Field Extensions - Remarks by Lovett - Page 326 .... ....

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Discussion Overview

The discussion revolves around understanding a specific remark from Stephen Lovett's "Abstract Algebra: Structures and Applications," particularly regarding field extensions and implications of certain algebraic expressions. Participants are exploring the relationship between the quotient ring and evaluations of polynomials at specific points.

Discussion Character

  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Peter seeks clarification on why the statement $$\overline{a(x) q(x)} = 1$$ implies that $$a(\alpha) q(\alpha) = 1$$.
  • One participant explains that if $$\overline{a(x)q(x)} = 1$$, then it follows that $$a(x)q(x) - 1 \in (p(x))$$, leading to the conclusion that evaluating at $$\alpha$$ gives $$a(\alpha)q(\alpha) = 1$$.
  • Another participant expresses curiosity about how Peter uploads snippets of literature, suggesting smartphone applications for better image quality.
  • Peter shares his method of scanning and processing images for clarity and size reduction.

Areas of Agreement / Disagreement

There is no explicit consensus on the initial question posed by Peter, but a participant provides a reasoning that aligns with Peter's inquiry. Other participants engage in a side discussion about image uploading methods, which does not relate to the main algebraic question.

Contextual Notes

The discussion includes assumptions about the properties of polynomials and their evaluations, which may depend on the definitions of the involved terms and the context of field extensions.

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I am reading "Abstract Algebra: Structures and Applications" by Stephen Lovett ...

I am currently focused on Chapter 7: Field Extensions ... ...

I need help with some remarks of Lovett following Theorem 7.1.12 and Example 7.1.13 on page 326 ...The remarks by Lovett read as follows:
https://www.physicsforums.com/attachments/6589
In the above remarks from Lovett, we read the following:

" ... ... In the quotient ring $$K$$, this implies that $$\overline{ a(x) q(x) } = 1$$. Thus in $$K, \ a( \alpha ) q( \alpha ) = 1$$. ... ... "My question is as follows:

Can someone please explain exactly why/how $$\overline{ a(x) q(x) } = 1$$ implies that $$a( \alpha ) q( \alpha ) = 1$$ ... ... ?Help will be appreciated ...

Peter
 
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I can't help with your question, but out of curiosity, how are you uploading your snippets of the various literature you upload? I'm going to guess they are photocopies.

I can recommend smartphone applications which will allow you to get a clear, more definitive black/white render of your image if you are interested! Also saves the headache of using a printer, if that's what you are doing..
 
Hi Peter,

If $\overline{a(x)q(x)} = 1$, then $a(x)q(x) - 1 \in (p(x))$, so then $a(x)q(x) - 1 = f(x)p(x)$ for some $f(x)\in F[x]$. Evaluating at $\alpha$, $a(\alpha)q(\alpha) - 1 = f(\alpha)p(\alpha) = f(\alpha)(0) = 0$. Hence, $a(\alpha)q(\alpha) = 1$.
 
Joppy said:
I can't help with your question, but out of curiosity, how are you uploading your snippets of the various literature you upload? I'm going to guess they are photocopies.

I can recommend smartphone applications which will allow you to get a clear, more definitive black/white render of your image if you are interested! Also saves the headache of using a printer, if that's what you are doing..
Hi Joppy,

I just scan the relevant textbook page and the select the relevant text ... then I use IrfanView to reduce the file size and convert to PNG format ... works Ok and is not very onerous ...

Peter

- - - Updated - - -

Euge said:
Hi Peter,

If $\overline{a(x)q(x)} = 1$, then $a(x)q(x) - 1 \in (p(x))$, so then $a(x)q(x) - 1 = f(x)p(x)$ for some $f(x)\in F[x]$. Evaluating at $\alpha$, $a(\alpha)q(\alpha) - 1 = f(\alpha)p(\alpha) = f(\alpha)(0) = 0$. Hence, $a(\alpha)q(\alpha) = 1$.
oh ... of course ...

Thanks Euge ...

Peter
 
Peter said:
Hi Joppy,

I just scan the relevant textbook page and the select the relevant text ... then I use IrfanView to reduce the file size and convert to PNG format ... works Ok and is not very onerous ...

Peter

Cool, just thought id mention it. ScannerPro is an app that syncs with Dropbox and allows you to easily crop and scale your images. The most important aspect being that you get much better coloration of your images (blacker blacks, whiter whites).
 

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