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Questions tagged [division-algebras]

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 algebra. A skewfield = a division ring is always a division algebra over its center. The quaternions form the best known non-commutative division algebra.

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Why are the solutions of polynomial equations so unconstrained over the quaternions?

An $n$th-degree polynomial has at most $n$ distinct zeroes in the complex numbers. But it may have an uncountable set of zeroes in the quaternions. For example, $x^2+1$ has two zeroes in $\mathbb C$,...
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linear algebra over a division ring vs. over a field

When I was studying linear algebra in the first year, from what I remember, vector spaces were always defined over a field, which was in every single concrete example equal to either $\mathbb{R}$ or $\...
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An example of a division ring $D$ that is **not** isomorphic to its opposite ring

I recall reading in an abstract algebra text two years ago (when I had the pleasure to learn this beautiful subject) that there exists a division ring $D$ that is not isomorphic to its opposite ring. ...
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Which number fields can appear as subfields of a finite-dimensional division algebra over Q with center Q?

I have some idle questions about what's known about finite-dimensional division algebras over $\mathbb{Q}$ (thought of as "noncommutative number fields"). To keep the discussion focused, let's ...
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What do we lose passing from the reals to the complex numbers?

As normed division algebras, when we go from the complex numbers to the quaternions, we lose commutativity. Moving on to the octonions, we lose associativity. Is there some analogous property that we ...
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What are some real-world uses of Octonions?

... octonions are the crazy old uncle nobody lets out of the attic: they are nonassociative. Comes from a a quote by John Baez. Clearly, the sucessor to quaterions from the Cayley-Dickson process is ...
21
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What is this 2D division algebra?

Consider the set $A$ of 2-tuples of real values $(a,b)$, equipped with an addition defined as $$ (a,b) + (c,d) = (a+c,b+d)$$ and multiplication defined as $$ (a,b) \times (c,d) = (ac+bd,ad-bc).$$ ...
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1answer
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what are the p-adic division algebras?

Is there a classification of division algebras over $\mathbb{Q}_p$? There are field extensions of $\mathbb{Q}_p$, but are there any others? In particular, I want to know if they are all commutative. ...
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Is my paper on a number system that allows arithmetic on 3D vectors useful?

I have constructed a number system similar to the quaternions, but with three dimensions, not four, ie vectors of the form $(x, y, z)$. It has fairly well-behaved multiplication and division and every ...
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Do Hopf bundles give all relations between these “composition factors”?

Write a fiber bundle $F\to E\to B$ in short as $E=B\ltimes F$ (in analogy with groups). (This is not necessary, but: given another bundle $X\to B\to Y$, we can write $E=(Y\ltimes X)\ltimes F$, but ...
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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. I ...
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Quaternion Rings

Let $R$ be a commutative ring. Define the Hamilton quaternions $H(R)$ over $R$ to be the free $R$-module with basis $\{1, i, j, k\}$, that is, $$H(R)=\{a_0+a_1i+a_2j+a_3k\;\;:\;\;a_l\in R\}.$$ and ...
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1answer
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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 $K=k[\sqrt{\delta}]$...
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1answer
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An example of noncommutative division algebra over $Q$ other than quaternion algebras

Could anyone please show me an example of finite dimensional noncommutative associative division algebra over the field of rational numbers $Q$ other than quaternion algebras?
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Brauer group of a field of rational numbers

Can we say anything about Brauer group of $\mathbb{Q}$? And how can we construct it?
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238 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. ...
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2answers
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What does $(a,b)_{\zeta}$ correspond to in $\mathrm{Br}(\mathbb{Q}_p)=\mathbb{Q}/\mathbb{Z}$

Let $p$ be a prime number, let $\mathbb{Q}_p$ be the local field, by Hensel's lemma, we know it has $p-1$-th roots of unity, let $\zeta$ be a fixed primitive $p-1$-th root of unity in $\mathbb{Q}_p$. ...
7
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1answer
261 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 $k$-...
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questions about double centralizer theorem

An important fact in the theory of central simple algebras is the double centralizer theorem, which says: if $k$ is a field, $A$ is a $k$-algebra, $V$ is a faithful semisimple $A$-algebra, then $C(C(A)...
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Nilpotent matrix over a division algebra

Suppose I have an $n\times n$ nilpotent matrix $A$. If the entries are from any field, then I can show that all eigenvalues are zero and the trace is zero. Indeed, if we consider the algebraic closure ...
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Finite dimensional central division $\mathbb K$-algebra as a subalgebra of a matrix $\mathbb K$-algebra

The question is as follows: A finite-dimensional central division $\mathbb K$-algebra $D$ is a $\mathbb K$-algebra isomorphic to a subalgebra of $M_r(\mathbb K)$ if and only if $\dim_{\...
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1answer
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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 ...
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1answer
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Extending valuation to division algebra over a non-Archimedean local field?

