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1 vote
Accepted

On reflexive bialgebras

There shouldn't be any relations in general. Given an algebra $A$, there can be multiple different co-multiplications on $A$ to make it a bialgebra, or even a Hopf algebra. For instance, let $M(G)$ be ...
Maximilien Péroux's user avatar
0 votes

What are hypergroups and hyperrings good for?

I will try to say something about hypergroups (more generally about the hyperstructures) and hopefully it be useful and can answer to the questions: How is a canonical hypergroup to be thought of as ...
Reza Ameri's user avatar
0 votes

Is a "separable" algebra over a field finite-dimensional?

That $A$ is finite-dimensional by condition 2 is proved in Algebra: Chapter VIII by Bourbaki, p.232 Theorem 1, where separable algebra in sense of condition 2 is called "absolutely semisimple ...
Wang Chao's user avatar
2 votes

Why should we study derivations of algebras?

Here’s a practical reason to look at derivations, from one algebraist’s perspective, broken down in a few steps. Suppose you would rather look at the group of automorphisms of the algebra, instead of ...
Skip's user avatar
  • 2,103
1 vote

Equivalent definition of Spin group in terms of automorphisms

$ \newcommand\R{\mathbb R} \newcommand\Cl{\mathrm{Cl}} \newcommand\tr{\mathrm{tr}} \newcommand\form[1]{\langle#1\rangle} \newcommand\Orthog{\mathrm O} \newcommand\Pin{\mathrm{Pin}} \newcommand\Spin{\...
Nicholas Todoroff's user avatar
6 votes
Accepted

Reference about cancellation property for semigroups

A semigroup with this property seems to be called a right reductive semigroup (Wikipedia, nLab).
Naïm Favier's user avatar
2 votes
Accepted

Finding $\mathbb{C}(u,v)$ such that $\mathbb{C}(u,v,x^p+y^p)=\mathbb{C}(x,y)$, for every prime number $p$

Not much can be said about $F$. The second conditions should hold for $\mathbb C(u,v)$ for $u$ and $v$ two "generic" algebraically independent polynomials. For an explicit construction, take ...
Will Sawin's user avatar
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4 votes
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Relationship between units of a ring and primitive characters of the ring under addition

The rings $\mathbb{Z}/n$ are the only examples. I assume that "primitive character" just means that it is faithful, or equivalently that it does not factor through a proper quotient; this is ...
Qiaochu Yuan's user avatar
0 votes

Is there a "natural" interpretation of the power function for octonions and for sedenions?

It seems that there cannot be a natural interpretation, because there is no canonical definition for the meaning of $x^y$. There is already no canonical definition when $x$ is a complex number (that ...
Skip's user avatar
  • 2,103
1 vote

A general form of a maximal totally isotropic subspace in the split octonion algebra

Yes to both questions! Every maximal totally isotropic $V$ is of this form by Theorem 3 on page 164 of van der Blij, F.; Springer, T. A., Octaves and triality, Nieuw Arch. Wiskd., III. Ser. 8, 158-169 ...
Skip's user avatar
  • 2,103
4 votes

How fast does the number of "fixed" points grow compared to the size of the ball in the following group?

One can describe $M/N$ by using basic commutative algebra. Thing of $M$ as a free module of rank 5 over the polynomial ring $\mathbf{Z}[t,t^{-1}]$, with basis $(e_j)_{1\le j\le 5}$. So what you denote ...
YCor's user avatar
  • 62.3k
5 votes

How fast does the number of "fixed" points grow compared to the size of the ball in the following group?

The subgroup $F$ (I have not checked this is indeed the set of fixed points of $\phi$) is isomorphic to $\mathbb Z$, the question is just whether it is distorted or not, and if it is how much? I'll ...
Corentin B's user avatar
  • 1,459
2 votes
Accepted

Infinite-dimensional, non-unital Frobenius algebras

It is studied under the name "nearly Frobenius algebras" in this paper. In Example 3.3, the algebra $\mathbb{C}[[x,x^{-1}]]$ of Laurent series can be endowed with (countably) many unital but ...
Qwert Otto's user avatar

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