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32

Let $F$ be the center of the division ring $D$. Then $D$ represents its class in the Brauer group $Br(F)$. The opposite ring $D^{opp}$ represents the inverse element. The reason why for example the quaternions are isomorphic to their opposite algebra is that the quaternions are an element of order 2 in $Br(\mathbf{R})$, and hence equal to its own inverse in ...

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Step 1: The characteristic of $A$ is $2$ (Credit for this observation goes to Jyrki Lahtonen) The mapping $x\mapsto -x$ is an involution on $A^\times$. Since $\lvert A^\times\rvert = 2^n - 1$ is odd, it has a fixed point. So $a = -a$ for an $a\in A^\times$. Multiplication with $a^{-1}$ yields $1 = -1$. Step 2: $\lvert A\rvert = 2^n$ From the ...

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In an earlier version of this post I caused an accident by giving a wrong answer. Thanks to those who pointed out the error! Here is a little update: The ring $M_2(\Bbb F_2)$ of $2 \times 2$-matrices with entries in $\Bbb F_2$ is a non-commutative ring with 16 elements, because $$\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 0 & 1 ... 17 Noncommutative algebra is filled with examples of this. For example, take A=\begin{bmatrix}0&1\\0&0\end{bmatrix} and B=\begin{bmatrix}0&0\\0&1\end{bmatrix}. You have AB\neq 0 but BA=0. Rings in which ab=0 implies ba=0 are called reversible rings. That is a particularly strong condition, and is pretty interesting to study. I ... 17 You have come across the quaternions. They are numbers of the form$$a+bi+cj+dk$$where a,b,c,d\in\mathbb{R} and i, j, and k are symbols satisfying$$i^2=j^2=k^2=ijk=-1ij=k,\quad jk=i,\quad ki=jji=-k,\quad kj=-i, \quad ik=-j$$Multiplication of quaternions is non-commutative in general, but it is still associative. The conjugate of a ... 16 I think I have one. Let k be the field with 2 elements. Let R be the k-algebra with generators x, y and z, modulo the relations$$zx=xz,\ zy=yz,\ yx=xyz.$$It is not hard to see that monomials of the form x^i y^j z^k are a basis for R. We will call these the standard monomials. For any f \neq 0 in R, write f = \sum f_{ij}(z) x^i ... 13 Because my monograph: S. Fujita, Symmetry and Combinatorial Enumeration in Chemistry (Springer 1991) has been referred to as an endeavor of dealing with topics of organic chemistry on the basis of mathematics, I would like to add two monographs aiming at organic reactions and stereochemistry coupled with mathematics (group theory): S. Fujita, ... 12 Let R = L(\ell^2) the linear continuous operators on \ell^2. Define a,b \colon \ell^2 \to \ell^2 by$$ a(x_1, x_2, \ldots) = (x_2, x_3, \ldots) $$and$$ b(x_1, x_2, \ldots) =(0, x_1, x_2, \ldots) $$Then ab = 1 is the identity, hence invertible, but neither a nor b are units as a is not one to one, where b is not onto. 11 Apparently such rings do exist. After trying to construct one without much success, I did some googling and found the following article: Maxson, C. J. 1979. Rigid rings. Proc. Edinburgh Math. Soc., 21(2): 95–101. In it, the author uses the following definition: a ring R (with non-zero multiplication, not necessarily possesing a multiplicative ... 10 Here's a construction of a non-commutative ring with no non-trivial automorphisms. It is not rigid though (rigid=no nontrivial endomorphisms. See Dejan's answer). First, define sets S_1,S_2\subset\mathbb{N}, where S_1 is the set of square-free numbers whose prime factors are equal to 1 mod 4 and S_2 is the set of square-free numbers equal to 3 mod 4. ... 9 Let K be a field. Consider the free K-algebra on two generators x,y. Its unit group is K^*, hence commutative, but x,y don't commute. Here is a more explicit example: Consider the ring of upper-triangular 2 \times 2-matrices over \mathbb{F}_2. It has 8 elements and it is in fact the smallest noncommutative ring. The unit group has just two ... 9 This question has been studied: Desmond MacHale, Community in Finite Rings, The American Mathematical Monthly, Vol. 83, No. 1 (Jan., 1976), pp. 30-32, online Theorem 1 states that if R is a non-commutative ring, then P(R) \leq 5/8, with equality if and only if (R:Z(R))=4. The proof is quite similar (but not equal) to the case of groups. Besides, ... 9 Organic chemistry S. Fujita's "Symmetry and combinatorial enumeration in chemistry" (Springer-Verlag, 1991) is one such endeavor. It mainly focuses on stereochemistry. Molecular biology and biochemistry A. Carbone and M. Gromov's "Mathematical slices of molecular biology" is recommended, although it is not strictly a book. R. Phillips, J. Kondev and J. ... 8 If you're interested in number theory, the commutative algebra will be much more helpful to you in the immediate future. Knowing some amount of noncommutative algebra (i.e. Wedderburn theory) will be needed later in your mathematical life, but right now you need to know things about Dedekind domains. This entails knowing what the words Krull dimension, ... 7 As indicated in the comments something is wrong with the isomorphism from (1,-4)_K to (1,1)_K. Would \phi:i_1\mapsto i_2, j_1\mapsto j_2(1+\sqrt5 i_2) work? Then you have$$ \phi(j_1i_1)=j_2(1+\sqrt 5 i_2)i_2=j_2i_2(1+\sqrt5 i_2)=-i_2j_2(1+\sqrt5 i_2)=\phi(-i_1j_1) $$as well as$$ \phi(-4)=\phi(j_1^2)=j_2(1+\sqrt5 i_2)j_2(1+\sqrt5 ...

