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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.
11
votes
Matrix representation of real *-algebras
You have to add the condition
$$\sum_{i} a_i^*a_i =0 \quad \Rightarrow \quad a_i =0 \quad \forall i.$$
If you consider $\ast$-algebras satisfying this condition, then $-1$ is not a sum of hermitean …
7
votes
Accepted
Infinite subfields of division algebras with finite center
It is not possible that $D^{\times}$ is a torsion group unless $D$ is a field.
Let $D$ be a division ring of characteristic $p$. Denote the center of $D$ by $Z(D)$.
Lemma 1: Every finite subgrou …
2
votes
Trace of the identity map in a projective module
For any homomorphism $\varphi : A \to \mathbb C$, $\varphi({\rm tr}({\bf 1}))$ will be the corresponding trace for the finite-dimensional vector space $M \otimes_A \mathbb C$, and hence be equal to th …
6
votes
Accepted
Is the set of polynomial sum of squares closed under limits?
The cone of sums of squares $\Sigma^2 \subset \mathbb R[x_1,\dots,x_n]$ is closed in the finest locally convex topology. This is equivalent to the assertion that the intersection of this cone with the …
6
votes
Ideals in a noncommutative ring such that their product is their intersection?
This is more a remark, since I do not directly answer the question. The statement in the question is true for all ideals $I,J$ (without the condition $I+J=R$) if and only if all ideals a idempotent, i …
8
votes
Infinite-dimensional normed division algebras
The associative case follows from Mazur's Theorem (see here). He proved that there are up to isomorphism precisely three Banach division algebras, namely $\mathbb R,\mathbb C$ and $\mathbb H$. This ap …
50
votes
Example for column rank $\neq$ row rank
It is a classical observation due to Nathan Jacobson that a division ring such that the set of invertible matrices over it is closed under transposition has to be a field, i.e. commutative.
The reaso …
17
votes
Accepted
Why does the Grothendieck group $K_0(R)$ of a ring not depend on our choice of using left mo...
First of all, right $R$-modules are the same as left $R^{op}$-modules. Hence you are asking whether the $K$-theory changes if you pass form $R$ to $R^{op}$. The answer is: It does not change.
The re …
7
votes
Is every polynomial a limit of polynomials in quadratic variables?
The dimension of the linear space of homogenous polynomials of degree $d$ is
$\binom{n+d-1}{d}$. The dimension of the space of homogenous polynomials of degree $d$ in squared linear variables is
$$\b …
1
vote
Commutative *-subrings of the noncommutative C*-algebra $B(l^2)$
A unital $*$-ring $A$ (commutative or not) is a subring of $B(H)$ if and only if for each $a \in A$, there exists a linear functional $\varphi \colon A \to \mathbb R$, such that
1) $\varphi(1)=1$ and …
0
votes
Commutative *-subrings of the noncommutative C*-algebra $B(l^2)$
Let $A$ be a (say finitely generated, unital) commutative complex $\star$-subalgebra of $B(H)$. Then, the self-adjoint elements form a real subalgebra $B:=A_h \subset A$, such that $B[i] = A$. Moroeve …
10
votes
Accepted
Generalization of finitely generated, finitely presented modules?
All these notions have been defined and studied long time ago. Serre called a module type $FL_n$ if it is finitely $n$-presented in your terminology. Type $FL_\infty$ and type $FL$ is used for finitel …
6
votes
Zero divisor conjecture for finite fields
The zero-divisor conjecture over $\mathbb Q$ is equivalent to the zero-divisor conjecture over the ring $\mathbb Z$ by clearing denominators. At the same time, the zero-divisor conjecture over $\mathb …
2
votes
Accepted
Doing Real Algebraic Geometry on *-Rings
Konrad Schmüdgen has set up a programme to develop analogues of the basic results in Real Algebraic Geometry in a setup of $\star$-algebras. See arxiv.org/abs/0709.3170 and for example arxiv.org/abs/0 …
12
votes
Accepted
Strong group ring isomorphisms
If $G$ is a finite abelian group, then $\mathbb C[G] = \lbrace f \colon \hat G \to \mathbb C \rbrace$, where $\hat G$ is the Pontrjagin dual of $G$. The isomorphism $g \mapsto g^{-1}$ translates into …