Kronecker's Theorem - Anderson and Feil, Theorem 42.1, Chapter 42 .... ....

[Math Amateur]

I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 42: Field Extensions and Kronecker's Theorem ...

I need some help with an aspect of the proof of Theorem 42.1 ( Kronecker's Theorem) ...

Theorem 42.1 and its proof read as follows:
https://www.physicsforums.com/attachments/6565
https://www.physicsforums.com/attachments/6566In the above text by Anderson and Feil we read the following:

" ... ... We show that there is an isomorphism from \(\displaystyle F\) into \(\displaystyle F[x] / <p>\) by considering the function \(\displaystyle \psi \ : \ F \longrightarrow F[x] / <p>\) defined by \(\displaystyle \psi (a) = <p> + a\), where \(\displaystyle a \in F\). ... ... "The authors show that \(\displaystyle \psi\) is one-to-one or injective but do not show that \(\displaystyle \psi\) is onto or surjective ...

My question is ... how do we know that \(\displaystyle \psi\) is surjective ...

... for example if a polynomial in \(\displaystyle F[x]\), say \(\displaystyle f\), is degree 5, and \(\displaystyle p\) is degree 3 then dividing \(\displaystyle f\) by \(\displaystyle p\) gives a polynomial remainder \(\displaystyle r\) of degree 2 ... then \(\displaystyle r + <p>\) will not be of the form \(\displaystyle <p> + a\) where \(\displaystyle a \in F\) ... ... and so it seems that \(\displaystyle \psi\) is not surjective ... since the coset of \(\displaystyle f\) is not of the form \(\displaystyle <p> + a\) where \(\displaystyle a \in F\) ...

... ?Obviously my thinking is somehow mistaken ...... can anyone help by demonstrating that \(\displaystyle \psi\) is surjective ... and hence (given that Anderson and Feil have demonstrated it is injective) an isomorphism ...Help will be appreciated ...

Peter

 

[Math Amateur]

I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 42: Field Extensions and Kronecker's Theorem ...

I need some help with an aspect of the proof of Theorem 42.1 ( Kronecker's Theorem) ...

Theorem 42.1 and its proof read as follows:

In the above text by Anderson and Feil we read the following:

" ... ... We show that there is an isomorphism from \(\displaystyle F\) into \(\displaystyle F[x] / <p>\) by considering the function \(\displaystyle \psi \ : \ F \longrightarrow F[x] / <p>\) defined by \(\displaystyle \psi (a) = <p> + a\), where \(\displaystyle a \in F\). ... ... "The authors show that \(\displaystyle \psi\) is one-to-one or injective but do not show that \(\displaystyle \psi\) is onto or surjective ...

My question is ... how do we know that \(\displaystyle \psi\) is surjective ...

... for example if a polynomial in \(\displaystyle F[x]\), say \(\displaystyle f\), is degree 5, and \(\displaystyle p\) is degree 3 then dividing \(\displaystyle f\) by \(\displaystyle p\) gives a polynomial remainder \(\displaystyle r\) of degree 2 ... then \(\displaystyle r + <p>\) will not be of the form \(\displaystyle <p> + a\) where \(\displaystyle a \in F\) ... ... and so it seems that \(\displaystyle \psi\) is not surjective ... since the coset of \(\displaystyle f\) is not of the form \(\displaystyle <p> + a\) where \(\displaystyle a \in F\) ...

... ?Obviously my thinking is somehow mistaken ...... can anyone help by demonstrating that \(\displaystyle \psi\) is surjective ... and hence (given that Anderson and Feil have demonstrated it is injective) an isomorphism ...Help will be appreciated ...

Peter

I probably should not be answering my own question ... but I am now of the opinion that \(\displaystyle \psi\) is not an isomorphism ... but is an embedding of \(\displaystyle F\) in \(\displaystyle F[x] / <p>\) ... I did not read the theorem/proof carefully enough ... :(

Peter