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Let $F$ be an algebraically closed field and $\mathbb{P}^1$ the projective line over $F$. Suppose $V_1, V_2$ are two 1-dimensional subvarieties of the 2-dimensional variety $\mathbb{P}^1\times\mathbb{P}^1$.
Now $V_1$ and $V_2$ do not necessarily intersect, as the simple example $V_i=\{x_i\}\times\mathbb{P}^1$ with $x_1\neq x_2$ shows. But are there any conditions on $V_1$ and $V_2$ known, under which they do intersect?

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    $\begingroup$ Look for "Picard group" on the web. The question is not appropriate for this site. $\endgroup$ – abx Jul 25 '18 at 17:34
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    $\begingroup$ I Disagree with abx. All algebraic geometers know this, but not all professional mathematicians. Algebraic geometers should want our field to be useful to the rest of mathematics, yes? $\endgroup$ – David E Speyer Jul 25 '18 at 17:57
  • $\begingroup$ @abx Whatever happened to MO being a place where researchers could get help with questions they suspect have answers well-known to experts, but where they don't have said experts at hand? (+1 to David Speyer) $\endgroup$ – Yemon Choi Jul 26 '18 at 16:19
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The curve $V_i$ is given by the vanishing of a polynomial $F_i(x_1,x_2,y_1,y_2)$ that is homogeneous in $x_1,x_2$ of degree $d_{i,1}$ and homogeneous in $y_1,y_2$ of degree $d_{i,2}$. Then counting intersection points with multiplicities, $$ V_1 \cdot V_2 = d_{1,1}d_{2,1} + d_{1,2}d_{2,2}. $$ So $V_1$ and $V_2$ always intersect except in the case that they are both horizontal slices or they are both vertical slices, as in the example that you give.

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    $\begingroup$ Typo: You mean where both are horizontal slices, or where both are vertical slices. $\endgroup$ – David E Speyer Jul 25 '18 at 17:29
  • $\begingroup$ @DavidESpeyer Oops, thanks for pointing that out, I fixed the text. $\endgroup$ – Joe Silverman Jul 25 '18 at 22:24
  • $\begingroup$ BTW this is [Hartshorne], Example V.1.4.3. $\endgroup$ – user19475 Jul 26 '18 at 6:29

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