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Results tagged with ra.rings-and-algebras
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user 12858
Non-commutative rings and algebras, non-associative algebras, universal algebra and lattice theory, linear algebra, semigroups. For questions specific to commutative algebra (that is, rings that are assumed both associative and commutative), rather use the tag ac.commutative-algebra.
6
votes
Accepted
Chirality of octonion algebras
Perhaps this is more a question about the Fano plane than about the (real) octonions. Notice that the automorphism group of the Fano plane is the simple group $\operatorname{GL}(3, \mathbb{F}_2) \cong …
- 12.9k
4
votes
Accepted
Subfields of division rings of degree $2$ which are not invariant
(This is basically a more detailed version of Eoin's comment.)
I assume that you are considering division algebras over a field $k$, i.e., $Z(A) = k$. If $B$ is a subalgebra of dimension $2$ of $A$, t …
- 6,614
7
votes
An algebra map between Hopf algebras that does not commute with the counit
Such a map can certainly exist. For instance, take the $k$-algebra $G = k \times k$, with
$$ \begin{aligned}
&\Delta(1,0) = (1,0) \otimes (1,0) + (0,1) \otimes (0,1), \\
&\Delta(0,1) = (1,0) \otimes ( …
- 6,614
18
votes
4
answers
2k
views
For which rings $R$ is $\mathrm{SL}_n(R)$ generated by transvections?
Let $R$ be a commutative ring $R$ with $1$, and $n \geq 2$ an integer.
Under which conditions is the group $\operatorname{SL}_n(R)$ generated by transvections?
(A transvection is a matrix with $1$ e …
- 63.9k
4
votes
Group representation with algebra structure
A classification is too much to hope for, but the representation theory tells you whether such an algebra structure can exist: if $V$ is your $G$-representation, then an algebra product corresponds to …
- 6,614
5
votes
3
answers
1k
views
adjoint of multiplication operator in a commutative algebra
Dan Popescu asked me the following question, and since I'm not an expert I'm throwing his question on MO.
Suppose that $A$ is a finite-dimensional vector space over an ordered field $k$ with $\operat …
- 3,738
1
vote
2
answers
229
views
Invertibility of all left multiplication maps in non-unital rings
Suppose that $R$ is a ring, not necessarily commutative nor associative. Assume that for every non-zero $a \in R$, the left multiplication map
$$ \lambda_a \colon R \to R \colon x \mapsto ax $$
is inv …
- 35.2k
12
votes
Applications of Jordan algebras
They turn up quite often in the study of (exceptional) linear algebraic groups. The most famous instance of this is the fact that algebraic groups of type $F_4$ are precisely the automorphism groups o …
- 6,614
2
votes
Tits-Kantor-Koecher construction for Jordan algebra of symmetric bilinear form
This is contained in Jacobson's Blue Book (Structure and Representations of Jordan Algebras, AMS Colloquium Publications, 1968), as Exercise 1 on p. 342, for arbitrary fields, and with no assumptions …
- 6,614
3
votes
2
answers
919
views
Skew fields inside quaternion division algebras
Suppose that $Q$ is a quaternion division algebra with center $k$, where $k$ is an arbitrary commutative field (let's say with $\operatorname{char}(k) \neq 2$ if necessary). Assume that $D$ is an arbi …
- 11.2k
12
votes
1
answer
1k
views
Divisibility and factorization in rings that are not integral domains
In my course notes for an undergraduate course "Algebra I", I wrote at the point when I'm introducing the notion of divisibility in rings (in a section on unique factorization):
We want to study f …
- 396
3
votes
Heisenberg-type groups over rings with involution
I wasn't aware of the paper by Abe that you mention, but I have used the group $A$ that you described in the case where $R$ is an octonion division algebra, in order to describe the rank one forms of …
- 6,614
2
votes
polarization/linearization as in jordan forms
The principle of polarizing is given by what you wrote yourself in the third paragraph. If $p$ is a homogeneous polynomial of degree $n$, then:
The clearest formulation is to take $p(x+ \lambda y) …
- 6,614
12
votes
Simplest examples of rings that are not isomorphic to their opposites
Here is an explicit example of a central simple algebra over $\mathbb{Q}$ not isomorphic to its opposite (which is merely a detailed example of what Pete explained).
First take a cubic cyclic Galois …
- 6,614
6
votes
Cayley-Dickson form of a quaternion
I believe there is a good reason why mathematicians don't use the terminology "simplex-part" and "perplex-part": they are not canonical! Indeed, algebraically there is no way to distinguish the elemen …
- 6,614