Let $k$ be a field of characteristic 0 and let $X$ and $Y$ be smooth, projective and geometrically integral $k$-schemes of finite type. Assume that both $X$ and $Y$ have 0-cycles of degree 1. Does $X\times_{k}Y$ have a 0-cycle of degree 1?
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asked Jun 5, 2016 at 21:32
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2$\begingroup$ will not the product of the cycles be the example? $\endgroup$– user42024Jun 5, 2016 at 22:05
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1$\begingroup$ @user42024 A 0-cycle of degree 1 on $X$ is a formal sum $(P_0)+\cdots+(P_n)-(Q_1)-\cdots-(Q_n)$ that is defined over $k$, i.e., such that each of the multi-sets $\{P_0,\ldots,P_n\}$ and $\{Q_1,\ldots,Q_n\}$ is Gal$(\bar k/k)$ invariant. Suppose that $Y$ has a similar 0-cycle $(P'_0)+\cdots+(P'_m)-(Q_1')-\cdots-(Q'_m)$. Then the "product" 1-cycle is (I guess) $\sum_{i,j} (P_i\times P_j') - \sum_{i,j} (Q_i\times Q_j')$. This has degree $(n+1)^2-n^2=2n+1$, so has degree 1 only if $n=0$. $\endgroup$– Joe SilvermanJun 5, 2016 at 22:41
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2$\begingroup$ @Joe Silverman No, my product is different. As far as I understand any 0-cycle can be represented as a sum $\sum_i P_i - \sum_j Q_j$ where $P_i=\mathrm{Spec} K_i$ for some extensions $K_i/k$ and $Q_j=\mathrm{Spec} L_j$ for some extensions $L_j/k$ . So by product I mean $\sum_{i,i'}P_i\times_k P'_{i'}-\sum_{i,j'}P_i\times_k Q'_{j'}-\sum_{j,i'}Q_j\times_k P'_{i'} + \sum_{j,j'}Q_j\times_k Q'_{j'}$ $\endgroup$– user42024Jun 5, 2016 at 23:39
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2 Answers
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A zero cycle of degree 1 on such an $X$ can be written as $D_1-D_2$ for two effective cycles with $\deg D_1=\deg D_2+1$. Similarly, we have $E_1-E_2$ for $Y$. Then the obvious product (somewhat loosely written) $D_1\times E_1-D_1\times E_2-D_2\times E_1+D_2\times E_2$ has degree 1.
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$\begingroup$ Dear Mohan, So you are saying that if $x$ and $x'$ are points on $X$ such that $\gcd([k(x):k], [k(x'):k])=1$ and $y$ and $y'$ are points on $Y$ such that $\gcd([k(y):k], [k(y'):k])=1$, then $\gcd([k(x,y):k], [k(x,y'):k], [k(x',y):k],[k(x',y'):k])=1$? $\endgroup$ Jun 6, 2016 at 16:14
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$\begingroup$ I am not sure I understand what you are writing. If $x\in X, y\in Y$ are closed points, letting $z=(x,y)\in X\times_k Y$, one has $[k(z):k]$ is a multiple of the lcm of $[k(x):k], [k(y):k]$ $\endgroup$– MohanJun 6, 2016 at 16:42
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$\begingroup$ My conditions mean there exist integers $a,b, c,d$ such that $D=ax+bx'=D_{1}-D_{2}$ and $E=cy+dy'=E_{1}-E_{2}$ are 0-cycles of degree 1 on $X$ and $Y$, respectively. Your argument then tells me, or so I think, that there exist integers e,f,g,h such that $e(x,y)+f(x,y')+g(x',y)+h(x',y')$ is a 0-cycle of degree 1 on $X\times_{k}Y$. So the gcd of those 4 degrees of field extensions ought to be 1 and I should be able to prove this in elementary terms, i.e., just playing around with gcd's, but I can't. I feel I should be able to do this to convince myself that your proof is correct. $\endgroup$ Jun 6, 2016 at 22:44
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The referee for one of my papers gave the following argument:
Let $k$ be any field and let $X$ and $Y$ be smooth, proper and geometrically integral k-schemes of finite type. Let $x$ be a $0$-cycle on $X$ and $y$ a $0$-cycle on $Y$. Then $x\times Y$ and $X \times y$ are cycles on $X \times Y$ of complementary dimensions, so that $z=(x \times Y).(X \times y)$ is a $0$-cycle on $X \times Y$. Then, by intersection theory, ${\rm deg}(z)={\rm deg}(x){\rm deg}(y)$. In particular, if both $X$ and $Y$ have $0$-cycles of degree $1$, then so does $X \times Y$.
answered Sep 8, 2017 at 17:49