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 Apr9 awarded Yearling Feb19 comment Problem 1.24, Introduction to representation theory, Etingof My first comment contains a typo which is clarified in my second comment. Regular representation means $A\oplus A$ acts on itself by multiplication. $(1,1)$ is a cyclic vector obviously. Subrepresentations include $A\oplus 0$ and $0\oplus A$. As for the downvote, you'll need to ask whoever did it. Jan30 comment Definitions of valuations in terms of totally ordered group Thank you! What are some of the 'mild hypotheses' referred to in para 2? Jan30 accepted Definitions of valuations in terms of totally ordered group Jan30 asked Definitions of valuations in terms of totally ordered group Nov19 revised Problem 1.25 of Etingof: Indecomposable rep which is not cyclic deleted 82 characters in body Nov18 comment Problem 1.25 of Etingof: Indecomposable rep which is not cyclic I see what you mean now, and you're right. My apologies. Nov18 revised Problem 1.25 of Etingof: Indecomposable rep which is not cyclic added 2 characters in body Nov18 revised Problem 1.25 of Etingof: Indecomposable rep which is not cyclic added 2 characters in body Nov18 revised Problem 1.25 of Etingof: Indecomposable rep which is not cyclic added 242 characters in body Nov3 asked Problem 1.25 of Etingof: Indecomposable rep which is not cyclic Oct28 asked Prime ideals in non-commutative ring Oct27 comment Problem 1.24, Introduction to representation theory, Etingof I mean the regular representation of $A\oplus A$ Oct27 comment Problem 1.24, Introduction to representation theory, Etingof The representation $A \oplus A$ of $A$ has cyclic vector $(1,1)$ but is decomposable... Oct21 comment Why is axiom of choice needed? (Equivalent conditions for Noetherian) I think I see what it is now. It is not induction at all as I'm not proving statements... I'm just abusing the term 'doing things by induction', meaning I can't actually write down the 'statement to prove' in the n-th step in first order or whatever unless (n-1)th step (ie picking the submodule) has already been done. Am I right? Oct19 comment Can the induced function of non-zero $f \in R[X]$ be zero, when $R$ is an infinite non-integral domain? $R = \mathbb{Z}_4^{\mathbb{N}}$ with $f(X) = 2X(X-1)$ by the same token. I was clearly overthinking... Oct19 accepted Can the induced function of non-zero $f \in R[X]$ be zero, when $R$ is an infinite non-integral domain? Oct19 revised Can the induced function of non-zero $f \in R[X]$ be zero, when $R$ is an infinite non-integral domain? added 242 characters in body Oct19 asked Can the induced function of non-zero $f \in R[X]$ be zero, when $R$ is an infinite non-integral domain? Oct19 revised What is a quick proof that $f \in \mathbb{C}[X_1,\dotsc,X_n]$ is determined by its induced function on $\mathbb{C}^n$? added 130 characters in body; edited tags