ has a root in $R$ if and only if $a$ is a square in $R.$ Therefore the result follows from part (a).

1.35. Show that $V(Y^2-X(X-1)(X-\lambda)\subset\mathbb{A}^2(k)$ is an irreducible curve for any algebraically closed field $k$ and any $\lambda\in k.$

Note $Y^2-X(X-1)(X-\lambda)$ has no root in $k[X],$ and therefore it is irreducible over $k[X,Y].$

1.36. Let $I=(Y^2-X^2,Y^2+X^2)\subset \mathbb{C}[X,Y].$ Find $V(I)$ and $\text{dim}_\mathbb{C}(\mathbb{C}[X,Y]/I).$

Note that $X^2,Y^2\in I.$ Thus since it is easy to see $(X^2,Y^2)\supset I,$ $I=(X^2,Y^2).$ Observe that $\mathbb{C}[X,Y]/(X^2,Y^2)$ is spanned by the linearly independent set $\{\overline{1},\overline{X},\overline{Y},\overline{XY}\}.$

1.37. Let $k$ be any field, $F\in k[X]$ of degree $n>0.$ Show the residues $\overline{1},\overline{X},\ldots,\overline{X}^{n-1}$ for a basis for $k[X]/(F).$

Let $F=\sum_{i=0}^na_ix^i.$ Then $\overline{X}^n = \sum_{i=0}^{n-1}\overline{-a_n^{-1}a_ix^i}.$ By induction, $\overline{X}^m$ is a linear combination of the linearly independent set $\{\overline{1},\overline{X},\ldots, \overline{X}^{n-1}\}.$

1.38.* Let $R=k[X_1,\ldots,X_n],$ $k$ algebraically closed, $V=V(I).$ Show that there is a natural one-to-one correspondence between algebraic subsets of $V$ and radical ideals in $R/I,$ and that irreducible algebraic sets (resp. points) correspond to prime ideals (resp. maximal ideals).

Let $U\subset V$ be an algebraic set. Then $I(U)$ is a radical ideal in $R$ such that $I\subset I(V)\subset I(U).$ Thus by Problem 22 (the Fourth Isomorphism Theorem in Dummit and Foote), $\overline{I(U)}$ is a radical ideal in $R/I.$

Similarly, if $\overline{J}$ is a radical ideal in $R/I$ then $J$ is a radical ideal in $R$ with $I\subset J.$ Thus $V(J)\subset V(I).$

We prove the final statements by noting that an ideal $J\supset I$ in $R$ is prime (resp. maximal) if and only if $\overline{J}$ is prime (resp. maximal) in $R/I.$

1.39. (a) Let $R$ be a UFD, and let $P=(t)$ be a principal proper prime ideal. Show that there is no prime ideal $Q$ such that $0\subset Q\subset P,$ $Q\ne 0,$ $Q\ne P.$

Suppose $Q\subset P$ is a prime ideal. Let $q\in Q,$ $q\ne 0.$ Let $q=p_0p_1\ldots p_k$ with $p_0,\ldots,p_k$ irreducible. For any $i$ note that if $p_i\in P$ then $p_i=ut$ for some unit $u.$

Suppose, with renumbering, that $p_0,\ldots, p_l\not\in P$ and $p_{l+1},\ldots,p_k\in P,$ were $0\le l\le k.$ Note that since $q\in P,$ $l<k.$ Moreover, since $p_0\ldots p_l\not\in Q,$ it must be that $p_{l+1}\ldots p_k\in Q.$ But $p_{l+1}\ldots p_k=ut^{k-l}$ for some unit $u.$ So it must be that $t^{k-l}\in Q.$ But $Q$ is prime, so therefore $t\in Q.$ Thus $Q=P.$

(b) Let $V=V(F)$ be irreducible. Show that there is no irreducible algebraic set $W$ such that $V\subset W\subset\mathbb{A}^n,$ $W\ne V,$ $W\ne\mathbb{A}^n.$

Observe if $F$ is reducible then, by Corollary 4, $F=G^n$ and $I(V(F))=(G)$ is prime. If $V\subset W\subset\mathbb{A}^n,$ then $0\subset I(W)\subset I(V)=(G).$ Therefore by part (a) $I(W)=0$ or $I(W)=I(V).$

Bonus (Exercise 1.9 from Gathman’s notes). Prove that every affine variety $X\subset\mathbb{A}^n$ consisting of finitely many points is the zero locus of $n$ polynomials.

