§ Weak and Strong Nullstllensatz

§ Weak Nulstellensatz: On the tin

For every maximal ideal mk[T1,,Tn]m \subset k[T_1, \dots, T_n] there is a unique akna \in k^n such that m=I({a}) m = I(\{ a \}). This says that any maximal ideal is the ideal of some point.

§ Weak Nullstellensatz: implication 1 (Solutions)

every ideal, since it is contained in a maximal ideal, will have zeroes. Zeroes will always exist in all ideas upto the maximal ideal.
  • It simply says that for all ideals JJ in C[X1,,Xn]\mathbb C[X_1, \dots, X_n], we have I(V(J))=sqrtJI(V(J)) = sqrt J
  • Corollary: II and VV are mutual inverses of inclusions between algebraic sets and radical ideals.

§ Weak Nullstellensatz: Implication 2 (Non-solutions)

If an ideal does not have zeroes, then it must be the full ring. Hence, 1 must be in this ideal. So if I=(f1,f2,,fn)I = (f_1, f_2, \dots, f_n) and the system has no solutions, then II cannot be included in any maximal ideal, hence I=C[X1,,Xn]I = \mathbb C[X_1, \dots, X_n]. Thus, 1I1 \in I, and there exist ciC[X1,,Xn]c_i \in \mathbb C[X_1, \dots, X_n] such that 1=sumifici1 = sum_i f_i c_i.

§ Strong Nullstellensatz: On the Tin

For every ideal JJ, we have that I(V(J))=0JI(V(J)) = |_0^\infty\sqrt J. I am adopting the radical (heh) notation 0x|_0^\infty \sqrt x for the radical, because this matches my intuition of what the radical is doing: it's taking all roots, not just square roots. For example, (8)=(2)\sqrt{(8)} = (2) in Z\mathbb Z.

§ Strong Nullstellensatz: Implication 1 (solutions)

Let J=(f1,,fm)J = (f_1, \dots, f_m). If gg is zero on V(J)V(J) , then gJg \in \sqrt J. Unwrapping this, rN,ciC[X1,,Xn],ifici=gr\exist r \in \mathbb N, \exists c_i \in \mathbb C[X_1, \dots, X_n], \sum_i f_i c_i = g^r.

§ Weak Nullstellensatz: Proof

  • Let mm be a maximal ideal.
  • Let KK be the quotient ring KC[X1,,Xn]/mK \equiv \mathbb C[X_1, \dots, X_n] / m.
  • See that KK is a field because it is a ring quotiented by a maximal ideal.
  • Consider the map α:C[X1,,Xn]K\alpha: \mathbb C[X_1, \dots, X_n] \rightarrow K, or α:C[X1,,Xn]C[X1,,Xn]/m\alpha : \mathbb C [X_1, \dots, X_n] \rightarrow \mathbb C[X_1, \dots, X_n] / m by sendingelements into the quotient.
  • We will show that α\alpha is an evaluation map, and K=CK = \mathbb C. So we will get a functionthat evaluates polynomials at a given point, which will have a single point as a solution.
  • Core idea: See that α(C)=CK\alpha(\mathbb C) = \mathbb C \subset K. Hence KK is a field that contains C\mathbb C. But C\mathbb C is algebraically closed, hence K=CK = C.
  • First see that CK\mathbb C \subset K, or that α\alpha preserves C\mathbb C [ie, α(C)=C\alpha(\mathbb C) = \mathbb C].note that no complex number can be in mm.If we had a complex number zz in mm, then we would need to have 1=1/zz1 = 1/z \cdot z in mm (since an ideal is closed under multiplication by the full ring), which means 1m1 \in m, due to whichwe get mm is the full ring. This can't be the case because mm is a proper maximal ideal.
  • Hence, we have CK\mathbb C \subseteq K or K=CK = \mathbb C.
  • Thus the map we have is α:C[X1,X2,,Xn]C\alpha: \mathbb C[X_1, X_2, \dots, X_n] \rightarrow \mathbb C.
  • Define zi=α(Xi)z_i = \alpha(X_i). Now we get that α(ijaijXij)=ijaijzij\alpha(\sum_{ij} a_{ij} X_i^j) = \sum_{ij} a_{ij} z_i^j. That is,we have an evaluation map that sends XiziX_i \mapsto z_i.
  • CLAIM: The kernel of an evaluation map α\alpha is of the form (X1z1,,Xnzn)(X_1 - z_1, \dots, X_n - z_n).
  • The kernel is also mm. Hence, m=(X1z1,,Xnzn)m = (X_1 - z_1, \dots, X_n - z_n), and point that corresponds to themaximal ideal is (z1,z2,,zn)(z_1, z_2, \dots, z_n).

