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July 20th, 2018, 01:32 PM   #41
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Quote:
 Originally Posted by zylo kernel of a homomorphism is just an example of an ideal, not a definition.
It might very well be a definition. All ideals show up as kernels of a homomorphism.

Last edited by skipjack; July 20th, 2018 at 02:54 PM.

July 20th, 2018, 03:37 PM   #42
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 Originally Posted by Micrm@ss Now, your observation that we mod out with kernels of homomorphisms is correct in rings and groups, but it is morally wrong.

Thanks Micrm@ss for your clarifying remarks on modding out by congruences.

Last edited by Maschke; July 20th, 2018 at 03:39 PM.

July 20th, 2018, 03:53 PM   #43
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 Originally Posted by zylo xI $\displaystyle \subset$ J if x+i=j, i ∈ I and j ∈ J (J:I) = {j - i | i ∈ I and j ∈ J}
No, $xI \subset J$ if for all $i \in I$, there exists $j \in J$ such that $xi = j$.. So $(J:I) = \{x \in R \mid \text{for all} \: i \in I, \: \text{there exists} \: j \in J\ \: \text{such that} \: xi = j \}$.

Quote:
 Originally Posted by zylo 2) J $\displaystyle \triangleleft$ (J:I) J is a sub-group of (J:I) If i=0, (J:I) = {j-0 | 0 ∈ I and j ∈ J} J is an ideal of (J:I) if jr ∈ J for j ∈ J and r ∈ (J:I): j(j'-i') = jj' - ji' and ii' ∈ J and ji' ∈ I because I is an ideal of R
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, 11:40 AM   #44
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 Originally Posted by zylo Thought I'd take a shot at part a) of OP from definitions: Let R be a commutative ring with 1. If I,J $\displaystyle \triangleleft$ R, then define (J:I) = {x ∈ R| xI $\displaystyle \subset$ J). a) Show (J:I) $\displaystyle \triangleleft$ R and J $\displaystyle \triangleleft$ (J:I) xI $\displaystyle \subset$ J if x+i=j, i ∈ I and j ∈ J *** (J:I) = {j - i | i ∈ I and j ∈ J} 1) (J:I) $\displaystyle \triangleleft$ R closed: (j-i) + (j'-i') = (j+j') - (i + i'), ∈ (J:I). additive identity: i=0, j=0 additive inverse: (-j) + i Ideal: r(j-i)=rj=ri. rj ∈ J and ri ∈ I because I and J are ideals. 2) J $\displaystyle \triangleleft$ (J:I) J is a sub-group of (J:I) If i=0, (J:I) = {j-0 | 0 ∈ I and j ∈ J} J is an ideal of (J:I) if jr ∈ J for j ∈ J and r ∈ (J:I): j(j'-i') = jj' - ji' and ii' ∈ J and ji' ∈ I because I is an ideal of R
*** xI $\displaystyle \equiv$ x+I= {x+i | i $\displaystyle \epsilon$ I} for any x in R. Definition of coset.

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, 01:29 PM #45 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,535 Thanks: 108 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...chapter3-1.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 01:36 PM.
July 26th, 2018, 02:52 PM   #46
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Quote:
 Originally Posted by zylo *** xI $\displaystyle \equiv$ x+I= {x+i | i $\displaystyle \epsilon$ I} for any x in R. Definition of coset.
No... As I said, $xI = \{xi \mid i \in I\}$. You've given the correct definition of $x + I$, but this is a different set.

Quote:
 Originally Posted by zylo 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.
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, 07:27 AM   #47
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Quote:
 Originally Posted by cjem No... As I said, $xI = \{xi \mid i \in I\}$. You've given the correct definition of $x + I$, but this is a different set. Ideals are not in general subrings. In fact, the only ideal of a ring $R$ that is also a subring is $R$ itself.
xI is composition of x with I. In this case the composition is x+I.

I have given definition of Ideal:
Subset I of Commutative Ring and
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 sub-group 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, 10:14 AM   #48
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Quote:
 Originally Posted by zylo xI is composition of x with I. In this case the composition is x+I.
You're free to use notation however you wish. If you want to write $xI$ for $x + I$, go right ahead. Just be aware that when other people write $xI$ in this context, they mean $\{xi \mid i \in I\}$ and not $x + I$.

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, 09:04 AM   #49
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Quote:
 Originally Posted by cjem ...... Just be aware that when other people write $xI$ in this context, they mean $\{xi \mid i \in I\}$ and not $x + I$. ......
You are right. Thanks.

Quote:
 Originally Posted by zylo The OP question: Let R be a commutative ring with 1. If I, J $\displaystyle \triangleleft$ R then we define (J : I) = {x $\displaystyle \epsilon$ R | xI $\displaystyle \subset$ J} a) Show that (J : I) $\displaystyle \triangleleft$ R and J $\displaystyle \subset$ (J : I) b) Show that $\displaystyle Hom_{R}(R/I,R/J) \cong (J : I)/J$ via the map $\displaystyle \phi \mapsto \phi (1+I)$
(J : I) $\displaystyle \triangleleft$ R:
(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, 05:40 AM #50 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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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