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

[Math Amateur]

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:

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 it is that $\overline{ a(x) q(x) } = 1$ implies that $a( \alpha ) q( \alpha ) = 1$ ... ... ?


Help will be appreciated ...

Peter

 

[andrewkirk]

In the equation $a(x)q(x)+b(x)p(x)=1$, substitute $\alpha$ for $x$. Since $p(\alpha)=0$ (we were told $\alpha$ is a root of $p(x)$) the equation collapses to the desired result.