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

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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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