Note. For this exercise I used more standard notation (lowercase $f,g,\ldots$ for polynomials, etc.) to match the notation used by Gathmann.

Let $X=\{p_1,\ldots,p_r\}$ with each $p_i=(a_{i1},\ldots,a_{in})\in\mathbb{A}^n.$ We will define $f_1,\ldots,f_n\in k[x_1,\ldots,x_n]$ such that $X=V(f_1,\ldots,f_n).$

Let $S=\{1,\ldots,r\}.$ Define $f_1=\prod_{i\in S}(x_1-a_{i1}).$

Let $i\in S.$ Define the polynomial $g_i(x_1)=\frac{\prod_{j\in S\setminus\{i\}}(x_1-a_{j1})} {\prod_{j\in S\setminus\{i\}}(a_{i1}-a_{j1})}.$ Suppose $f_1=0,$ that is, $x_1=a_{k1}$ for some $k.$ Then $g_i(x_1)=0$ for all $i\ne k$ and $g_k(x_1)=1.$

Let $d\in \{2,\ldots,n\}.$ Define the polynomial $f_d(x_1,\ldots,x_n)=\prod_{i\in S}(x_d-a_{id})-\sum_{i\in S}\left[ g_i(x_1)(x_d-a_{id})\left( \prod_{j\in S\setminus\{i\}}(x_d-a_{jd}) + 1 \right) \right].$ Suppose $f_1=0,$ that is, $x_1=a_{k1}$ for some $k.$ Then $g_i(x_1)=0$ for all $i\ne k$ and $g_k(x_1)=1.$ Therefore $f_d(x_1,\ldots,x_n) = \prod_{i\in S}(x_d-a_{id}) - (x_d-a_{id})\left( \prod_{j\in S\setminus\{i\}}(x_d-a_{jd}) + 1 \right) = (x_d-a_{kd}).$ Thus $X=V(f_1,\ldots,f_n).$

1.40. Let $I=(X^2-Y^3, Y^2-Z^3)\subset k[X,Y,X].$ Define $\alpha:k[X,Y,Z]\to k[T]$ by $\alpha(X)=T^9,$ $\alpha(Y)=T^6,$ and $\alpha(Z)=T^4.$

(a) Show that every element of $k[X,Y,Z]/I$ is the residue of an element $A+XB+ YC+XYD,$ for soem $A,B,C,D\in K[Z].$

Observe that $\overline{X}^2=\overline{Y}^3$ and $\overline{Y}^2=\overline{Z}^3.$ Thus for $n,m\ge 2$ we may write $\overline{X}^n$ and $\overline{Y}^m$ as the residue of a polynomial in $k[Z].$

(b) If $F=A+XB+YC+XYD,$ $A,B,C,D\in k[Z]$ such that $\alpha(F)=0,$ show $F=0.$

We may factor out $T$ ’s to find $\alpha(F)=\alpha(A)+T^9\alpha(B)+T^6\alpha(C) +T^15\alpha(D).$ Observe $4n=6+4m$ and $9+5n=15+4m$ have no integer solutions. Thus it must be that $A=B=C=D=0.$

By (b) we know $\ker\alpha\subset I.$ It remains to show $I\subset\ker\alpha.$ Observe $\alpha(X^2-Y^3)=T^{18}-T^{18}=0$ and $\alpha(Y^2-Z^3)=T^{12}-T^{12}=0.$ Thus if $F\in I,$ $\alpha(F)=0,$ and $F\in\ker\alpha.$

Since $\ker\alpha=I,$ by the first isomorphism theorem $k[X,Y,Z]/I\cong\text{Im }\alpha\subset k[T].$ Since $k[T]$ is an integral domain, so is $k[X,Y,Z]/I$ ; therefore $I$ is prime. Thus $V(I)$ is irreducible and $I(V(I))=I.$