Fields and Field Extensions - Dummit and Foote, Ch. 13 .... ....

[Math Amateur]

I am reading Dummit and Foote, Chapter 13 - Field Theory.

I am currently studying Theorem 3 [pages 512 - 513]

I need some help with an aspect the proof of Theorem 3 ... ...

Theorem 3 on pages 512-513 reads as follows:
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View attachment 6581
In the above text from Dummit and Foote, we read the following:

" ... ... We identify \(\displaystyle F\) with its isomorphic image in \(\displaystyle K\) and view \(\displaystyle F\) as a subfield of \(\displaystyle K\). If \(\displaystyle \overline{x} = \pi (x)\) denotes the image of \(\displaystyle x\) in the quotient \(\displaystyle K\), then

\(\displaystyle p( \overline{x} ) = \overline{ p(x) }\) ... ... (since \(\displaystyle \pi\) is a homomorphism)

... ... "My question is as follows: ... where in the proof of \(\displaystyle p( \overline{x} ) = \overline{ p(x) }\) does it depend on \(\displaystyle \pi\) being a homomorphism ...

... indeed, how does one formally and rigorously demonstrate that \(\displaystyle p( \overline{x} ) = \overline{ p(x) }\) ... ... and how does this proof depend on \(\displaystyle \pi\) being a homomorphism ...
To make my question clearer consider the case of \(\displaystyle p(x) = x^2 - 5\) ... ...

Then ...

\(\displaystyle p( \overline{x} ) = \overline{x}^2 - 5_K \)

\(\displaystyle = ( x + ( p(x) ) ( x + ( p(x) ) - ( 5 + ( p(x) )\)

\(\displaystyle = ( x^2 + ( p(x) ) - ( 5 + ( p(x) )\)

\(\displaystyle = ( x^2 - 5 ) + ( p(x) ) = 0\)

\(\displaystyle = \overline{ p(x) } \)
... ... in the above case, my question is ... where does the above calculation depend on \(\displaystyle \pi\) being a homomorphism ... ?
Hope someone can help ...

Peter

 

[Euge]

... indeed, how does one formally and rigorously demonstrate that \(\displaystyle p( \overline{x} ) = \overline{ p(x) }\) ... ... and how does this proof depend on \(\displaystyle \pi\) being a homomorphism ...

The homomorphism property of $\pi$ allows for multiplication of cosets, as in your example with $p(x) = x^2 - 5$. In this general case, since $\pi$ is a homomorphism of rings, $\pi(ax^k) = \pi(a)\pi(x)^k=a\pi(x)^k$ for all scalars $a$ and for all $k \ge 0$, i.e., $\overline{ax^k} = a\bar{x}^k$ for all $a$ and $k$. Hence, by additivity of $\pi$, $p(\bar{x}) = \overline{p(x)}$.

 

[Math Amateur]

The homomorphism property of $\pi$ allows for multiplication of cosets, as in your example with $p(x) = x^2 - 5$. In this general case, since $\pi$ is a homomorphism of rings, $\pi(ax^k) = \pi(a)\pi(x)^k=a\pi(x)^k$ for all scalars $a$ and for all $k \ge 0$, i.e., $\overline{ax^k} = a\bar{x}^k$ for all $a$ and $k$. Hence, by additivity of $\pi$, $p(\bar{x}) = \overline{p(x)}$.

Thanks Euge ... most helpful as usual ...

Peter