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Yes. A lot of interesting number theory is involved. The Brauer group classifies division algebras with a given center, and in class field theory that plays a big role (when the center field is a number field). See for example this question and this. In addition to number theory, the topic is interesting on its own merits. Over more complicated center ...

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Even simpler: Take any direct product of two nonzero rings $R\oplus S$. Then $R\oplus 0$ and $0\oplus S$ are nonzero ideals multiplying to zero. One definition for prime ring is a ring in which this does not happen, so any non-prime ring will suffice. Also, if you believe in nonzero nilpotent ideals, then you would easily find an example: If $I^n=\{0\}$ ...

6

At the moment there is a problem with your question: $\text{GL}_n(\Bbb{C})$ is not a field. The notion of being algebraically closed only applies to fields. In fact $\text{GL}_n(\Bbb{C})$ is not even a ring because the ring axioms mean that we need it to be closed under addition. But then $diag(1,\ldots 1) + diag(-1,\ldots,-1) = 0$ so it is not a ring.

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In a recent paper, I adapted some of the terminology of Kochen and Specker's original paper to a more ring-theoretic context. I would refer to your first type of morphism as a morphism of partial $\mathbb{Z}$-algebras from $R$ to $S$, or even better as a morphism of partial rings. The point is that every ring has the underlying structure of a partial ring ...

6

As said Robert Israel, they are called skew fields. But the study of skew fields is very different from commutative fields. For example, if your field is the Quaternions $\mathbb H$, and you consider the polynomial with real coefficients $\rm X^2 + 1$, it has more than 2 roots in $\mathbb H$ !

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Take the polynomial ring over a field in two noncommuting variables: $\mathbb{F}\langle x,y\rangle$. I've put them in langle/rangle brackets to remind everyone that they do not commute with each other. Then $x+y$ has your property. Generally any free object on two generators is going to do something like that, because free objects don't satisfy any ...

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There is no function $f$ such that $f(A^B)=B^A$ (for all real numbers $A,B$ such that $A^B$ and $B^A$ are defined). If there were such a function $f$, then for example we would have $$36=6^2=f(2^6)=f(4^3)=3^4=81,$$ which is clearly not the case. Now, for any positive $A\neq 1$, there is a function $f_A$ such that $f_A(A^B)=B^A$ for any $B$. In particular, ...

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There are a lot of ways to see this. In $\S 1.9.2$ of my noncommutative algebra notes, I discuss the Invariant Basis Number property of a ring $R$, namely that the rank of a finitely generated free (say left) $R$-module is a well-defined invariant. I show the following: 1) The ring $R = \operatorname{End}_k(V)$ does not satisfy IBN. (This is probably ...

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No, the universal enveloping algebra can have a nontrivial center even if the Lie algebra itself has only $0$ as center. The (or at least a) concept you want to look up is "Casimir element; see for example http://en.wikipedia.org/wiki/Casimir_element .

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When one thinks of a noncommutative ring with unity (at least I), tend to think of how I can create such a ring with $M_n(R)$, the ring of $n \times n$ matrices over the ring $R$. The smallest such ring you can create is $R=M_2(\mathbb{F}_2)$. Of course, $|R|=16$. Now it is a matter if you can find a even smaller ring than this. Of course, the subring of ...

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The answer in general is no (see francis-jamet's answer) but there is at least one important case where the answer is affirmative. If $S\subset R$ is an integral extension and $\frak m$ is a maximal ideal in $R$ then ${\frak m}\cap S$ is a maximal ideal in $S$. This situation is very relevant for instance in algebraic number theory where $R$ and $S$ can be ...

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We assume $R$ and $S$ are unital. Hence there exists the canonical homomorphism $k \rightarrow R$. Hence by restricting the actions of $R$ on $M$ to $k$, $M$ can be a $k$-module. Similarle $M$ can be a $k$-module through $S$. We assume the both $k$-module structures on $M$ coincide. Let $T = End_k(M)$ be the $k$-algebra of $k$-endomorphisms on $M$. For $r ... 4 If we fix a basis$e_1,…,e_d$for$V$, then$V^{\otimes n}$has a basis given by "words" of length$n$in the letters$e_1,…,e_d$. This allows you to construct an isomorphism between$T(V)$and the free non-commutative algebra in$n\$ variables. This construction is analagous to the (basis independent) construction of commutative polynomial rings as the ...

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