A division algebra $D$ is a vector spaces over a field $F$ equipped with a bilinear product and a multiplicative neutral element $1$. All the non-zero elements of $D$ have a multiplicative inverse. Associativity is often assumed but not always. Any field is a commutative, associative division ...

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4
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33 views

4-D lattices and quaternions

It is easy to prove that there are only 2 extensions $\mathbb{Q}(a)$, with $|a|=1$, of $\mathbb{Q}$ where $\mathbb{Z}[a]$ becomes a lattice(discrete free abelian subgroup of rank 2) in the complex ...
1
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0answers
62 views

Existence of a division ring on a field.

Suppose that $F$ is a field. Show that there exists a $F$-division algebra $D$ with two elements $a\neq b\in D$ such that $a^2-2ab+b^2=0$. In the field extensions we know that $a^2-2ab+b^2=0$ if and ...
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votes
0answers
18 views

concept of conjugacy class in a ring

Can we think of a similar concept of a conjugacy class in a ring which satisfies two three properties like conjugacy classes. I think of a set $M_x={xyx^{-1}-y}$ for $x\in R$ and $R$ is a division ...
1
vote
1answer
59 views

Parallelization of a Sphere gives Division Algebra

Is there an elementary proof of the fact, that a parallelization of $S^n$ can turn $\mathbb{R}^{n+1}$ into a division algebra? My guess was something like this: Let $v_1(x),\dots, v_{n}(x)$ denote ...
4
votes
1answer
43 views

Brauer group of cyclic extension of the rationals

I am trying to compute the relative Brauer group of the cyclic Galois extension $L=\mathbb Q[x]/(x^3-3x+1)$ of $\mathbb Q$. I know that $$ \mathrm{Br}(L/\mathbb Q)\cong H^2(G,L^*)\cong\mathbb Q^*/N(L^*...
6
votes
2answers
73 views

Why do division algebras always have a number of dimensions which is a power of $2$?

Why do number systems always have a number of dimensions which is a power of $2$? Real numbers: $2^0 = 1$ dimension. Complex numbers: $2^1 = 2$ dimensions. Quaternions: $2^2 = 4$ dimensions. ...
1
vote
2answers
79 views

Confusion about division in Clifford Algebra

On page 202 of The Road to Reality, Penrose claims that if we want to generalize Quaternions to n dimensions using Clifford Algebra, we must abandon the division property. I have a hard time believing ...
2
votes
0answers
29 views

Is the generalized hopf map of an alternative finite-dimensional real division algebra continuous?

Let $A$ be an $n$-dimensional alternative real division algebra (not necessarily associative). Is the map $$ \eta \colon \bigl\{(x,y) \in A \times A : |x|^2+|y|^2=1\bigr\} \to A \sqcup \{\infty\}, \...
2
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2answers
52 views

Finite dimensional division algebra over C

Another abstract algebra question from my university days that has me stumped at where to start! I know what a division ring is and I think I understand what a division algebra over $\mathbb C$ is. (...
4
votes
0answers
39 views

Crossed products and division algebras

I am currently reading some introductory material on Brauer groups ("Noncommutative Algebra", by Farb and Dennis) and the following two questions came to my mind: 1) Are all crossed products algebras,...
0
votes
0answers
18 views

Cyclic division ring

Suppose that $D$ is a division ring with center $F$ and with index $p$ prove that $D$ is cyclic if and only if there exists $x$ $\notin$$F$ which $x$$^p$ $\in$$F$.
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2answers
53 views

Central division algebras and splitting fields

Let $K$ be a field and $D$ be a central division algebra over $K$ of degree $n$. Suppose that $L\subset D$ is a maximal subfield, so that $[L:K]=n$. Then we know that $L$ is a splitting field, so ...
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votes
2answers
31 views

Minimal projections on von Neumann Algebras

A projection $p \neq 0$ in a von Neumann Algebra $A$ is called minimal, if for every projection $0\neq q\in A$ with $q \leq p$ already $q=p$. I want to prove the following theorem: For a minimal ...
3
votes
1answer
59 views

Field extension whose tensor product with itself over $\mathbb{Q}$ is not a field

An old qual problem reads Let $D$ be a 9-dimensional central division algebra over $\mathbb{Q}$ and $K \subset D$ be a field extension of $\mathbb{Q}$ of degree $>1$. Show that $K \otimes_\...
2
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0answers
41 views

Motivating the Cayley-Dickson construction by proving Hurwitz's theorem

To me it seems the way to motivate the Cayley-Dickson construction is to prove Hurwitz's theorem, which is done over at Wikipedia. The theorem states the only real division algebras equipped with a ...
0
votes
1answer
31 views

Polarization identity $2(a,b)(c,d)=(ac,bd)+(ad,bc)$

I am interested in following along this Wikipedia article's derivation of properties of composition algebras (in particular, Euclidean Hurwitz algebras). Let $A$ be a unital, not necessarily ...
1
vote
1answer
58 views

