Exercise. Why P is projective?

So, the main Idea is to get a map like in this lemma.

2 Main Results

Lemma 2.2 Let R=R'[X] be polynomial ring over a noetherian commutative ring R' with dim R' =4 (so dim R=5). Let I be a locally complete intersection ideal of height 2 in R that contains a monic polynomial and A= R/I. Assume that all rank 2 projective A=R/I modules have cancellation property. Then

1. I/I² $\equiv$ A ^-1
2. Consider the following conditions
   1. $\omega $ $$ is generated by two elements.
   2. I/I² $\equiv \; \omega \; \oplus \; \omega$^-2
   3. $\omega$ ^-2 is generated by 2 elements.
   Then a) $\Rightarrow$ b) $\Rightarrow$ c).

Proof.

Here we have dim A =3. We have an exact sequence


0 $\to$ P $\to$ Rn $\to$ I $\to$ 0


where P is a projective R-module of rank n-1. Since we can assume that rank P = n-1 > dim R and because I has a monic polynomial f it follows that P_f $\equiv $ R$$_f^n-1.

Hence, from theorem of Quillen and Suslin it follows P $\equiv $ Rn-1.$$

Tensoring the above sequence with A we get the following exact sequence:


0 $\to$ L $\to$ An-1 $\to$ An $\to$ I/I2 $\to$ 0


So, [I/I²] = [L]+[A] in K₀(A). Since rank 2 projective dim A-modules have cancellation property, we have I/I² $\equiv$ L $\oplus$ A. We also have $\Lambda$² I/I² = L = $\omega$ ^-1. So, I/I²$\equiv$ A $\oplus$$\omega$ ^-1.

Now we prove the second statement:

a) $\Rightarrow$ b): By 1) we have I/I² $\equiv$ $\omega$^-1 $\oplus$ A and also A² $\equiv$ $\omega$ $\oplus \; \omega$^-1. So,


I/I2 $\oplus$ A $\equiv$ (A $\oplus \; \omega$-1) $\oplus$ A $\equiv$ $\omega$-1 $\oplus$ A2 $\equiv$ $\omega$-1 $\oplus$ ($\omega$ $\oplus \; \omega$-1) $\equiv$ $\omega$ $\oplus$($\omega$ -1 $\otimes$ A2) $\equiv$ $\omega$ $\oplus \; \omega$-1 $\otimes$ ($\omega$ $\oplus \; \omega$-1) $\equiv$ $\omega$ $\oplus$ A $\oplus \; \omega$-2.

Since all rank 2 projective A-modules have cancellation property, we have I/I²$\equiv$ $\omega$ $\oplus \; \omega$^-2.

b)$\Rightarrow$ c): By 1) and b) it follows that


A $\oplus \; \omega$-1 $\equiv$ $\omega$ $\oplus \; \omega$-2.

By dualizing we get


A $\oplus \; \omega$ $\equiv$ $\omega$-1 $\oplus \; \omega$2.

Now adding A to both sides and using 1) we have:


A2 $\oplus \; \omega$ $\equiv$ (A $\oplus \; \omega$-1) $\oplus \; \omega$2 $\equiv$I/I2 $\oplus \; \omega$2 $\equiv$($\omega$ $\oplus \; \omega$-2) $\oplus \; \omega$2


Since we have assumed that rank 2 projective A-modules have cancellation property, by canceling $\omega$ we get A² $\equiv$ $\omega$^-2 $\oplus \; \omega$². So, $\omega$² is generated by 2 elements.

Following theorem relates the cancellation property of rank 2 projective modules with set theoretic complete intersection property of height 2 locally complete intersection ideals.

Theorem 2.1 Let R=R'[X] be a polynomial ring over a noetherian commutative ring R' with dim R' =4 (so dim R=5). Let I $\subseteq$ R be a locally complete intersection ideal of height 2 that contains a monic polynomial and A = R/I. Assume that projective A-modules of rank 2 have the cancellation property. Let $\omega$ = Ext²(A,R). Assume that $\omega$^r is generated by 2 elements, for some integer r ≥1. Then I is a set-theoretic complete intersection ideal.

Proof.

By the above theorem we have, I/I²$\equiv$ A $\oplus \; \omega$^-1. So, we have an exact sequence


0 $\to$ A $\to$ I/I2 $\to$ $\omega$-1 $\to$ 0,


where the map A $\to$ I/I² is given by an element f $\in$ I. So, I=Rf + I' , where I² I' I is an ideal and I'/I² $\equiv$ $\omega$^-1. Write J=I^r+Rf. We will see that J is locally complete intersection ideal. For p $\in$ Spec(R) containing I , let I_p = (f,g), where image of g generate $\omega$^-1. So, J_p= (f,g)^r + R_pf. Hence J_p=(g^r,f), is complete intersection.

Let L be the cokernel of the map given by f as follows


f: R/J $\to$ J/J2 $\to$ L $\to$ 0.


Since, locally J_p is generated by (f,g^r), as in the above paragraph, we have L is locally generated by one element. So, L is a line bundle.

