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April 20th, 2018, 12:14 PM   #1
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Isomorphism

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 April 20th, 2018, 02:29 PM #2 Senior Member   Joined: Jan 2015 From: usa Posts: 101 Thanks: 0 Please help me to prove that $$h: (J:I)\rightarrow \text{Hom}_R(R/I, R/J)$$ $$x\mapsto h_x$$ where $$h_x:R/I\rightarrow R/J$$ $$r+I\mapsto xr+J$$ is well defined
April 20th, 2018, 02:57 PM   #3
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Quote:
 Originally Posted by mona123 Please help me to prove that $$h: (J:I)\rightarrow \text{Hom}_R(R/I, R/J)$$ $$x\mapsto h_x$$ where $$h_x:R/I\rightarrow R/J$$ $$r+I\mapsto xr+J$$ is well defined
What work have you done? Where are you stuck? It's inappropriate to ask questions at this level without showing that you've done any work of your own.

FWIW, relative to the ongoing moderation discussions, I'll take a thousand cranks and nuts rather than people repeatedly asking upper-division questions without showing any work.

 April 20th, 2018, 02:59 PM #4 Senior Member   Joined: Jan 2015 From: usa Posts: 101 Thanks: 0 we need to show that $$r+I=r'+I \implies xr+J=xr'+J$$ for all $x\in (J:I)$ but i don't manage to do that
April 20th, 2018, 03:26 PM   #5
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 Originally Posted by Maschke FWIW, relative to the ongoing moderation discussions, I'll take a thousand cranks and nuts rather than people repeatedly asking upper-division questions without showing any work.
Seriously? You compare "integration" with THIS?

April 20th, 2018, 04:56 PM   #6
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 Originally Posted by Micrm@ss Seriously? You compare "integration" with THIS?
I did not make a comparison. I said that I prefer one thing to another thing. A statement of preference can't be wrong, it's nothing more than my preference. I did not ask anyone else to share my personal preference.

I'm not a fan of integration, but I'm REALLY not a fan of posters who repeatedly ask upper-division or early grad-level question and NEVER show any work of their own.

That's my personal preference. It's not an endorsement of integration's posts. It's a statement of what frosts my butt. One thing frosts my butt more than another thing. Your butt frosting may vary. You should click through the OP's posts to get a sense of what I mean. I'm not basing my remarks on one post.

For the record I don't mean to bash one particular poster, this OP just happened to show up the same day that there's a discussion of moderation policies. And again, why is the Mariga thread still going on? That one frosts my butt too. Damn my butt is getting cold.

Last edited by Maschke; April 20th, 2018 at 05:01 PM.

April 20th, 2018, 05:01 PM   #7
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Quote:
 Originally Posted by Micrm@ss Seriously? You compare "integration" with THIS?
I'm not sure I would go so far as to prefer the crackpots over this, but I agree 99% with Maschke on this. Its absolutely absurd to be working at this level of math and still be unable to ask a more pointed questions than simply repeating the exercise. Mona's reply to his question is also completely unsatisfactory as it is nothing more than a restatement of the original question.

May 11th, 2018, 12:38 PM   #8
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Quote:
 Originally Posted by mona123 we need to show that $$r+I=r'+I \implies xr+J=xr'+J$$ for all $x\in (J:I)$ but i don't manage to do that
Try thinking about it for more than 3 seconds before asking on a forum. This is practically immediate from the definition $(J:I)$...

 May 14th, 2018, 03:11 PM #9 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,437 Thanks: 106 What's so advanced about the OP? It's no more difficult than Pinochle. It's just a matter of definitions. Commutative ring is a straightforward set of elementary definitions you could teach in grammar school. What is the definition of the triangle? Is that what makes something advanced, an undefined symbol?
May 14th, 2018, 08:57 PM   #10
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Quote:
 Originally Posted by cjem Try thinking about it for more than 3 seconds before asking on a forum. This is practically immediate from the definition $(J:I)$...
How does definition of (J:I) answer question b), which involves many more complicated definitions?

Definitions:

$\displaystyle \triangleleft$ Ideal: Ideal -- from Wolfram MathWorld

R/I is a quotient ring: Couldn't find a coherent definition.
From: Quotient Ring -- from Wolfram MathWorld
"A quotient ring (also called a residue-class ring) is a ring that is the quotient of a ring A and one of its ideals a, denoted A/a."
But qotient in this context is undefined.

Add to this the notation in b) of OP, which I couldn't google, and you are in the bowells of abstract pinochle.

Drilling down through all the definitions I find uninstructive, and impossible to remember and create a coherent picture.

But just out of curiousity, I would like to see a clean proof of OP b) with the steps referenced to approbriate definitions.

No matter what the topic, if the problem can be completely ensconced in a small set of intelligible definitions, it is interesting and educational, in spite of what Maschke says. I thought this might be the case here, but I see now that it would have been impossible for OP to even describe the problem in anything other than symbolism.

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