This is part of a homework problem I am working on. In light of this, please do not post an entire solution. I've been told that we do not have to use anything but Fermat's little theorem and a bit of group theory to prove this result. And I'm honestly out of ideas on how to do it. Every resource I've found seems to suggest that polynomial roots and field theory is necessary.
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What is $U(p)$? $\mathbb Z/p\mathbb Z$ or the group of $p^{\mathrm{th}}$ root of unity in $\mathbb C$? – Can I play with Mathness Nov 09 '18 at 15:22
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It is $(Z/pZ)^*$ – farleyknight Nov 09 '18 at 15:23
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We can try to "hide" the polynomial part, for example: look at the Vandermonde determinant (the polynomial is hidden here) $$ \det\begin{bmatrix}1&x_1&x_1^2&\dots&x_1^m\\1&x_2&x_2^2&\dots&x_2^m\\1&x_3&x_3^2&\dots&x_3^m\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&x_{m+1}&x_{m+1}^2&\dots&x_{m+1}^m\end{bmatrix}=\prod_{1\leq i<j\leq m}(x_j-x_i) $$ If $x_1,\dots,x_{m+1}$ are solutions to $x^m\equiv 1\pmod{p}$, then the matrix has two identical columns (mod $p$) so LHS is $\equiv 0\pmod{p}$. So $p$ must divide one of the factor on the RHS.
user10354138
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Holy cow! Yes, maybe we could use linear algebra to prove this.. I'm not sure. However, this problem came up in the first few weeks of my abstract algebra course and my prof insisted that we didn't need to use anything advanced. In spite of that, it's interesting that there is a way to avoid the polynomials, in a sense. Thanks for your response! – farleyknight Nov 09 '18 at 17:00
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I don't think that repeating one common proof of Lagrange's theorem should count as avoiding use of it. The exercise clearly intends other methods. – Bill Dubuque Nov 09 '18 at 17:33
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Rule Number 1. $(\mathbb Z/p\mathbb Z)^\times$ is cyclic, generated by some element $a$ of order $p-1$. So whatever solution you are after, it must be a power of $a$.
Richard Martin
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Can you prove that $(Z/pZ)^*$ is cyclic without using polynomials? – farleyknight Nov 09 '18 at 15:22
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2@QuangHoang: That uses polynomials (with the use of Lagrange's polynomial congruence theorem) – user10354138 Nov 09 '18 at 15:45
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@user10354138 My bad. Here is another one in Elementary number theory, it's gotta be polynomial-free :-). – Quang Hoang Nov 09 '18 at 15:53
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1@QuangHoang: It uses IV.4.c which is Lagrange's polynomial theorem :-) I don't believe a truly "polynomial-free" proof exists. – user10354138 Nov 09 '18 at 16:26