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Let $\mathbb{F}$ be a field and let $A$ be an associative unital $\mathbb{F}$-algebra. Is there a criterion to let me know if $A$ is isomorphic to the algebra $\mbox{End}(\mathbf{V})$ of endomorphisms for some $\mathbb{F}$-vector space $\mathbf{V}$? Generalizations to $A$ being an associative unital ring and $\mathbf{V}$ an Abelian group or similar are welcome. Answers for particular cases $\mathbb{F}\in\{\mathbb{R},\mathbb{C}\}$ are also appreciated. Thank you.

This is a repetition of the following question I posed on MathSE days ago which has no answers so far. As I am no expert in algebra, so I thought maybe it is a research level question.

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    $\begingroup$ If $A$ is finite-dimensional the condition is that $A$ is semisimple and has a unique simple module $M$ such that $\text{End}_A(M) \cong F$. Are you interested in the infinite-dimensional case? Here the endomorphism algebras are a bit weird. What sort of $A$ are you interested in, and what do you know about them? $\endgroup$
    – Qiaochu Yuan
    Jul 27, 2016 at 22:36
  • $\begingroup$ Thanks for the comment. The question is as stated, in any dimension. But if you provide a satisfactory answer in finite dimensions then it is already something. The point is, I don't know a priori what class of algebras can appear as endomorphism algebras, so I don't know anything about them. Moreover, I am looking for a criterion in terms of the algebra itself rather than in terms of (the category of) its modules. Otherwise I could have just tautologically required the existence of a surjective faithful representation. $\endgroup$
    – Bedovlat
    Jul 27, 2016 at 23:00
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    $\begingroup$ If your actual question is how to understand the algebras $\text{End}(V)$, this is straightforward. If you pick a basis of $V$, then with respect to that basis, the elements of $\text{End}(V)$ can be written as column-finite square matrices (matrices indexed by the basis each of whose columns contain finitely many entries). $\endgroup$
    – Qiaochu Yuan
    Jul 28, 2016 at 0:15
  • $\begingroup$ Suppose you have a full algebra of column finite matrices. First thing you notice is that $\dim A\in\{\mathbb{N}^2,\infty\}$. And for every given dimension (including cardinality of infinity) you have only 1 isomorphism class. But then, what are the abstract algebraic properties of this algebra? Ideals, automorphisms.. How uniquely they determine $A$ for a given dimension? $\endgroup$
    – Bedovlat
    Jul 28, 2016 at 6:39
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    $\begingroup$ If you want a characterization without using modules, let $A$ be finite dimensional. The condition is that $A$ is simple (and therefore semi-simple, since $A$ is artinian) and has trivial Bruaer Class. The later condition can't really be simplified, but is cohomological in nature (it is a class in $H^2(F; \mathbb{G}_m)=H^3(F; \mathbb{Z})$. For your last comment, I think you are confused, since $|\mathbb{N}^2|=|\mathbb{N}|=\infty$. Infact this is true for any infinite cardinal. For ideals and automorphisms, the information is very classical: $\endgroup$
    – Pax
    Jul 28, 2016 at 14:05

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If $V$ is a vector space, the two-sided ideals of $E:=End(V)$ form a chain. The nonzero ideals are exactly those of the form $I_{\alpha}=\{\varphi\in E : {\rm dim}({\rm im}(\varphi))<\alpha \}$, where $\alpha$ is an infinite cardinal.

Thus, for instance, if $V$ is a vector space of countable dimension, then there are three ideals in $E$. Viewing $E$ as the ring of column-finite matrices, these ideals are exactly the zero ideal, the ideal consisting of matrices with only finitely many nonzero crows, and $E$ itself.

The one-sided ideals are much more complicated. In particular, the Jacobson radical is the zero ideal.

The ring $E$ is von Neumann regular; meaning that for every $x\in E$ there exists some $y\in E$ with $xyx=x$.

The ring $E$ is clean, meaning that for every $x\in E$ there exists some unit $u$ and idempotent $e$ such that $x=e+u$.

There are many other purely ring theoretic properties that such endomorphism rings have. A good source of information is Lam's "A First Course in Noncommutative Rings".

I know of no simple criterion for being an endomorphism ring of a vector space over a field. (That's not to say there isn't one.)

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    $\begingroup$ One additional thought. The base field of the vector space is equal to the center of $E$. If you view $E$ as a topological ring (with the finite topology) you can decompose the identity into a summable family of indecomposible, orthogonal idempotents. This should help in viewing $E$ as an endo-ring. $\endgroup$
    – Pace Nielsen
    Aug 25, 2016 at 13:46

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