# Doubt in this proof of Horrocks theorem

I'm beginning to study some research papers and I need right now to understand the solution of Vaseršteĭn of Serre's theorem (simplest proof of this theorem), to do so, I'm beginning to understand Horrocks theorem:

I can understand that there are some $g_1, g_2$ such that $f_1g_1+f_2g_2$ has leading coefficient 1 and degree $\le d-1$ but why by row operations we can suppose for some $i\neq 1,2$, $f_i=f_1g_1+f_2g_2$?

If this question is not suitable to Mathoverflow site, let me know and I will delete it.

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Repost of math.stackexchange.com/questions/729081/… . As I said in another of your threads, the proof is from Serge Lang's "Algebra", and you might like to try a different book instead, since Lang's proofs often lack detail and precision (I don't understand this one either). – darij grinberg Mar 27 '14 at 22:41
@darijgrinberg yes, I've already searched in another books, I couldn't find any other books/articles with the same proof, do you know some references? – user26832 Mar 27 '14 at 22:44
Proposition V.2.6 in T. Y. Lam's "Serre's problem on projective modules" is a generalization of the theorem. See also his Proposition III.2.6. – darij grinberg Mar 27 '14 at 22:58
@darijgrinberg Thank you for the reference. However, it's another kind of proof. I'm still curious why this part is true in Lang's book. I think this part is more complicated than I thought. – user26832 Mar 27 '14 at 23:15
Why downvoted?? – user26832 Mar 28 '14 at 22:35

Let $F=f_1g_1+f_2g_2$ of degree at most $d-1$ and leading coefficient 1. Replacing $g_i$ with $x^kg_i$ if necessary, we may assume $\deg F=d-1$. Now, consider $f_3$ (which exists, since $n>2$). We have assumed that $\deg f_3<d$. If it's $\deg d-1$ coefficient is a unit, there is little to do. So, assume that it is in the maximal ideal. Then add $F$ to it to get a degree $d-1$ polynomial with unit leading term.