Clarification and intuition request for rationally equivalent algebraic cycles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:56:23Z http://mathoverflow.net/feeds/question/108681 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108681/clarification-and-intuition-request-for-rationally-equivalent-algebraic-cycles Clarification and intuition request for rationally equivalent algebraic cycles OMYAC 2012-10-03T02:16:07Z 2012-10-03T03:25:23Z <p>I am having some difficulty lining up the definition and my intuition for rational equivalence of cycles. My intuition is based off of the idea that two cycles being rationally equivalent is analogous to the two cycles being homotopic. </p> <p>If it is requested of me to state the definition of rational equivalence, I will; for the moment I will refrain and simply link to the wikipedia page for the <a href="http://en.wikipedia.org/wiki/Chow_ring" rel="nofollow">chow ring</a> and <a href="http://en.wikipedia.org/wiki/Equivalence_relations_on_algebraic_cycles" rel="nofollow">adequate equivalence relation</a>. </p> <p>Here are two examples that I have thought of that are giving me trouble.</p> <ol> <li><p>In $\mathbb{k}[x,y]$, consider the line given by $x = 0$ and the unit circle. By taking the rational function $\frac{x^2+y^2 - 1}{x}$ on $\mathbb{k}[x,y]$, we get that these two are rationally equivalent. These two being homotopic is only semi-okay with me. I can imagine making the circle larger and larger, making it look more and more like a line. But it seems to me that the final step of taking it to the line is not allowed by homotopy.</p></li> <li><p>Consider the 2-torus $\mathbb{T}$, thought of as sitting in $\mathbb{A}^3$ with center at the origin. Let $S_1$ be (one of the) circles obtained by intersecting $\mathbb{T}$ with the $xy$-plane, and let $a$ be the regular function defining it. Let $S_2$ be one of the circles obtained by intersecting with the $xz$-plane, and $b$ the regular function defining it. Then the rational function $\frac{a}{b}$ gives a rational equivalence of $S_1$ and $S_2$. This goes very much against my intuition. </p></li> </ol> <p>Hopefully somebody will tell me where I am going wrong in my thinking.</p> http://mathoverflow.net/questions/108681/clarification-and-intuition-request-for-rationally-equivalent-algebraic-cycles/108686#108686 Answer by Steven Landsburg for Clarification and intuition request for rationally equivalent algebraic cycles Steven Landsburg 2012-10-03T03:25:23Z 2012-10-03T03:25:23Z <p>Any codimension 1 cycle that is defined by a regular function is already zero in the Chow group. So your question should not be "Why are the line $x=0$ and the unit circle rationally equivalent to each other?" but rather "Why is the line $x=0$ rationally equivalent to zero?" (and ditto for the unit circle).</p> <p>The answer is that either of these divisors can be "moved to infinity" and hence out of the affine plane altogether. (E.g. you can think of the graph of the regular function $x$ as a path that takes the line at $x=0$ to the (missing) line $x=\infty$.</p>