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

Gold Member
MHB
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 $$\displaystyle K$$, this implies that $$\displaystyle \overline{ a(x) q(x) } = 1$$. Thus in $$\displaystyle K, \ a( \alpha ) q( \alpha ) = 1$$. ... ... "

My question is as follows:

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

Help will be appreciated ...

Peter

Joppy
MHB
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..

Gold Member
MHB
POTW Director
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, $a(\alpha)q(\alpha) = 1$.

Gold Member
MHB
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 - - -

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

Joppy
MHB
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).