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 Apr18 accepted Equivalence Of Definitions Of Prime Ideal In Ring Without $1$ Apr17 comment Equivalence Of Definitions Of Prime Ideal In Ring Without $1$ @rschwieb Right ok, I'll have a play around with that to check that I get everything. Thanks everyone for the help! Apr17 comment Equivalence Of Definitions Of Prime Ideal In Ring Without $1$ @rschwieb Cool, could you possibly explain the set where we assume that $(a)^2(b)^3\subset RaRbR$? I can't see why this follows? Thanks Apr17 comment Equivalence Of Definitions Of Prime Ideal In Ring Without $1$ That is if $1\notin R$ then surely we do not need to have $a\in aR$ or $b\in bR$? Apr17 comment Equivalence Of Definitions Of Prime Ideal In Ring Without $1$ How did you conclude that $(a)^2(b)^3\subseteq RaRbR$? Apr17 revised Can these two quotient groups be isomorphic? added 792 characters in body Apr17 answered Can these two quotient groups be isomorphic? Apr17 comment Equivalence Of Definitions Of Prime Ideal In Ring Without $1$ @ah11950 It does indeed! Apr17 comment Equivalence Of Definitions Of Prime Ideal In Ring Without $1$ @rschwieb No sorry the question is not online, I shall email my lecturer and hope that he gets back to me. Apr17 comment Equivalence Of Definitions Of Prime Ideal In Ring Without $1$ @ah11950 I believe we are/have taking/taken the same course! Apr15 asked Equivalence Of Definitions Of Prime Ideal In Ring Without $1$ Apr14 comment Modern book on Gödel's incompleteness theorems in all technical details I think that Peter Smyth's book is pretty good, logicmatters.net/igt Apr11 comment Proving The Diamond Lemma @fgp Sorry that was meant to be $P$. Yeah I think it's an induction step with a subtlety about making sure chains stop after nice numbers of steps or something, however the way I am asked to answer this question is using this $Q(a)$ and I have no idea where to proceed with it, every time I do I just end up back at the induction. Apr11 revised Proving The Diamond Lemma edited body Apr11 comment Proving The Diamond Lemma @fgp but condition 2) only applies if $a\rightarrow b_1$ and $a\rightarrow b_2$. If we have a chain $a\rightarrow a_1 \ldots \rightarrow b$ then we would need to apply condition 2) to each step of the chain right? I think that is what this $Q(a)$ is attempting to avoid? Apr11 comment Proving The Diamond Lemma @fgp but why does $b_1\twoheadrightarrow c_1$ give that $b_1\rightarrow c_1$? Apr11 comment Proving The Diamond Lemma @fgp why if two elements have a common predecessor does that violate 2)? Apr11 asked Proving The Diamond Lemma Apr8 comment functors on Zero-Object in $_RMod$-category @NajibIdrissi Sorry my notation was that $0$ was the zero morphism not the zero object Apr8 comment functors on Zero-Object in $_RMod$-category @ZhenLin Ok, thanks. Is there some sort of restriction I can put on the functor $F$ so that this is true. If I could say that $F$ maps the zero morphism to the zero morphism then I could show that $Id_{F(0)}=0_{F(0)}$ and I would be done right?