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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
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