Version of Wedderburn's theorem on central simple algebras

Suppose that $A$ be a central simple algebra over a field $k$. Then by Wedderburn's theorem $A\cong M_n(D)$ for some division $k$-algbera $D$. But to define the 'Brauer equivalence' I need that $D$ is ...
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0answers
24 views

Unwind quaternion multiplication

I am trying to understand quaterions division. Imagine I have the following equation, where every member is a quaternion: $$Q = (qq_1)(qq_2)...(qq_n)$$ I suppose that, if I maintain the order of ...
1
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2answers
54 views

$\Bbb{H}_{\Bbb{Q}}$ is only four dimensional division algebra over rationals.

How to prove that only four dimensional division algebra (noncommutative) over $\Bbb{Q}$ is rational quaternions? After a bit of internet research, I am very sure about the above statement, if not ...
3
votes
1answer
45 views

Show 3D-division algebra over the reals cannot exist using linear algebra

There is a great comment by Jyrki Lahtonen here: Why is quaternion algebra 4d and not 3d? It is not too difficult to show that a 3D-division algebra over the reals cannot exist. If $D$ were such a ...
2
votes
1answer
51 views

Is there any construction of infinite dimensional algebraic division ring?

I know that there is a division algebra over $\mathbb{Q}$ such that it is algebraic and infinite dimensional over it's center i.e. $\mathbb{Q}$. But for construct this division algebra. we can use ...
0
votes
1answer
37 views

Show that a finite dimensional algebra D with identity over a skewfield F is a semifield if and only if it has no zero divisors.

I'm struggling with a proof of the next lemma. Show that a finite dimensional algebra $D$ with identity over a skewfield $F$ is a semifield if and only if it has no zero divisors. EDIT: Actualy I ...
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votes
0answers
12 views

How much do we know about (univariate) polynomials over division rings?

Is there any algorithm that finds the factors of a (left) polynomial, say $P(t)$ over a divison ring, say $K$. If there is, is there characterization for finiteness of factorizations and for such ...
2
votes
1answer
59 views

How to understand that the left regular representation of a division algebra is irreducible?

In Weyl's book The classical groups, it is said the regular representations of a division algebra are faithful and irreducible. The key step is to show the ideal of the division algebra is $\{0\}$ ...
3
votes
3answers
95 views

algebraically closed field in a division ring?

Is it possible to have $K \subset D$ where $K$ is algebraically closed field and $D$ is a division ring such that $K \subseteq Z(D)$?
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0answers
33 views

Involutions on endomorphisms over division rings

Let $D$ be a division ring, and let $M $ be a free left $D$-module of finite rank. Assume that $x\mapsto x^*$ is an involution on the ring $\operatorname{End}_D(M)$ (which in this case means: ${}^*$ ...
2
votes
1answer
66 views

Finite dimensional central division algebras over a finite extension of $\mathbb{F}_q(T)$

Over number fields, finite dimensional central division algebras are always cyclic algebras. So the construction of cyclic algebras is a nice recipe to create algebras, which exhausts all finite ...
3
votes
2answers
56 views

On Nilpotent Elements of $M_n (F)$

For a field $F$, I have proved that $A \in M_n(F)$ is nilpotent iff $A^n=0$. Now I am curious about Division Rings. If we consider $F$ as a division ring then what happens? Does the result remain true?...
1
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0answers
34 views

Division algebra over 2-adic fields

Let $D$ be the quaternion division algebra and $O$ be a maximal $\mathbb{Z}$-order in $D$, say the Hurwitz quaternion integers. It can be proved that $D$ and $O$ split at odd primes, that is $$D\...
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vote
0answers
118 views

Has the Riemann Hypothesis been generalized to the Octonions and the Quaternions?

I've noticed that it uses imaginary numbers. I know that sometimes when I have too few dimensions like (-1)^n, dots show where I might expect lines due to imaginary numbers. So perhaps there is a ...
1
vote
2answers
38 views

Finding basis and dimension based on definition of space

I've got two vector spaces $U$ and $V$ over division ring $\mathbb{T}$ . Space $W$ over division ring $\mathbb{T}$ is defined as $W =\{( u, v ); u \in U, v \in V \}$ with operations $(u_1, v_1) + (...
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0answers
43 views

A corollary of Niven

Please proof corollary of Niven: For $a \in D\backslash R$, the equation ${t^n} = a$ has exactly $n$ solutions in $D$, all of which lie in $R\left( a \right)$, in there $R$ is a real-closed field and $...
4
votes
3answers
236 views

Prove that the division ring is commutative if for every $x$, $x^7=x$

I'm trying to solve a problem and I'm stuck. Here is the original problem: Let $A$ be a finite-dimensional algebra over a field $K$, such that for every $a\in A$, $a^7=a$. Show that $A$ is a ...
5
votes
1answer
288 views

