1
vote
0answers
48 views

Tensor product of two simple modules

Let $k$ be a field of arbitrary characteristic. Suppose that $A$ and $B$ are finite dimensional $k$-algebras. Let $S$ be a finite dimensional simple $A$-module and let $T$ be a finite dimensional ...
1
vote
0answers
28 views

Tensoring and retaining projectiveness

Let $A$be a unital associative ring, If $A$ is not a projective $A$-bimodule, however $A\otimes A$ is a projective $A$-bimodule, can we conclude that $A\otimes A \otimes A$ is also projective?
5
votes
1answer
66 views

Monoidal categories and tensor products

Does the multiplication $-\square-$ biendofunctor in a Monoidal category, $\mathfrak{C}$ necessarily commute with coproducts? This is true in some familiar categories, such as $_RMod$, $Grp$ $CRings$ ...
3
votes
1answer
86 views

Criterion for Simpleness of Modules over an Algebra

Let $k$ be a (if necessary algebraically closed) field. Let $A$ be a finite-dimensional $k$-algebra and $M$ be an $A$-left-module. Consider $M$ to be a right-module over $B = \mathrm{End}_A(M)$. ...
8
votes
1answer
275 views

Tensor product of simple modules

Let $M$ a right simple module and $N$ be a left simple module over a ring $R$. My questions are: How can we describe $M \otimes_R N$ explicitly? Well, I guess that it is a quotient of $R$ by a sum of ...
3
votes
1answer
194 views

An $(R,S)$-bimodule is a left $R \otimes_k S^{\text{op}}$-module

Let $k$ be a commutative ring, and let $R,S$ be $k$-algebras. To me "$R$ is a $k$-algebra" means that $R$ is a $k$-module such that $a(rs)=(ar)s=r(as)$ for all $a\in k$ and $r,s \in R$. Let $M$ be a ...
2
votes
1answer
86 views

Question concerning an isomorphism between a tensor product and a finite-dimensional K-algebra.

In the book Skew Fields, by P.K.Draxl, at page 60, states there as lemma 2: Let $A,B,C$ be finite-dimensional $K$-algebras such that $|C:K|\leq|A:K||B:K|$ and let $f: A\rightarrow C$, and $g: ...
1
vote
1answer
74 views

Relation between free algebras and polynomials

Given a vector space $V=\mathbb{F}^{d}$, the free algebra, or tensor algebra, of $V$ is $T\left(V\right)=\oplus_{n\geq0}V^{\otimes n}$. Now, it is stated everywhere, that this is exactly the algebra ...