Let $D$ be a division algebra over a non-Archimedean local field $K$. I would like to extend the discrete valuation on $K$ to $D$. For any $x \in D$, the subfield $K(x)$ of $D$ has a unique ...
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1answer
161 views

Hasse invariants under extension of scalars

Let $K\subset L$ be finite extensions of $\Bbb{Q}$. Background. Let $D$ be a finite dimensional division algebra with center $K$. Its class in the Brauer group $Br(K)$ then maps injectively into the ...
5
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1answer
316 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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1answer
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Dimension of $End(V)$ with $V$ countable dimension irreducible module over a complex algebra

Let $A$ be a $\mathbb{C}$-algebra and $V$ be an irreducible $A$-module with countable dimension. What is the dimension of $End(V)$ as $A$-module? Note that every endomorphism must be injective and ...
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Galois theory for (non-commutative) division rings

Is there a 'Galois theory' with fields replaced by (non-commutative) division rings? I have googled this, and it seems that there are known results in that direction, for example, this paper which ...
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Degree $n$ extension of local field splits degree $n$ division algebra

I am trying to write an article which is pretty self-contained on the number theory side, and would like to use the following result: Let $K$ be a local field, $n > 1$ a natural number, $D$ a ...
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1answer
626 views

Is every endomorphism of the quaternion ring surjective?

Is the quaternion ring an EAS Division ring? An EAS Division ring is a ring $D$ such that each endomorphism of $D$ is surjective. I know that $\mathbb{R}$ and $\mathbb{Q}$ are EAS Division rings.
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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. (...
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3answers
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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 ...
4
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1answer
95 views

Division ring as a $K$-algebra.

I want to solve the following question: Suppose that the division ring $\Delta$ is a $K$-algebra with $(\Delta:K)$ finite. Prove that $\Delta=K$ if $K$ is algebraically closed. Deduce that if $K$ ...
4
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1answer
250 views

normed division algebra

Can we prove that every division algebra over $R$ or $C$ is a normed division algebra? Or is there any example of division algebra in which it is not possible to define a norm? Definition of normed ...
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1answer
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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 ...
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1answer
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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^*...
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2answers
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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_\...
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1answer
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About a proof of the Frobenius Theorem on Division Algebras

I'm trying to understand a proof of the Frobenius Theorem on Division Algebras, but my knowledge of the relevant mathematics isn't really up to scratch. The proof I'm reading is this one. I'm not ...
4
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1answer
97 views

Central simple algebras give rise to algebraic groups

Let $D$ be a central division algebra over a field $k$ of dimension $n^2$. I have heard that the functor $$R \mapsto (D \otimes_k R)^{\ast}$$ going from commutative $k$-algebras to groups is ...
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2answers
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Alternative proof of Wedderburn's little theorem

I have this exercise where I'm proving: "Every finite division ring is a field". I need only a part (c) of it: (a) show that a subalgebra of a finite dimensional central division algebra is a finite ...
4
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1answer
114 views

Extension of $\mathbb{Q}[i]$

In this question, a ring is not supposed to be commutative. Let $K=\mathbb{Q}[i]$, and $A$ a subalgebra of $\mathcal{M}_n(K)$ (the algebra of $n \times n$ matrices). We suppose $A$ is a division ring....
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77 views

finite division algebras over a field

A theorem of Wedderburn says that finite division ring is field. Here, ring means "ring with unity and is also associative". In particular, finite division (associative) algebras are fields. I was ...
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0answers
75 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 ...
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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,...
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2answers
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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?...
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3answers
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Sub division rings of division rings

Below $^\ast$ denotes "nonzero elements of". There is a problem in Jacobson's Basic Algebra 1, there is a problem to this effect: if $S$ is a subdivision ring of $\mathbb{H}$ such that $S^\ast$ is a ...
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1answer
283 views

Do the octonions form a field?

The octonions are a noncommutative nonassociative normed division algebra over $\mathbb{R}$. Multiplication distributes over addition. Somehow, the existence of a norm implies the existence of ...
3
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2answers
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Tensor product of fields

Suppose $D$ is a finite dimensional skew field over the field $K$. Futher, take $x \in D\setminus K$ and let $L=K(x)$. My question: is $D\otimes_K L$ a field? I think not. However I can't seem to ...
3
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2answers
492 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 ...
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The dimension of a division ring over its center is square. [closed]

Let $D$ be a division ring and let $K$ be the center of $D$. Assume $\dim_K(D)<\infty$. Why is $\dim_K(D)$ a square?
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A proper subring of an 4-dimensional division F-algebra is a field.

I'm stuck with this thing: Let $F$ be a field of characteristic $\neq 2$, let $D$ be a 4 dimensional noncommutative division algebra over $F$. For $x\in D\smallsetminus F, F[x]$ is a field of ...