§ Strong Nullstellensatz: Proof

We use the Rabinowitsch trick.
  • Suppose that wherever f1,,fmf_1, \dots, f_m simultaneously vanish, then so does gg. [that is, gI(V(J))g \in I(V(J))where J=(f1,,fm)J = (f_1, \dots, f_m)].
  • Then the polynomials f1,,fm,1Ygf_1, \dots, f_m, 1 - Yg have no common zeros where YY is a new variable into the ring.
  • Core idea of why they can't have common zeros: Contradiction. assume that 1Yg1 - Yg, and all the fif_ivanish at some point.Then we need 1Yg=01 - Yg = 0 which mean Y=1/gY = 1/g, so gg cannot vanish, so g0g \neq 0.However, since all the fif_i vanish, gg also vanishes as g(V(J))g \in (V(J)). This is contradiction.
  • Now by weak Nullstellensatz, the ideal J=(f1,,fm,(1Y)g)J = (f_1, \dots, f_m, (1-Y)g) cannot be containedin a maximal ideal (for then they would simultaneously vanish). Thus, J=RJ = R and 1J1 \in J.
  • This means there are coefficients ci(Y,x)C[X1,,Xn,Y]c_i(Y, \vec x) \mathbb C[X_1, \dots , X_n, Y] such that
1=c0(Y,x)(1Yg(x))i=1mci(Y,x)fi(x) 1 = c_0(Y, \vec x) (1 - Yg(\vec x)) \sum_{i=1}^m c_i(Y, \vec x) f_i(\vec x)
Since this holds when x,Y\vec x, Y are arbitrary variables, it continues to hold on substituting Y=1/gY = 1/g, the coefficient c0(1Yg)=c0(1g/g)=c0(11)=0c_0(1-Yg) = c_0(1 - g/g) = c_0(1 - 1) = 0 disappears. This gives: 1=i=1mci(Y,x)fi(x)1 = \sum_{i=1}^m c_i (Y, \vec x) f_i(\vec x) since Y=1/gY = 1/g, we can write ci(Y=1/g,x)=ni(x)/gir(x)c_i(Y=1/g, \vec x) = n_i(\vec x)/g^r_i(\vec x). By clearing denominators, we get:
1=i=1mni(x)fi(x)/gR(x) 1 = \sum{i=1}^m n_i(\vec x) f_i(\vec x)/ g^R(\vec x)
This means that gR(x)=i=1mni(x)fi(x) g^R(\vec x) = \sum_{i=1}^m n_i(\vec x) f_i(\vec x)

§ Strong Nullstellensatz: algebraic proof

  • We have gI(V(J))g \in I(V(J)).
  • We want to show that gJg \in \sqrt{J} in RR.
  • This is the same as showing that g0g \in \sqrt{0} in R/JR/J. (J0J \mapsto 0 in the quotient ring).
  • If gg is nilpotent in R/JR/J, then the (R/J)g(R/J)_g becomes the trivial ring {0}\{ 0 \}. [Intuitively, if gg is nilpotent and a unit, then we will have gn=0g^n = 0, that is unit raised to somepower is 0, from which we can derive 1=01 = 0].
  • Localising at gg is the same as computing R[Y]/(1Yg,J)R[Y]/(1 - Yg, J).
  • But we have that V(1Yg,J)=V(1 - Yg, J) = \emptyset. Weak Nullstellensatz implies that (1Yg,J)=(1)(1 - Yg, J) = (1).
  • This means that R[Y]/(1Yg,J)=R[Y]/(1)={0}R[Y]/(1 - Yg,J) = R[Y]/(1) = \{ 0 \}. Thus, (R/J)g(R/J)_g has gg as nilpotent, or gJg \in \sqrt J in RR.

§ Relationship between strong and weak

Strong lets us establish what functions vanish on a variety. Weak let us establish what functions vanish at a point.

§ Strong Nullstellensatz in scheme theory

  • Same statement: I(V(J))=JI(V(J)) = \sqrt J.
  • V(J)V(J) is the set of points on which JJ vanihes. Evaluation is quotienting. So it's going to beset of prime ideals pp such that JR/p0J \xrightarrow{R/p} 0. So JpJ \subset p. This means thatV(J)={pprime ideal in R,Jp}V(J) = \{ p \text{prime ideal in } R, J \subseteq p \}.
  • I(V(J))I(V(J)) is the set of functions that vanish over every point in V(J)V(J). The functions that vanishat pV(J)p \in V(J) are the elements of pp. So the functions that vanish over all points is I(V(J))=V(J)I(V(J)) = \cap V(J).
  • Unwrapping, this means that I(V(J))I(V(J)) is the intersection of all ideals in V(J)V(J), which is the intersectionof all primes that contains JJ, which is the radical of JJ.
Holy shit, scheme theory really does convert Nullstellensatz-the-proof into Nullstellensatz-the-definition! I'd never realised this before, but this.. is crazy. Not only do we get easier proofs, we also get more power! We can reason about generic points such as (x)(x) or (y)(y) which don't exist in variety-land. This is really really cool.