Examples of division algebras

Together with the Grunwald–Wang theorem, the Albert–Brauer–Hasse–Noether theorem implies that every central simple algebra over an algebraic number field is cyclic, i.e. can be obtained by an explicit ...
7
votes
1answer
126 views

A central division algebra is not its commutator

In looking at old qualifying exam questions, I've come upon a question that has me stumped. Let $A$ be a central division algebra (of finite dimension) over a field $k$. Let $[A,A]$ be the $k$-...
1
vote
1answer
49 views

Ideal in a certain algebra over a field

Let $K|k$ be a finite field extension. Define $D$ to be a finite dimensional $k$ division algebra. If $J$ is a nonzero two-sided ideal of $D\otimes_k K$ then by considering $K$-dimensions, I see that ...
2
votes
1answer
77 views

Show that $D \cong {\rm End}_A(D^n)$ where $D$ is a division algebra and $A\cong M_n(D)$

Define $A$ to be a finite dimensional simple algebra over a field $k$. $D$ is a $k$-division algebra (not necessarily commutative) such that $A\cong M_n(D)$ for some integer $n$. Let $L$ be a minimal ...
1
vote
1answer
78 views

Simple $M_n(D)$-module with $D$ a division ring

Define $D$ to be a division algebra over a field $k$ and $R=M_n(D)$ the $n\times n$ matrix ring over $D$. A simple $R$-module $M$ is the quotient of $R$. I can write $R=\bigoplus_j I_j$ where $I_j$ is ...
6
votes
0answers
116 views

Involutions of the second type in a division algebra

I'm trying to figure out some details about involutions of division algebra, thought maybe someone here might have a better insight. Let $k$ be a $p$-adic or number field, and let $K=k[\sqrt{\delta}]$...
5
votes
1answer
247 views

Proving that $\mathbb R^3$ cannot be made into a real division algebra (and that extending complex multiplication would not work)

I am trying to solve the following exercise: Prove that complex multiplication does not extend to a multiplication on $\mathbb R^3$ so as to make $\mathbb R^3$ into a real division algebra. ...
2
votes
1answer
74 views

Ring Structures On $\mathbb {R} ^n$

In the book of Musili it is written that $\mathbb{R}^n$ is a division ring under usual addition and multiplication for $n=1,2,4$. I have understood this. But after that he said, in those cases we ...
2
votes
1answer
73 views

The reals as an algebra over the rationals

R, the real numbers, is an infinite dimensional commutative division algebra over the rationals Q. Is there an example of an infinite dimensional noncommutative division algebra over the rationals Q?
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1answer
50 views

finite integral domain algebras are division algebras [duplicate]

If R is a finite-dimensional algebra over a field k, and if R is also an integral domain, show that R is a division algebra over k.
2
votes
1answer
371 views

Endomorphisms of direct sum and division ring

How to prove that $$\operatorname{End}_R(V^{ \oplus n }) \cong M_n(D),$$ where $V$ is a simple left $R$-module and $D=\operatorname{End}_R(V)$. This is part of the proof finally leads to prove ...
1
vote
1answer
150 views

Why are there no Dual-octonions?

In the case of quaternions, we can define the traditional quaternions setting the imaginary components equal to root negative one, the hyperbolic quaternions by using root positive one, and the dual ...
2
votes
1answer
65 views

Example of an ordered, noncommutative division ring

Does there exist a noncommutative division ring $D$ (i. e. a field except that commutativity of multiplication is violated, e. g. the quaternions) which is also an ordered ring? Since most examples ...
1
vote
1answer
73 views

Simple $R$-module

Let $M$ be a simple $R$-module and $N=M\bigoplus M$. Then which one is true: 1) $N$ has a finite number of submodules. 2) $\operatorname{Hom}_R(N,N)$ is a division ring. 3) $\operatorname{Hom}_R(N,...
2
votes
1answer
58 views

direct products and direct sums of skew fields

If F is a skew field then are the arbitrary direct sums of F-modules isomorphic to their direct products (over the same index). I mean, if R is a division ring, and $\{M_i\}_{i\in I}$ some familly ...
5
votes
1answer
164 views

“Vector spaces” over a skew-field are free?

Are modules over a skew field free? That is, if $F$ is a skewfield then can any module $M$ be written as $\underset{i \in I}{\bigoplus} F$ for some indexing set $I$?
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150 views

Indefinite quaternion algebra over Q

Let $D$ be an indefinite quaternion algebra over $\Bbb Q$. We have a chosen isomorphism $\iota \colon D \otimes_{\Bbb Q} {\Bbb R} \cong {\mathrm{M}}_2({\Bbb R})$. Q: If we choose another isomorphism $...