All Questions
5 questions
9
votes
0
answers
366
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A characterisation of symmetric algebras using Hochschild (co)homology
A finite dimensional (connected if needed) $K$-algebra $A$ over a field $K$ is called symmetric when $A \cong Hom_K(A,K)$ as $A$-bimodules. Symmetric algebras are Frobenius algebras and include for ...
6
votes
1
answer
448
views
Hochschild cohomology of certain local algebras
Let $K$ be a field and $A$ the algebra $K\langle x_1,...,x_n\rangle/J^m$ for $n \geq 1$ and $m \geq 2$, where $K\langle x_1,...,x_n\rangle$ is the non-commutative polynomial ring in $n$ variables over ...
2
votes
1
answer
257
views
First Hochschild cohomology of $A=K[x]/(x^n)$
Given the algebra $A=K[x]/(x^n)$ for some field $K$ and natural number $n \geq 2$ with enveloping algebra $A^e=A \otimes_K A$.
It is easy to see that the 1. Hochschild cohomology of $A$ is nonzero ...
2
votes
0
answers
74
views
Units in the (stable) center of a Frobenius algebra [duplicate]
Let $A$ be a Frobenius algebra with center $Z(A)$ and $I\subset Z(A)$ the ideal of elements in the image of some $A$-bimodule map $A\rightarrow A\otimes A\rightarrow A$, where the second map is ...
1
vote
0
answers
141
views
Question on vanishing Hochschild cohomology
Recall that for an $K$-algebra $A$ with $A^e:=A^{op} \otimes_K A$ the Hochschild cohomology is defined as $HH^n(A,M):=Ext_{A^e}^n(A,M)$.
Question:
Is there a finite dimensional selfinjective ...