Sorry but I’m quite new to group algebras and even Latex so if this is all wrong I apologize. By $\mathbb CC_3$ I mean the group algebra of the cyclic group of order 3 in the complex numbers A group algebra is a ring so is it best to say $C_3$ has elements $e$, $a$ […]

The following is well-known. Proposition. Let $R$ denote a field and $A$ denote an $R$-algebra (not necessarily commutative). Then the set of $R$-algebra homomorphisms $A \rightarrow R$ form a linearly independent subset of the $R$-module of all $R$-module homomorphisms $A \rightarrow R$. Can this be generalized so that $R$ is only assumed to be a […]

Let $A,B,$ and $R$ be $k$-algebras. There is a functor from the category of $k$-algebras to the category of sets, called $\operatorname{Spec}A$, defined by $R\mapsto\operatorname{Hom}_{k\text{-alg}}(A;R)$. If $f: A\rightarrow B$ is a $k-$algebra homomorphism, then there is also the induced map $\operatorname{Spec}B(R)\rightarrow\operatorname{Spec}A(R)$ given by $\phi\in\operatorname{Spec}B(R)=\operatorname{Hom}_{k\text{-alg}}(B;R)\mapsto\phi\circ f\in \operatorname{Spec}A(R)=\operatorname{Hom}_{k\text{-alg}}(A;R).$ I’m trying to work out the following example. Let […]

Edit As Kimball point out, in the following question for me an ideal $I$ is a full $\mathbb{Z_p}$-lattice of $M_2(\mathbb{Q}_p)$ such that $$\lbrace\alpha \in M_2(\mathbb{Q}_p)\vert \alpha I \subset I \rbrace=M_2(\mathbb{Z}_P). $$ Let $p$ be a prime number and consider the ring, formed by the elements $$ \begin{pmatrix} a & b \\ pc & d \end{pmatrix}$$ […]

Let $K$ be a field and $A$ be a $K$-algebra. I know, if $A$ is artinain algebra, then by Krull-Schmidt Theorem $A$ , as a left regular module, can be written as a direct sum of indecomposable $A$-modules, that is $A=\oplus_{i=1}^n S_i$ where each $S_i$ is indecomposable $A$-module Moreover, each $S_i$ contains only one maximal […]

This question already has an answer here: Proving that $\mathbb R^3$ cannot be made into a real division algebra (and that extending complex multiplication would not work) 1 answer

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