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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.