Help Understanding Proof of Eisenstein's Criterion











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The proof begins like this.



So we suppose $f(x)= a_0+a_1x+...+a_nx^n$ is reducible and thus



$f(x) = g(x)h(x) = (b_0+b_1x+...+b_rx^r)(c_0+c_1x+...+c_sx^s$).



Then $a_0=b_0c_0$. By the definition of Eisenstein's criterion $pmid a_0$ and so $pmid b_0$ or $pmid c_0$. Lets say $pmid b_0$. "Since $p^2$ does not divide $a_0$ (by definition of Eisentein's criterion), we see that $c_0$ is not divisible by $p$."



I don't understand that last statement. How does $c_0$ not being divisible by $p$ follow from the fact that $p^2$ does not divide $a$?










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  • 2




    If $p| c_0$ and $p|b_0,$ then $p^2(c_0/p)(b_0/p)=a_0$, hence $p^2| a_0$. By contrapositive if $p|b_0$, then $pnot|c_0.$
    – Melody
    2 days ago












  • thank you so much!
    – s_healy
    2 days ago















up vote
0
down vote

favorite












The proof begins like this.



So we suppose $f(x)= a_0+a_1x+...+a_nx^n$ is reducible and thus



$f(x) = g(x)h(x) = (b_0+b_1x+...+b_rx^r)(c_0+c_1x+...+c_sx^s$).



Then $a_0=b_0c_0$. By the definition of Eisenstein's criterion $pmid a_0$ and so $pmid b_0$ or $pmid c_0$. Lets say $pmid b_0$. "Since $p^2$ does not divide $a_0$ (by definition of Eisentein's criterion), we see that $c_0$ is not divisible by $p$."



I don't understand that last statement. How does $c_0$ not being divisible by $p$ follow from the fact that $p^2$ does not divide $a$?










share|cite|improve this question




















  • 2




    If $p| c_0$ and $p|b_0,$ then $p^2(c_0/p)(b_0/p)=a_0$, hence $p^2| a_0$. By contrapositive if $p|b_0$, then $pnot|c_0.$
    – Melody
    2 days ago












  • thank you so much!
    – s_healy
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











The proof begins like this.



So we suppose $f(x)= a_0+a_1x+...+a_nx^n$ is reducible and thus



$f(x) = g(x)h(x) = (b_0+b_1x+...+b_rx^r)(c_0+c_1x+...+c_sx^s$).



Then $a_0=b_0c_0$. By the definition of Eisenstein's criterion $pmid a_0$ and so $pmid b_0$ or $pmid c_0$. Lets say $pmid b_0$. "Since $p^2$ does not divide $a_0$ (by definition of Eisentein's criterion), we see that $c_0$ is not divisible by $p$."



I don't understand that last statement. How does $c_0$ not being divisible by $p$ follow from the fact that $p^2$ does not divide $a$?










share|cite|improve this question















The proof begins like this.



So we suppose $f(x)= a_0+a_1x+...+a_nx^n$ is reducible and thus



$f(x) = g(x)h(x) = (b_0+b_1x+...+b_rx^r)(c_0+c_1x+...+c_sx^s$).



Then $a_0=b_0c_0$. By the definition of Eisenstein's criterion $pmid a_0$ and so $pmid b_0$ or $pmid c_0$. Lets say $pmid b_0$. "Since $p^2$ does not divide $a_0$ (by definition of Eisentein's criterion), we see that $c_0$ is not divisible by $p$."



I don't understand that last statement. How does $c_0$ not being divisible by $p$ follow from the fact that $p^2$ does not divide $a$?







elementary-number-theory polynomials






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edited 2 days ago









Carmeister

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asked 2 days ago









s_healy

103




103








  • 2




    If $p| c_0$ and $p|b_0,$ then $p^2(c_0/p)(b_0/p)=a_0$, hence $p^2| a_0$. By contrapositive if $p|b_0$, then $pnot|c_0.$
    – Melody
    2 days ago












  • thank you so much!
    – s_healy
    2 days ago














  • 2




    If $p| c_0$ and $p|b_0,$ then $p^2(c_0/p)(b_0/p)=a_0$, hence $p^2| a_0$. By contrapositive if $p|b_0$, then $pnot|c_0.$
    – Melody
    2 days ago












  • thank you so much!
    – s_healy
    2 days ago








2




2




If $p| c_0$ and $p|b_0,$ then $p^2(c_0/p)(b_0/p)=a_0$, hence $p^2| a_0$. By contrapositive if $p|b_0$, then $pnot|c_0.$
– Melody
2 days ago






If $p| c_0$ and $p|b_0,$ then $p^2(c_0/p)(b_0/p)=a_0$, hence $p^2| a_0$. By contrapositive if $p|b_0$, then $pnot|c_0.$
– Melody
2 days ago














thank you so much!
– s_healy
2 days ago




thank you so much!
– s_healy
2 days ago















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