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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##. ... ... "Te 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 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:
" ... ... 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##. ... ... "Te 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
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