Timeline for Is Gauss's generalization of Wilson's theorem non-superficially related to the classification of moduli for which primitive roots exist?
Current License: CC BY-SA 4.0
19 events
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| Apr 8 at 10:22 | comment | added | Thrash | With Gauß' generalisation, we have: $(\mathbb Z/(n))^{\times}$ is cyclic $\implies n \in \{1,2,4,p^{\alpha},2p^{\alpha} \}$. I'm curious if we can get the converse implication in a similar way. | |
| Apr 8 at 10:01 | comment | added | Thrash | Does this also work the other way round, i.e., does Gauß' generalisation of Wilson's theorem imply the classification of moduli for which the respective unit group is cyclic? | |
| Apr 7 at 12:03 | comment | added | Thrash | @KConrad: Based on your comment mathoverflow.net/questions/425047/…, I have found something: sites.math.washington.edu/~morrow/336_09/papers/Andrew.pdf | |
| Jun 21, 2022 at 22:22 | comment | added | KConrad | To illustrate Will's 2nd comment, $(\mathcal O_K/\mathfrak p)^\times$ is cyclic for all prime ideals $\mathfrak p$, but for $\alpha \geq 2$ the group $(\mathcal O_K/\mathfrak p^\alpha)^\times$ can fail to be cyclic for infinitely many $\mathfrak p$, such as when $p$ is an odd prime and is inert in $K$ (that happens for 1/2 the primes if $K/\mathbf Q$ is quadratic): $(\mathcal O_K/\mathfrak p^\alpha)^\times \cong \mathbf F_{p^n}^\times \times (\mathbf Z/p^{\alpha-1}\mathbf Z)^n$ where $n := [K:\mathbf Q] \geq 2$. | |
| Jun 21, 2022 at 21:46 | comment | added | Will Sawin | This gives the logical relationship between the two results. The other implications can be proved, but the proofs are not necessarily shorter than the proofs of the two results independently. Mathematically this is the same as what you said, just a different perspective! Maybe another thing to say is that for a general number ring, the moduli with a primitive root are a subset of the moduli where the product of modular units is $-1$, and this subset may be proper for rings other than $\mathbb Z$. | |
| Jun 21, 2022 at 21:42 | comment | added | Will Sawin | I would formulate the relationship in a different way. For each one, we can break the result into two halves: Some property holds for this list of moduli, and it doesn't hold for another list of moduli. There is an easy implication from one half of Gauss's generalization (that the product is $1$ outside this list) to one half of the classification of moduli with primitive roots (no primitive root outside this list), and a corresponding easy implication from the other half of the classification to the other half of Gauss's generalization. | |
| Jun 21, 2022 at 20:54 | comment | added | KConrad | I rewrote Step 3(ii) to avoid mentioning isomorphisms, but they're implicitly there since the reduction map $(\mathbf Z/(n))^\times \to (\mathbf Z/(n'))^\times$ is onto with a trivial kernel: each unit mod $n'$ is the reduction of one unit mod $n$. That's why the exponent on the right side of the displayed congruence in Step 3(ii) is $1$ instead of an even number like in all other similar-looking steps. | |
| Jun 21, 2022 at 20:53 | history | edited | KConrad | CC BY-SA 4.0 |
added 144 characters in body
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| Jun 21, 2022 at 19:03 | comment | added | Favst | I have gone through the proof in detail and there is just one part I don't understand and would like to modify: In Step 3(ii), what is the purpose of the isomorphism discussion and the final sentence "The only unit... displayed congruence is $1\mod n$."? Could one not proceed as in Step 3(iii) with $\beta=1$ instead and taking the product modulo $2$, which makes the product $1$ modulo $2$ (since all the elements $u$ in the left product are odd), and then piece modulo $n'$ together with modulo $2$ via Chinese remainder theorem? | |
| Jun 21, 2022 at 1:18 | comment | added | KConrad | Yes, in kconrad.math.uconn.edu/blurbs/ugradnumthy/carmichaelkorselt.pdf the proof of Korselt's criterion uses the fact that the units mod $p$ form a cyclic group when $p$ is prime. | |
| Jun 21, 2022 at 1:15 | comment | added | Favst | Yes, I proved a version of Euler's criterion in every case where there is a primitive root, calling it a generalized Euler's criterion (which actually came in handy in an article I recently wrote that develops a formula for counting all $k^{\text{th}}$ power residues modulo $n$, based on less general methods of a fellow named Stangl). The generalized Euler's criterion is essentially a result on page 104 of Niven's book. Also, there is a proof of Korselt's criterion that applies part of the primitive root theorem, though not the full power of it (I think one of your articles has it). | |
| Jun 21, 2022 at 0:33 | comment | added | KConrad | What other applications of the primitive root theorem do you have? In many places I've seen people appeal to $(\mathbf Z/(p))^\times$ being cyclic to prove Euler's criterion $a \equiv \Box \bmod p \Longleftrightarrow a^{(p-1)/2} \equiv 1 \bmod p$ when $(a,p) = 1$ for odd primes $p$. I feel like that is overkill, since it can be shown using the fact that a polynomial of degree $d$ over a field (like $\mathbf Z/(p)$) has at most $d$ roots in the field, which is more intuitive and simpler than bringing in a generator of the unit group mod $p$. | |
| Jun 21, 2022 at 0:02 | comment | added | Favst | Thanks. While there is simplicity in Ore's exposition, I actually prefer your method for a couple of reasons. Firstly, it illuminates a reason for why the two results have similarities. Secondly, although the primitive root theorem is a strong result to be using, I feel that your proof is less ad hoc than Ore's. Also, it provides a nice application of the primitive root theorem in my modular exponentiation chapter. | |
| Jun 20, 2022 at 23:51 | comment | added | KConrad | That's fine, but while it's nice to see Gauss' generalization of Wilson's theorem is a consequence of the classification of $n$ for which the units mod $n$ are cyclic (that's what you asked about here, and I didn't known it before), I think it makes Gauss' generalization seem harder than it really is. All you need to know about the units mod $n$ to get Gauss' result is the number of solutions to $x^2 \equiv 1 \bmod n$, which can be done with the Chinese remainder theorem and some simple calculations modulo prime powers, just as in Ore's book. That is much simpler than the method I wrote above. | |
| Jun 20, 2022 at 22:17 | vote | accept | Favst | ||
| Jun 20, 2022 at 22:17 | comment | added | Favst | Beautiful. It's a pleasure to read your detailed response. More so because I believe you are Keith Conrad, whose expository papers on elementary number theory helped me as a high school olympiad student and later as an undergraduate. I actually wondered if you might respond, as this seemed to be your kind of a problem! Would it be alright if I included a version of your proof here in some books that I have written (currently in editing stage)? I would write up my own rendition and give you credit, of course. | |
| Jun 20, 2022 at 16:35 | history | edited | KConrad | CC BY-SA 4.0 |
Added details to proof to cover the case of even n.
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| Jun 20, 2022 at 15:24 | history | edited | KConrad | CC BY-SA 4.0 |
added 17 characters in body
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| Jun 20, 2022 at 1:24 | history | answered | KConrad | CC BY-SA 4.0 |