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I am motivated by this: Let $V$ be a vector space over a field of $q$ elements with $\dim V\geq 2$ then $V$ can be written as union of $q+1$ proper subspaces.

I've written a constructive proof here.

Now, I let $\mathbb F$ be a finite field having $q$ elements and consider the vector space $\mathbb F^2$.

$\mathbb F^2$ can be written as union of $q+1$ number of one-dimensional subspaces, say, $V_0$, $V_1$, $\ldots$, $V_q$. Corresponding to each $V_i$, there is a non-zero linear functional $f_i(x_1,x_2) = a_ix_1+b_ix_2\in\mathbb F[x_1,x_2]$ s.t. $\ker f_i =V_i$ i.e., $f_i$ vanishes exactly on $V_i$.

It follows that the product of these functionals i.e., $f_0f_1\ldots f_q$ vanishes everywhere on $\mathbb F^2$. This is a $q+1$ degree homogeneous polynomial in $\mathbb F[x_1,x_2]$.

If I did everything correctly, we come to the conclusion that:

Let $n\in\mathbb Z_{\geq 2}$. Given a field $\mathbb F$ of $q$ elements, you can find a homogeneous polynomial of degree $q+1$ in $\mathbb F[x_1,\ldots,x_n]$ which vanishes everywhere on $\mathbb F^n$.

Is that right? I have never seen this result anywhere. Can anyone provide a reference to this? I would also want to see other proofs of this and related results.

Nothing special
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3 Answers3

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I've seen this result being mentioned as an aside in algebra classes, but don't have a concrete textbook reference:

There are infinitely many polynomials but only finitely many (polynomial) functions on a finite set. By the pigeonhole principle this implies that there must be two different polynomials which induce the same function on a given finite field. Their difference thus induces the zero function.

Or you could simply take the product $\prod_{a\in F}(x-a)$ of linear factors over all field elements as a concrete example.

ahulpke
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    This doesn't prove that the polynomial has degree $q+1$ or is homogeneous. Also your concrete example is zero over $\mathbb{F}$, not $\mathbb{F}^n$. Incidentally, it is equal to $x^q-x$. – Adayah Oct 02 '24 at 09:00
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Your reasoning looks correct. A more direct way to reach the same conclusion would be:

Since $x=x^q$ for every $x\in\mathbb{F}_q$, we have $$ x^q y = xy = x y^q $$ so $x^q y - x y^q$ is zero everywhere, and clearly homogeneous of degree $q+1$.

By the way, this factors as $xy(x^{q-1}-y^{q-1})$. With that factorization, being identically zero can alternatively be understood as: Either $x$ is zero or $y$ is zero, or both are nonzero -- and in the latter case their $(q-1)$th powers are both $1$.

Troposphere
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    And in fact $x^{q-1} - y^{q-1}$ factors as $\prod_{c \in \mathbb{F}_q^*} (x - cy)$, making this answer closely related to the one by Adayah. – Daniel Schepler Oct 02 '24 at 15:57
  • Yeah, with two variables there's only that solution, up to constant factors. Namely: The $x^{q+1}$ coefficient must vanish; otherwise we wouldn't get zero at (1,0). But now the coefficient of $x^qy$ must determine the entire polynomial: If we had two distinct solutions with the same $x^qy$ coefficient, then set $y=0$ and subtract them; the result would be an everywhere zero polynomial in $x$ of degree $<q$, which is impossible. – Troposphere Oct 02 '24 at 16:18
  • This would have been my choice for the answer as well :-) – Jyrki Lahtonen Oct 05 '24 at 21:00
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Your result is correct. When applied to the decomposition from the Hagen von Eitzen's answer in the linked thread, it yields a concrete example:

$$f(x_1, x_2, \ldots, x_n) = x_1 \cdot \prod_{c \in \mathbb{F}} (x_2 - cx_1).$$

Adayah
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  • I can also see the subspaces corresponding to each factor... $V_0={(x_1,x_2,\ldots,x_n):x_1=0}$, $V_c={(x_1, x_2,\ldots, x_n): x_2=cx_1}$. These are hyperspaces of $\mathbb F^n$. – Nothing special Oct 02 '24 at 08:43
  • I think an alternate representation of the same polynomial should be $x_1 x_2^q - x_1^q x_2$. Which evaluates to zero everywhere since $x_2^q = x_2$ and $x_1^q = x_1$. – Daniel Schepler Oct 02 '24 at 15:51
  • @DanielSchepler That's correct, nice find. – Adayah Oct 03 '24 at 06:58