7

The classification is trivial for a $\,C_1$-field $K$ (that is, such that any homogeneous polynomial in $K[x_1,\ldots ,x_n]$ of degree $<n$ has a nontrivial zero): the only such division algebras are the finite extensions of $K$. For the proof you just need the notion of reduced norm, which can be explained in a reasonably elementary way (see Central ...

5

The augmentation ideal is finitely generated as a left (or right) ideal if and only if the group is finitely generated. It is obvious the augmentation ideal is generated as an abelian group by all elements of the form $g-1$ with $g\in G$. Then from the computation $ab-1=a(b-1)+a-1$ one easily deduces by induction on word length that if $S$ generates $G$, ...

answered Mar 30 at 0:23

Benjamin Steinberg

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4

In
Belitskii, Genrich; Lipyanski, Ruvim; Sergeichuk, Vladimir V., Problems of classifying associative or Lie algebras and triples of symmetric or skew-symmetric matrices are wild, Linear Algebra Appl. 407, 249-262 (2005). ZBL1159.17304.
it is proved that the classification of local commutative associative algebras $A$ over an algebraically closed field of ...

3

Here is an example if we interpret all direct sums as internal direct sums.
Example. Let $R$ be a discrete valuation ring with uniformiser $\pi$ and fraction field $K$. Let $X = R^{(\mathbf N)}$, and let $T_i$ be the free rank $1$ submodule with basis $\pi e_{i+1}-e_i$. Then the natural map
$$\bigoplus_{i=1}^n T_i \to X$$
is injective with image $T_{\leq n} ...

answered Mar 12 at 23:53

R. van Dobben de Bruyn

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3

See "Seven Lectures on the Universal Algebraic Geometry" by
Boris Plotkin (Hebrew University). The text is in arXiv. Since the OP does not explain what "algebraic geometry" means, here are some explanations. The point is that there are several statements in the classical algebraic geometry which make sense and are even true in much more general situations. ...

3

There are already infinitely many isomorphism classes of finite-dimensional commutative associative (nilpotent) $\mathbb{C}$-algebras of rank $\geq 6$. See for example Suprunenko and Tueshkevich's book on commutative matrices (it's in Russian, I'm not sure if there is an English translation of it) for the construction. So, the classification question is ...

2

i) A monoid
ii) Representations of a quasi-Hopf algebra

2

Yes.
Let $A=kQ/I$ be a quiver algebra with $n$ simple modules and finite global dimension such that $\text{rad}^nA\neq0$. For a vertex $i$ of the quiver $Q$, I'll use $e_i$ to denote the corresponding primitive idempotent of $A$, and $S_i$ to denote the corresponding simple (right) $A$-module.
Since $\text{rad}^nA\neq0$, there is a path of length $n$ that ...

2

The problem is still ill-posed, because the set of nilpotent bounded operators is not a vector space. (The sum of two nilpotents need not be nilpotent.) Let's interpret the question as: which algebras admit faithful representations as algebras of nilpotent bounded operators?
Note that "nilpotent" usually means $T^n = 0$ for some $n$, not $T^2 = 0$. But ...

2

I put in a remark about this question back in 2013. I'm going to expand on that for emphasis. Let me shamelessly quote myself, from the expository paper http://www.math.uchicago.edu/~may/PAPERS/mayi.pdf
where I describe the coining of the word operad.
``To these ends, I consciously sacrificed all-embracing generality: many types of algebras defined by ...

1

$\newcommand{\Ann}{\operatorname{Ann}}$Let $A=K[x]/(x^n)$ and $I=\langle x^a\rangle$ and $J=\langle x^b\rangle$ for $1 \leq a \leq b \leq n-1$. Their intersection is nonzero.
$\Ann(I)=\langle x^{n-a}\rangle, \Ann(J)=\langle x^{n-b}\rangle$ and their radicals are $\langle x^1\rangle$, which is maximal.
$I+J=\langle x^{\min(a,b)}\rangle$ is indecomposable.

1

For $k=\infty$, this algebra appears when studying integrable representations of level 1 of the Lie algebra $\widehat{\mathfrak{sl}}_2$, see, for example, discussion in Section 2 of
A. V. Stoyanovsky, B. L. Feigin. Functional models for representations of current algebras and semi-infinite Schubert cells. Functional analysis and its applications, 28 (1994),...

1

The answer is wrong, one needs the ring to be noetherian, it's false in general.
For instance, take a local ring $A$ with a finite type maximal ideal, then one has that $\dim(gr(A))=\dim(gr(\hat{A}))$ and for such a ring the completion is noetherian, so then you get that $\dim(gr(A))=dim(\hat{A})$.
But at the same time, by considering valuation rings, you ...

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