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Feb 4, 2016 at 10:47 comment added JoS I think the morphism $\mathbb{P}^{n-1} \to \mathbb{P}^{m-1}$ should be given by $[f_1, \ldots, f_m]$. The fact that they only vanish simultaneously at $0$ shows that the morphism is well defined. Here, the $f_i$ can be interpreted as sections of $\mathcal{O}(d)$. Taking linear projections means that we take $m-1$ linear combinations of the $f_i$ to define the morphism to $\mathbb{P}^{m-2}$. The fact that they are still well-defined means that these linear combinations still only have $0$ as common zero set.
Feb 4, 2016 at 10:05 comment added Max Horn Thanks for your reply, but I am not quite sure I understand. Which morphism $\mathbb{P}^{n-1}\to\mathbb{P}^{m-1}$ do we get here? And how does the morphism you obtain after projecting to $\mathbb{P}^{m-2}$ (and finally down to $\mathbb{P}^{n-1}$, I assume) help to find a suitable subset of the span of $A$? (Presumably this is evident as soon as it clear which morphism you have in mind?)
Feb 3, 2016 at 15:57 history answered Mohan CC BY-SA 3.0