## Question and background

I have started reading the paper Frontiers of Reality in Schubert Calculus. On page 5 near the bottom of the page, the author picks a subspace of a Grassmannian and a basis of polynomials for that subspace. He then constructs a ramification at a point $s\in\mathbb{C}$. The part of this discussion that I am wondering about is

Subtracting an appropriate multiple of $f_0$ from each of the other polynomials, we may assume that they vanish to order greater than $a_0$ at $s$.

After some time, I was able to understand this statement, and confirm that this process works(the proof I found is below), but I am not quite happy with this.

In particular, I am wondering if this is a well known result that I don't know of, or if the argument is simpler than I made it. Or if perhaps there is something deeper than I am missing.

This is mostly because he dismisses it as so obvious. If my proof is the intended argument, ok, I just wanted to be sure I was not missing something here. I hope this question isn't too pointless :/

## My Proof

Since $f_0$ vanishes at $s$ with order $a_0$, then $f_0\left(x\right)=\left(x-s\right)^{a_0}g\left(x\right)$ such that $g\left(s\right)\neq0$. If $\hat{f}_{1}$ vanishes at s with order greater than $a_{0}$ then let $f_{1}:=\hat{f}_{1}$. Otherwise, $\hat{f}_{1}=\left(x-s\right)^{a_0}h\left(x\right)$ for $h\left(s\right)\neq0$ since $a_{0}$ was minimal. Now define $f_{1}:=\hat{f}_{1}-\dfrac{h\left(s\right)}{g\left(s\right)}\left(f_{0}\right)$. Continuing in this manner we can generate $f_{i}$ for all $i$.

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 This seems pretty clear to me. What's the confusion? (Maybe it would help if you looked at the case s = 0.) – Qiaochu Yuan Feb 16 2010 at 6:59