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

My question is ... how do we know that $$\psi$$ is surjective ...

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

... ?Obviously my thinking is somehow mistaken ...... can anyone help by demonstrating that $$\psi$$ 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 $$F$$ into $$F[x] / <p>$$ by considering the function $$\psi \ : \ F \longrightarrow F[x] / <p>$$ defined by $$\psi (a) = <p> + a$$, where $$a \in F$$. ... ... "The authors show that $$\psi$$ is one-to-one or injective but do not show that $$\psi$$ is onto or surjective ...

My question is ... how do we know that $$\psi$$ is surjective ...

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

... ?Obviously my thinking is somehow mistaken ...... can anyone help by demonstrating that $$\psi$$ 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 $$\psi$$ is not an isomorphism ... but is an embedding of $$F$$ in $$F[x] / <p>$$ ... I did not read the theorem/proof carefully enough ... :(

Peter