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This looks very much like something which is a duplicate, but I couldn't find the "desired duplicate". Sorry if this has been asked before.

Let $F$ be a field of $q$ elements and $F\setminus \{0\} = \{a_1,\ldots,a_{q-1}\}$. Show that $\prod a_i = -1$.

If $F$ has characteristic $2$ then this is trivial. So let's assume it has characteristic $p>2$. I tried to use the fact that $F\setminus \{0\}$ is a cyclic group with no avail.

Any hint?

Cauchy
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    https://en.wikipedia.org/wiki/Wilson%27s_theorem – Chappers Jul 14 '17 at 17:32
  • Use the fact that $F^\times$ is cyclic with $#F^\times \equiv -1\pmod{p}$. – anomaly Jul 14 '17 at 17:32
  • Our search engine is not the best. But, yeah, it's a duplicate all right :-/ – Jyrki Lahtonen Jul 14 '17 at 17:38
  • @JyrkiLahtonen I used the best search engine here :( (namely, the suggested questions which pop out after you type in the title, and I typed in several titles) – Cauchy Jul 14 '17 at 17:40
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    Good job Cauchy! That actually is often the best search method. I searched within the tag [tag:finite-fields], added the buzzwords product of elements, and sorted the results by votes. The best fit was hit #4. – Jyrki Lahtonen Jul 14 '17 at 17:42
  • @JyrkiLahtonen thanks, this makes me feel less sinful. Should I accept an answer or just leave the question as it is? – Cauchy Jul 14 '17 at 17:46
  • It is probably simpler to accept one. Otherwise the question may be bumped periodically. It's closed now, so no new answers can come. The alternative would be to merge this with the older one. Meaning that we would need to check that notations match etc. Not fun :-) – Jyrki Lahtonen Jul 14 '17 at 17:55

3 Answers3

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Since you are working with a finite field with $q$ elements, anyone of them is a root of the following polynomial

$$ x^q-x=0.$$

In particular if we rule out the 0 element, any $a_i\neq 0$ is a root of

$$x^{q-1}-1=0.$$

This polynomial splits completely in $\Bbb F_q$ so we find

$$(x-a_1)\cdots (x-a_{q-1})=0$$

in particular $$x^{q-1}-1=(x-a_1)\cdots (x-a_{q-1})$$

Thus $a_1\cdots a_{q-1}=-1$.

InsideOut
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  • I think this requires one more piece of argument. By Vieta's formulas, $a_1\cdots a_{q-1}=(-1)^{q-1}(-1)=(-1)^q$. Now if $q=p^n$ for some prime $p$, proof is done. Otherwise, we are in field of characteristics $2$, and $1=-1$. – Nikola Ubavić Jun 02 '23 at 16:55
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    @NikolaUbavić, I believe I was thinking the Identity Theorem of polynomial that says that if two polynomials, say $p(x)$ and $q(x)$, of the same degree have the same roots then they are the same. This implies $a_1,\cdots,a_{q-1}=-1$ because we would have $x^{q-1}-1=x^{q-1}+\cdots+a_1,\cdots,a_{q-1}$. Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. So once you know $a_1,\dots,a_{q-1}$ are all the roots then your argument applies and we are done one again. Thanks a lot for your comment. – InsideOut Jun 05 '23 at 14:41
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Hint:

Match every element of $F$ with its inverse. Note that the equation $x^2-1=0$ has exactly two solutions in any field of characteristic $>2$.

ajotatxe
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Hint: You can pair up elements $(x,1/x)$, unless an element is its own inverse, i.e., $x^2=1$.

Aaron
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