Let $f_1, f_2, \ldots, f_m \in \mathbb{Z}[x_1, \ldots, x_n]$. Assume $f_1(X) = f_2(X) = \ldots = f_m(X) = 0$ have no solutions over $\mathbb{C}^n$, then by Hilbert's Nullstellensatz, there exists polynomials $g_1, \ldots, g_m \in \mathbb{C}[x_1, \ldots, x_n]$ such that $1 = f_1 g_1 + \dots f_m g_m$.

In this case, can we always find $g_1, g_2, \ldots, g_m \in \mathbb{Q}[x_1, \ldots, x_n]$ such that $1 = f_1 g_1 + \dots f_m g_m$? In words, if the coefficients of $f_1, \ldots, f_m$ are all integers, can $g_1, g_2, \ldots, g_m$ be taken as polynomials with integer coefficients, such that $f_1 g_1 + \ldots +f_m g_m \in \mathbb{Z}^{+}$?

Added later: I checked Wikipedia on Hilbert's Nullstellensatz. Sorry, it seems to be a stupid question, and the answer is YES.


2 Answers 2


Yes. Given the degrees of the $g_i$, the equation $1 = \sum_i f_i g_i$ is tantamount to a system of linear equations in the coefficients of the $g_i$, and those linear equations have rational coefficients. Once such a system has a complex solution it automatically has a rational solution.

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    $\begingroup$ nice argument ! $\endgroup$
    – jsliyuan
    Commented Feb 7, 2013 at 0:27

The answer is definitely yes. The argument is very simple: the ring extension $$\mathbb Q[X_1,\dots,X_n]\subset \mathbb C[X_1,\dots,X_n]$$ is faithfully flat.

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    $\begingroup$ Whoa $-$ "very simple" perhaps, but much more advanced than is necessary; this is just linear algebra in the coefficients. $\endgroup$ Commented Feb 2, 2013 at 5:44
  • $\begingroup$ Please don't delete this. (For what it's worth this is the first argument that came to my mind, too.) $\endgroup$ Commented Feb 2, 2013 at 9:57
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    $\begingroup$ I agree with Dan Petersen: Please do not remove your answer. There are often very different answers to the same question, and it can be instructive to see and compare them. I just didn't want people to get the idea that they must understand concepts like "faithfully flat" to prove this result. $\endgroup$ Commented Feb 2, 2013 at 15:36
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    $\begingroup$ @YACP: Sorry, that was not my intent; I guess I should remember that not everybody here speaks English fluently. Looking in m-w.com, "strong reaction of ... astonishment" is the closest to what I meant, since your answer up to "The argument is very simple" led me to think you were going to give the elementary linear-algebra argument, and then suddenly I was whisked into graduate commutative algebra. @G.Myerson: actually (again going by m-w.com) "subtile" and "subtilities" are correct albeit old-fashioned alternatives to "subtle" and "subtleties", and closer to the original Latin subtilis. $\endgroup$ Commented Feb 3, 2013 at 4:46
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    $\begingroup$ Actually the two answers coincide, when you write down the proofs in detail. $\endgroup$ Commented Feb 3, 2013 at 14:51

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