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$(A)$ If $K$ is algebraically closed field, then consider $m$ polynomials $f_i$ for $i=\{1,\dots,m\}$ in $K[x_1,\dots,x_n]$ of degree $d_i$ each with no common root. We know there exists $g_i$ for $i=1,\dots,m$ such that $$f_1g_1+\dots+f_mg_m=1$$

What can we say about the minimum degrees of $g_i$? ($(A)$ has been answered below).

$(B)$ If we relax the $1$ requirement or any fixed constant requirement and ask just for polynomials such that $$f_1g_1+\dots+f_mg_m\neq 0 \mbox{ (but not a constant)}$$ which seems same as $$Z(f_1g_1+\dots+f_mg_m)\bigcap Z((0))=\emptyset$$ can we say anything about the minimum degrees of $g_i$?

Is anything meaningful possible if $K$ is not closed in the relaxed problem $(B)$? Again what are the minimum degrees of $g_i$ here (both lower and upper bounds)?

I am only interested in $K=\Bbb C\mbox{ or }\Bbb R$ here.

related:bound on degree of certain polynomials

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    $\begingroup$ ams.org/journals/jams/1988-01-04/S0894-0347-1988-0944576-7/… $\endgroup$ Nov 20, 2014 at 16:59
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    $\begingroup$ To expand on Felipe's comment, the link is to the article: JOURNAL OF THE AMERICAN MATHEMATICAL SOCIETY Volume I, Number 4, October 1988: SHARP EFFECTIVE NULLSTELLENSATZ by JANOS KOLLAR $\endgroup$ Nov 20, 2014 at 17:10
  • $\begingroup$ @FelipeVoloch the references here seem to talk about upper bound for minimum degree. Please correct me if I am wrong? Is there a reference for lower bound? Thankyou for the link. $\endgroup$
    – Turbo
    Nov 20, 2014 at 22:31

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The bounds in the article by Kollár as refered in the comment of Felipe Voloch are optimal, see Example 2.3 in that article.

To see this concretely, assume that $m \leq n$, and let $f_1 = x_1^{d_1},f_2 = x_1 x_n^{d_2-1}-x_2^{d_2},...,f_{m-1} = x_{m-2}x_n^{d_{m-1}-1}-x_{m-1}^{d_{m-1}},f_m = 1-x_{m-1} x_m^{d_m-1}$.

By the article of Kollár, one can find $g_1,...,g_m$ such that $f_1 g_1 + ... + f_m g_m = 1$ and $\deg f_i g_i \leq d_1 \cdots d_m$. In the above example, these bounds are optimal, which can be seen as follows: If we assume that $f_1 g_1 + ... + f_m g_m = 1$, and apply this equality along the curve $x = (t^{d_2... d_m - 1},...,t^{d_m-1},1/t)$, one sees that $\deg_{x_m} g_1 \geq d_1\cdots d_m - d_1$, and thus the bound is optimal. (cf. http://www.encyclopediaofmath.org/index.php/Masser-Philippon/Lazard-Mora_example)

(Edit: removed claim with incorrect proof regarding (B))

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  • $\begingroup$ Could you clarify the "relaxed problem" answer better? $\endgroup$
    – Turbo
    Nov 23, 2014 at 11:27
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    $\begingroup$ I mean that when $K = \mathbb{C}$, or more generally, when $K$ is algebraically closed, then any non-vanishing polynomial is constant, so there are no non-constant polynomials $f_1 g_1 + ... + f_m g_m \neq 0$. (If you by $\neq 0$ mean nowhere vanishing.) On the other hand, if $K = \mathbb{R}$, there certainly are non-constant non-vanishing polynomials, and I don't know whether the answers to (A) and (B) would be different in that case. $\endgroup$ Nov 23, 2014 at 12:01
  • $\begingroup$ thank you.... can you expect any reasonable answer to this question as well mathoverflow.net/questions/187652/… $\endgroup$
    – Turbo
    Nov 23, 2014 at 18:33

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