Fulton 1.7

Posted on November 15, 2017

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.

(c) Show that \ker\alpha=I, so I is prime, V(I) is irreducible, and I(V(I))=I.

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.