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Let $k$ be a field, $f_1$, $\ldots$, $f_m \colon k^n \to k$ affine functions without common zeroes ( that is, there does not exist $x \in k^n$ with $f_1(x) = \cdots = f_m(x) = 0$). Then there exist $(\alpha_1, \ldots, \alpha_m) \in k^m$ such that

$$\sum_{i=1}^m \alpha_i f_i = 1$$

Notes:

  1. This should be a standard result, a kind of Nullstellensatz. Meaning : if $f_i$ have no common zero, it is for a reason.

  2. Posted as reference/challenge.

Any feedback would be appreciated!

$\bf{Added:}$ Here is a simple proof: the system $f_1(x) = \ldots = f_m(x)$ is incompatible. Transforming into $A x = b$, the RHS $b$ is not in the column space of the matrix $A$. This implies there exists a linear functional on $k^m$ that is $0$ on the column space and $-1$ on $b$, done!

orangeskid
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  • The kernels define hyperplanes. For hyperplanes, having nonempty intersection is equivalent to being equal. Also, hyperplanes intersect unless they are parallel. So in order to not all intersect, one of these hyperplanes is parallel to one other. The difference between the corresponding affine functions is a constant – praeseo Jul 22 '23 at 04:36
  • @Praneet Srivastava: not in general, think of the three sides of a triangle. – orangeskid Jul 22 '23 at 05:21
  • @orangeskid -- could you sketch out what you've tried so far? – Annika Jul 22 '23 at 13:16
  • @User5678: I will add the solution I have on mind soon. But want to see if there are other ideas. It is not a difficult problem... – orangeskid Jul 22 '23 at 21:37
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    @orangeskid gotcha -- posted my attempt below. Used quotient spaces to reduce the linear combination of functions to a 1-D vector space problem. Might be some errors, but I tried to check it over. – Annika Jul 22 '23 at 21:51
  • @orangeskid -- ok, double checked and I think this works. – Annika Jul 22 '23 at 22:53
  • I think my answer boils down to a similar set of observations as Praneet. Parallel nullspaces imply parallel affine functions, so they differ only by a constant for any $x \in K^n$ – Annika Jul 22 '23 at 23:37

1 Answers1

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Let $\widetilde{f_i} : k^{n + 1} \to k$ be linear such that $f_i(x) = \widetilde{f_i}(x, 1)$. The fact that $f_i$ do not have a common $0$ corresponds to the common solutions to $\widetilde{f_i}(x, z)$ being contained in $z = 0$. Dualizing, $z$ must be in the span of $\widetilde{f_i}(x, z)$, ie there are $\alpha_i \in k$ such that $\sum \alpha_i \widetilde{f_i}(x, z) = z$. Now set $z = 1$.

I think this is not so different from your linear algebra proof, possibly equivalent to considering the augmented matrix $[A \mid b]$, but the homogenization trick is often useful.

ronno
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  • Yes, a very nice solution. If you think about it, it is similar to the homogenization trick used for the standard Nullstellenatz. Great! – orangeskid Aug 23 '23 at 20:16