rschwieb
Reputation
65,541
98/100 score
 Apr 26 awarded Nice Question Apr 22 comment are Orthogonal vectors superior to independent non Orthogonal ones? @Travis OK, thanks for answering. Apr 22 comment are Orthogonal vectors superior to independent non Orthogonal ones? @Travis Just curious: what sense of 'natural' are you thinking of? Apr 22 comment Principal Ideal using coordinates? @p.l Your confusion is palpable because I have no idea what you are talking about :) For a commutative ring with 1, $(a)=\{ar\mid r\in R\}$ there is no "under addition" about it. Why do you think $(a)$ is a summation? An ideal is a set, not a "summation" Apr 22 comment Principal Ideal using coordinates? What do you mean by "...using coordinates"? Do you mean "... In a product ring"? If you know the operations in the product ring then everything works exactly the same way as if you were talking about a single ring ( because we are talking about a single ring). Apr 21 awarded Revival Apr 21 comment Question about finite ring with more than one element; division ring The poster here is asking about a particular step, but not the entire question. The solution to the question itself actually appears elsewhere, though math.stackexchange.com/q/591725/29335 , for future reference. Apr 21 answered Question about finite ring with more than one element; division ring Apr 21 revised Finite rings without zero divisors are division rings. added 224 characters in body; added 18 characters in body Apr 20 comment Projective Modules, Wedderburn Rings, a course in ring theory The question also apparently does not deal with projective modules, nor Wedderburn rings directly. And "a course in ring theory" adds nothign to the title. Please consider making a more direct title as you edit. Apr 20 revised Motivating the Cross-Ratio and 'the ratio of ratio's' in $\mathbb{R}\mathbb{P}^2$ added 902 characters in body Apr 20 revised Motivating the Cross-Ratio and 'the ratio of ratio's' in $\mathbb{R}\mathbb{P}^2$ added 902 characters in body Apr 20 answered Motivating the Cross-Ratio and 'the ratio of ratio's' in $\mathbb{R}\mathbb{P}^2$ Apr 20 comment Reduced ring are SI? Cool, so it is not too hard. Apr 19 awarded Revival Apr 17 awarded Yearling Apr 16 comment Let $D$ be a principal ideal domain. Show that every proper ideal of $D$ is contained in a maximal ideal of $D$. Are you working with rings without identity or in the absence of the axiom of choice or something? Assuming identity and the axiom of choice, this is true for all rings, not just PIDs... Apr 16 comment $\mathbb{Z}_p \hookrightarrow R$ is it necessary $R$ commutative? @Chris Which part is not easy? Apr 15 answered Reduced ring are SI? Apr 14 comment Each element is invertible @MaryStar Both $baR$ and $(1-ba)R$ are proper, and would be contained in the unique maximal right ideal. Clearly the could not add up to $R$, so this is a contradiction.