# 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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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 $\... 1answer 5k views ### An example of a division ring$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$that is not isomorphic to its opposite ring. ... 1answer 729 views ### Which number fields can appear as subfields of a finite-dimensional division algebra over Q with center Q? I have some idle questions about what's known about finite-dimensional division algebras over$\mathbb{Q}$(thought of as "noncommutative number fields"). To keep the discussion focused, let's ... 2answers 2k views ### What do we lose passing from the reals to the complex numbers? As normed division algebras, when we go from the complex numbers to the quaternions, we lose commutativity. Moving on to the octonions, we lose associativity. Is there some analogous property that we ... 5answers 6k views ### What are some real-world uses of Octonions? ... octonions are the crazy old uncle nobody lets out of the attic: they are nonassociative. Comes from a a quote by John Baez. Clearly, the sucessor to quaterions from the Cayley-Dickson process is ... 1answer 516 views ### What is this 2D division algebra? Consider the set$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 $$(a,b) \times (c,d) = (ac+bd,ad-bc).$$ ... 1answer 831 views ### 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. ... 1answer 329 views ### 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 ... 0answers 137 views ### 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 ... 2answers 1k 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. I ... 1answer 3k views ### 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 ... 1answer 364 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$K=k[\sqrt{\delta}]$... 1answer 1k views ### 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? 3answers 2k views ### Brauer group of a field of rational numbers Can we say anything about Brauer group of$\mathbb{Q}$? And how can we construct it? 2answers 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. ... 2answers 109 views ### 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$\mathbb{Q}_p$. ... 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$k$-... 1answer 390 views ### 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)...
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 ...