I am reading Dummit and Foote, Chapter 13 - Field Theory.(adsbygoogle = window.adsbygoogle || []).push({});

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:

In the above text from Dummit and Foote, we read the following:

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

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

... ... "

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

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

To make my question clearer consider the case of ##p(x) = x^2 - 5## ... ...

Then ...

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

##= ( x + ( p(x) ) ( x + ( p(x) ) - ( 5 + ( p(x) )##

##= ( x^2 + ( p(x) ) - ( 5 + ( p(x) )##

##= (x^2 - 5) + ( p(x) ) = 0##

##= \overline{ p(x) }##

... ... in the above case, my question is ... where does the above calculation depend on ##\pi## being a homomorphism ... ?

Hope someone can help ...

Peter

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

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