Consider $X = \mathbb{P}^2_k$ and $\mathcal{O}_X(2).$ If we consider the linear system $\mathcal{L}$ of conics passing through the point $[0:0:1]$ we can see that this is spanned by the sections $\{x^2,y^2,xy,yz,xz\}$ and that this linear system defines a rational map $f:X \dashrightarrow \mathbb{P}^5_k.$ Blowing up at the point $[0,0,1]$ we get a surface $\tilde{X} \rightarrow X.$ One can, without too much trouble, show that if we let $x,y,z$ be the standard coordinates on $X$ then $\tilde{X} \cong \dfrac{\text{Proj } k[x,y,z] \times_k \text{Proj } k[w_1,w_2]}{ w_1y-w_2x}$ where we interpret the relation $w_1y-w_2x$ as a bihomogenous relation. Then, there is an extension of $f$ to a map $g: \tilde{X} \rightarrow \mathbb{P}^2_k.$ I'm curious, is there a way to get an explicit description of what this map is, in terms of a linear system on $\tilde{X}?$ How do I get a linear system extending the map?
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1$\begingroup$ Geometrically, it is clear which curves appear in the system (pullbacks of conics minus the exceptional divisor). Do you want something more explicit? (More coordinates?) $\endgroup$– t3sujiCommented Mar 31, 2016 at 18:51
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$\begingroup$ Yes, I would like coordinates in response to @t3suji's comment. $\endgroup$– AGLearnerCommented Mar 31, 2016 at 20:00
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$\begingroup$ If you like more coordinates: $\tilde X$ is covered by two charts: $w_1\ne 0$ and $w_2\ne 0$. On the former, the map is given by $\{w_1x,w_2y,w_1y,w_2z,w_1z\}$ (I just multiplied $\{x^2,y^2,xy,yz,xz\}$ by $w_1$, and then cancelled out $x$ using the identity $w_1y=w_2x$). On the other chart, it is a similar formula: multiply by $w_2$ and cancel $y$, getting $\{w_1x,w_2y,w_2x,w_2z,w_1z\}$. You now see the formula on the two charts is essentially the same, except that the middle term is written in two different ways. $\endgroup$– t3sujiCommented Apr 1, 2016 at 16:52
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First, your rational map $f$ is actually to $\mathbb{P}^4$ and not 5-space. Then you take the graph in $\mathbb{P}^2-\{[0:0:1]\}\times \mathbb{P}^4$ of this map and its closure in $\mathbb{P}^2\times \mathbb{P}^4$. One checks that the graph is $\widetilde{X}$ and the projection to $\mathbb{P}^2$ is your $g$. The projection to $\mathbb{P}^4$ is just the map from $\widetilde{X}$ to 4-space. These can be used to make most of the calculations you want to make in general.
