July 20th, 2018, 02:32 PM  #41 
Senior Member Joined: Oct 2009 Posts: 733 Thanks: 247  It might very well be a definition. All ideals show up as kernels of a homomorphism.
Last edited by skipjack; July 20th, 2018 at 03:54 PM. 
July 20th, 2018, 04:37 PM  #42  
Senior Member Joined: Aug 2012 Posts: 2,157 Thanks: 631  Quote:
Thanks Micrm@ss for your clarifying remarks on modding out by congruences. Last edited by Maschke; July 20th, 2018 at 04:39 PM.  
July 20th, 2018, 04:53 PM  #43  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 307 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry  Quote:
This doesn't make sense. $(J:I)$ is not a ring, so it doesn't have any ideals. The question is just to show that $J$ is a subset of $(J:I)$ (i.e. to show if $j \in J$, then $j \in (J:I)$). It is true that $J$ is actually a subgroup of $(J:I)$, but this is immediate once you show it's a subset. If you don't see why this is immediate, have a think about it.  
July 26th, 2018, 12:40 PM  #44  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
Ideal: Subring of R closed wrt left and right multiplication by r (if a in I and r in R, ar and ra in I). Needed for second part of proof.  
July 26th, 2018, 02:29 PM  #45 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
b) Show that Hom$\displaystyle _{R}$(R/I,R/J) $\displaystyle \cong$ (J:I)/J) via the map $\displaystyle \phi \mapsto \phi(1 + I)$ Hom$\displaystyle _{R}$(R/I,R/J) is, by definition, an A module. http://www.ucl.ac.uk/~ucahjlo/teachi...chapter31.pdf def 3.1.21 or Hom$\displaystyle _{R}$(M,N) = {$\displaystyle \alpha $: M $\displaystyle \rightarrow$ N  $\displaystyle \alpha$ is an A module homomorphism} $\displaystyle (\alpha + \beta)m = \alpha(m) + \beta(m)$ and $\displaystyle a\alpha(m) = \alpha(am)$ make Hom$\displaystyle _{R}$(M,N) an A module.** Now the baffling question: how do you map a map to a quotient ring, ie given $\displaystyle \phi \epsilon$ Hom$\displaystyle _{R}$(R/I,R/J), what the hell does $\displaystyle \phi \mapsto \phi(1 + I)$ mean? Definition of the terms should have been brought up at the beginning. 5 pages later we still don't know what $\displaystyle \phi \mapsto \phi(1 + I)$ means. It could mean (1+I) times whatever $\displaystyle \phi$ maps m to, or whatever $\displaystyle \phi$ maps (1+I) to. I have grown to detest Abstract algebra. But I can see a way to make a fortune out of it. 1) Short sell a private school for 100mil. 2) Make Abstract algebra a requirement for all incoming freshman. This takes out the freshman class and hence the school, which you can then buy for 5 to meet the short for a net profit of 100mil  5. (the dollar sign screwed things up). Seriously, I could never establish a simple set of definitions and a structure which would act as a foundation to understand Abstract algebra. It was definition upon definition upon definition, which I can't deal with but obviously others can. I am not a pianist. I thought if I learned Twinkle Twinkle Little Star I could be a pianist. But I did answer part a) of OP, quite beautifully actually. ** 25:00 into lecture Last edited by zylo; July 26th, 2018 at 02:36 PM. 
July 26th, 2018, 03:52 PM  #46  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 307 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry  Quote:
Ideals are not in general subrings. In fact, the only ideal of a ring $R$ that is also a subring is $R$ itself.  
July 27th, 2018, 08:27 AM  #47  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
I have given definition of Ideal: Subset I of Commutative Ring and 1) Additive subgroup 2) For all a in I and r in R, ar in I. There are corresponding definitions for left (ar in R) and right (ra in R) Ideals Ideal  from Wolfram MathWorld Earlier I gave definition of normal subgroup N of G if gN=Ng. To elaborate, and shed more light on the standard definition. gN = {gn  n $\displaystyle \epsilon$ N} any g in G Ng = {ng  n $\displaystyle \epsilon$ N} any g in G gN=Ng true if G is Abelian. Also true if for any gn there is an n' in N st gn=n'g, or gng$\displaystyle ^{1}$ is an element of N for any g in G and any n in N, conjugation.  
July 27th, 2018, 11:14 AM  #48  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 307 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry  Quote:
In particular, this is how $xI$ is meant when $(J:I)$ is defined as $\{x \in R \mid xI \subset J\}$. If they were using $xI$ to mean $x + I$ instead, this would be a useless definition: $(J:I)$ would the empty set (and so not even an ideal) whenever $I \not \subseteq J$ and it would be $J$ whenever $I \subseteq J$.  
August 5th, 2018, 10:04 AM  #49  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
Quote:
(J : I) is a subgroup: If x $\displaystyle \epsilon$ (J : I) and y $\displaystyle \epsilon$ (J : I), then xi and yi $\displaystyle \epsilon$ J $\displaystyle \rightarrow$ xi+yj $\displaystyle \epsilon$ J $\displaystyle \rightarrow$ (x+y)i $\displaystyle \epsilon$ j $\displaystyle \rightarrow$ x+y $\displaystyle \epsilon$ (J : I) For any a in R, if xi $\displaystyle \epsilon$ J, a(xi) $\displaystyle \epsilon$ J $\displaystyle \rightarrow$ ax $\displaystyle \epsilon$ (J:I) because J is an Ideal. J $\displaystyle \subset$ (J : I): Substitute j for x in (J : I) = {x $\displaystyle \epsilon$ R  xI $\displaystyle \subset$ J} ji $\displaystyle \epsilon$ to J because J is an Ideal. $\displaystyle \rightarrow$ j $\displaystyle \epsilon$ (J:I) Definition: Ideal I of a commutative ring R. I is a subgroup of R. For any a $\displaystyle \epsilon$ R and any i $\displaystyle \epsilon$ I, ai $\displaystyle \epsilon$ I.  
August 10th, 2018, 06:40 AM  #50 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 
It is certainly true that in a group we might write the operation as "+", or "*", or simply by "juxtaposition" but in a ring we have two operations already given as "+" and "*". An ideal uses the multiplicative operation, not the additive.


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