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May 13, 2020 at 21:33 history edited user157838 CC BY-SA 4.0
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May 13, 2020 at 21:32 comment added user157838 @AlexM. I have removed that source
May 13, 2020 at 16:36 comment added Alex M. @user157838: No, Hindawi is not ok. Their peer reviewing process is... "problematic", to say the least. I constantly find emails from them in my "spam" folder, in which they promise very quick acceptance for publication. Such aggressive self-promotion should be condemned; and the promise of such quick reviewing tells a lot about its quality. Regarding the authors' English, I simply cannot understand the quoted paragraph "Regarding this [...] predicted groups": it simply makes no sense in English. Anyway, your question was about something else, so let us end this off-topic discussion here.
May 13, 2020 at 14:35 comment added user157838 @AlexM. Is Hindawi considered a reputable publisher? I looked into it briefly and found they were generally considered to be ok. Furthermore, I didn't want to discard a paper solely because of the English.
May 13, 2020 at 9:10 comment added Alex M. @user157838: Just a side note: you cite some authors, but their English is so bad, that I wouldn't even bother reading their papers. But then, they got published by Hindawi, so... this should tell a lot.
May 12, 2020 at 22:19 comment added Lev Soukhanov I'm currently reading arxiv.org/pdf/1709.02079.pdf this, hopefully it is related. As @AchimKrause points out, currently the problem is still trivial because considered over Q. You should probably work with polynomials over H (similar to standard NTRU) to obtain interesting lattice problem.
May 12, 2020 at 17:40 comment added Achim Krause If I am not mistaken, the question as written now follows by doing the linear algebra approach over $\mathbb{Q}$ and then scaling $z$ to get integer coefficients.
S May 12, 2020 at 17:24 history suggested Steven Stadnicki CC BY-SA 4.0
Small further revision, cleaning up problem statements in line with a previous comment and closing a dangling parenthetical expression I left.
May 12, 2020 at 17:23 review Suggested edits
S May 12, 2020 at 17:24
S May 12, 2020 at 17:03 history suggested Steven Stadnicki CC BY-SA 4.0
Cleared up the first paragraph
May 12, 2020 at 16:49 review Suggested edits
S May 12, 2020 at 17:03
May 12, 2020 at 16:43 comment added user157838 @LevSoukhanov Not exactly, I am just interested in the applicability of a quaternion ring to non-commutative crypto.
May 12, 2020 at 13:26 comment added Lev Soukhanov Excuse me, I would like you to clarify something. I've looked up few papers on this theme. Is it some non-commutative version of NTRU?
May 12, 2020 at 12:25 history edited user157838 CC BY-SA 4.0
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May 12, 2020 at 12:23 comment added user157838 @StevenStadnicki I think you are right, I do mean conjugacy in rings, and I realize now that my question was not formulated correctly. I apologize, I am still very shaky with this kind of math. I will edit the question to be more specific
May 11, 2020 at 23:21 comment added John Voight @user157838: yes, the linear algebra approach works in the unit group of any quaternion algebra over a (computable) field.
May 11, 2020 at 20:26 comment added Steven Stadnicki Presuming that the question here is about conjugacy in these rings, then the problem is likely to be Complicated, by analogy with similar problems in $SL_2(\mathbb{Z})$ for instance. I think there's an interesting question here but you need to be more specific about it.
May 11, 2020 at 20:25 comment added Steven Stadnicki (Also, if you do mean 'as a group' then the problem is trivial, because the group structure on Hurwitz or Lipschitz quaternions is the additive one, which is still commutative.)
May 11, 2020 at 20:24 comment added Steven Stadnicki When you say 'the group' do you mean 'the ring'? Neither the Hurwitz nor Lipschitz quaternions form a field, so you have to be very careful using expressions like $z^{-1}$ in such a case. The question of whether there's a $z$ such that $zy=xz$ is very different from the question of whether there's an invertible such $z$ (which is nearly trivial; very few Hurwitz or Lipschitz quaternions are invertible multiplicatively, after all...)
May 11, 2020 at 18:05 comment added LSpice I removed the irrelevant [algebraic-groups] and [free-groups] tags.
May 11, 2020 at 18:03 history edited LSpice CC BY-SA 4.0
Names of papers; removed irrelevant [algebraic-groups] and [free-groups] tags
May 11, 2020 at 17:23 history edited user157838 CC BY-SA 4.0
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May 10, 2020 at 22:14 comment added user157838 @JohnVoight Is that true for all quaternion groups, like the Lipschitz or Hurwitz?
May 10, 2020 at 20:05 comment added John Voight mathoverflow.net/questions/358830/…
May 10, 2020 at 19:25 review Close votes
May 14, 2020 at 2:58
May 10, 2020 at 19:18 answer added Vít Tuček timeline score: 4
May 10, 2020 at 19:01 comment added Watson Ladd I'm going to assume you are discussing the quaternions over $\mathbb{R}$. In this case what you are asking is essentially the conjugacy structure of $\mathrm{SO}(3)$, and it's easiest to see geometrically. Every rotation in three dimensions is a rotation by some angle around an axis, and two rotations by the same angle are conjugate via any rotation that takes one axis to the to other. The structure of the double cover doesn't change much about the conjugacy.
May 10, 2020 at 15:57 review First posts
May 10, 2020 at 19:38
May 10, 2020 at 15:53 history asked user157838 CC BY-SA 4.0