Since rad(I) = rad(J), all rank 2 projective R/J-modules also have cancellation property. By the above lemma, we have J/J²$\equiv$ R/J $\oplus \; \omega$_J^-1 and L $\equiv$ $\omega$_J^-1. It follows that

1. J/J² $\equiv$ (R/J)f $\oplus \; \omega$_J^-1.
2. $\omega$_J^-1 $\equiv$ J/(J²+Rf).

Consider the exact sequence


0 $\to$ R/J$\to$ J/J2 $\to$ $\omega$J-1$\to$.


By tensoring the sequence with R/I, we get


0 $\to$ R/I $\to$ J/IJ $\to$ $\omega$J-1 $\otimes$ R/I $\to$ 0


So,


$\omega$J-1 $\otimes$ R/I $\equiv$ J/(IJ,f) $\equiv$ (Ir,f)/(Ir+1,f) $\equiv$ $\omega$I-r,


which is 2 generated. So, $\omega$_J^-1 is also 2 generated. Hence A² $\equiv$ $\omega$_J^-1 $\oplus \; \omega$_J. By lemma 2.2 2a) $\Rightarrow$ 2b), it follows that there is a surjective map


J/J2 $\to$$\omega$J.




Now it follows from lemma 2.1 that J is set theoretically generated by 2 elements. This completes the proof of this theorem.

3 Smooth 3-folds in A⁵

Notation 3.1 Let A be a smooth affine algebra over an algebraically closed field k and X=Spec(A).

1. Recall CH^r(X) denotes the Chow group of cycles of codimension r.
2. For a projective A-module P, C_i(P) $\in$ CHⁱ(X) will denote the i-th Chern class of P.

Following is a restatement of the theorem of Murthy and Mohan Kumar ([MKM]) in our context of LRC-question.

Theorem 3.1 Let X =Spec(A) be a smooth affine 3-fold over an algebraically closed field k. Assume that projective A-modules of all rank ( equivalently of rank 2) have cancellation property. Then, for any two projective modules P and Q of same rank, P $\equiv$ Q if and only if C_i(P) = C_i(Q) for i=1,2,3.

Proof.

($\Rightarrow$) This implication is obvious.

($\Leftarrow$) The theorem of Mohan Kumar and Murthy ([MKM]) implies that P and Q are stably isomorphic. Now P $\equiv$ Q by the cancellation property.

Theorem 3.2 Let X =Spec(A) be a smooth affine algebra over an algebraically closed field k. Assume that projective A-modules of rank 2 have cancellation property. Let L be a projective A-module of rank one. Then the following conditions are equivalent:

1. L is generated by 2 elements.
2. C₁(L)² =0 in A²(X).
3. L $\oplus$ L^-1 $\equiv$ A².

Proof.

1) $\Rightarrow$ 2): Since L is two generated, we have L $\oplus$ L^-1 $\equiv$ A². So, C(L $\oplus$ L^-1) = 1. Hence (1+C^-1(L))(1-C^-1(L)) =1 and C^-1(L)² =0.

2) $\Rightarrow$ 3): It follows that the total Chern class C(L $\oplus$ L^-1) =1. Hence [L] $\oplus$ [L^-1] = [A²] in K₀(A) and by cancellation L $\oplus$ L^-1 $\equiv$ A².

3) $\Rightarrow$ 1): This implication is obvious.

Theorem 3.3 Suppose k is an algebraically closed field and R=R'[X] a polynomial ring over an affine algebra R' with dim R' =4. Let I be a locally complete intersection ideal in R of height 2 that contains a monic polynomial. Assume A =R/I is smooth and rank 2 projective A-modules have cancellation property. Then the following conditions are equivalent:

1. K² =0 in A²(X) (where K =C₁($\omega$_I)).
2. $\omega$_I is two generated.

In all these cases, we have I/I² $\equiv$ $\omega$_I $\oplus \; \omega$_I^-2 and that I is a set-theoretic complete intersection ideal.

Proof.

Obvious from above.

Theorem 3.4 Suppose k is an algebraically closed field and R=R'[X] is a polynomial ring over an affine algebra R' with dim R' =4. Let I be a locally complete intersection ideal of R with height(I) = 2 and A=R/I is smooth over k. Assume that all rank 2 projective A-modules have cancellation property. Suppose rK² = 0 for some r > 0 where K=C₁( $\omega$_I). Then I is set-theoretic complete intersection.

Proof.

We can assume that r²K²=0. Let L = $\omega$^r. Then the total Chern classes


C(L $\oplus$ L-1) = (1+rK)(1 - rK) = 1.

So, L $\oplus$ L^-1 $\equiv$ A². So, $\omega$_I ^r is two generated. By Theorem 2.1, I is a set-theoretic complete intersection.

Following corollary follows from the above theorem.

Corollary 3.1 Let k be an algebraically closed field and R=k[X₁, ... ,X₅] be polynomial ring and $\phi$ : R $\to$ A =A'[X] be a surjective map onto a polynomial algebra over a smooth algebra A' of dimension 2. Let I be a the kernel of $\phi$ and K=C₁ ($\omega$_I). If rK² =0 for some positive integer r then I is set theoretic complete intersection.

Proof.

Note that rank 2 projective A-modules have cancellation property. Now the corollary follows from the above theorem.

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