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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:
View attachment 6580
View attachment 6581
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
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:
View attachment 6580
View attachment 6581
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
Last edited: