Small question about Gauss's lemma (polynomial)












0












$begingroup$


There is a beautiful proof for Gauss's lemma on Wikipedia here.



There is just the last bit I don't understand. It says: "This sum contains a term $a_r b_s$ which is not divisible by p (by Euclid's lemma, because p is prime), yet all the remaining ones are (because either $i < r$ or $j < s$), so the entire sum is not divisible by $p$."



Now I can't understand this. I understand that $a_r b_s$ is now divisible by $p$ since $a_r$ and $b_s$ are both not divisible by $p$ and $p$ was prime. But why is the sum of all the other elements and $a_r b_s$ also not divisible? Is there a lemma or theorem for the fact that?






abstract-algebra polynomials proof-explanation irreducible-polynomials







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$endgroup$












  • $begingroup$
    "yet all the remaining ones are (because either $i < r$ or $j < s$)".
    $endgroup$
    – darij grinberg
    Jan 11 at 14:37






  • 1




    $begingroup$
    It becomes obvious viewed $bmod p,,$ namely: the product of two nonzero polynomials is nonzero because this holds true for their leading coef's, i,e. $p$ remains prime in $R[x]$ because $R[x]/p cong (R/p)[x]$ is a domain - see my answer.
    $endgroup$
    – Bill Dubuque
    Jan 11 at 15:09


















0












$begingroup$


There is a beautiful proof for Gauss's lemma on Wikipedia here.



There is just the last bit I don't understand. It says: "This sum contains a term $a_r b_s$ which is not divisible by p (by Euclid's lemma, because p is prime), yet all the remaining ones are (because either $i < r$ or $j < s$), so the entire sum is not divisible by $p$."



Now I can't understand this. I understand that $a_r b_s$ is now divisible by $p$ since $a_r$ and $b_s$ are both not divisible by $p$ and $p$ was prime. But why is the sum of all the other elements and $a_r b_s$ also not divisible? Is there a lemma or theorem for the fact that?






abstract-algebra polynomials proof-explanation irreducible-polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    "yet all the remaining ones are (because either $i < r$ or $j < s$)".
    $endgroup$
    – darij grinberg
    Jan 11 at 14:37






  • 1




    $begingroup$
    It becomes obvious viewed $bmod p,,$ namely: the product of two nonzero polynomials is nonzero because this holds true for their leading coef's, i,e. $p$ remains prime in $R[x]$ because $R[x]/p cong (R/p)[x]$ is a domain - see my answer.
    $endgroup$
    – Bill Dubuque
    Jan 11 at 15:09
















0












0








0


0



$begingroup$


There is a beautiful proof for Gauss's lemma on Wikipedia here.



There is just the last bit I don't understand. It says: "This sum contains a term $a_r b_s$ which is not divisible by p (by Euclid's lemma, because p is prime), yet all the remaining ones are (because either $i < r$ or $j < s$), so the entire sum is not divisible by $p$."



Now I can't understand this. I understand that $a_r b_s$ is now divisible by $p$ since $a_r$ and $b_s$ are both not divisible by $p$ and $p$ was prime. But why is the sum of all the other elements and $a_r b_s$ also not divisible? Is there a lemma or theorem for the fact that?






abstract-algebra polynomials proof-explanation irreducible-polynomials







share|cite|improve this question











$endgroup$




There is a beautiful proof for Gauss's lemma on Wikipedia here.



There is just the last bit I don't understand. It says: "This sum contains a term $a_r b_s$ which is not divisible by p (by Euclid's lemma, because p is prime), yet all the remaining ones are (because either $i < r$ or $j < s$), so the entire sum is not divisible by $p$."



Now I can't understand this. I understand that $a_r b_s$ is now divisible by $p$ since $a_r$ and $b_s$ are both not divisible by $p$ and $p$ was prime. But why is the sum of all the other elements and $a_r b_s$ also not divisible? Is there a lemma or theorem for the fact that?






abstract-algebra polynomials proof-explanation irreducible-polynomials




abstract-algebra polynomials proof-explanation irreducible-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 14:43









José Carlos Santos

174k23133242




174k23133242










asked Jan 11 at 14:32









KingDingelingKingDingeling

1857




1857












  • $begingroup$
    "yet all the remaining ones are (because either $i < r$ or $j < s$)".
    $endgroup$
    – darij grinberg
    Jan 11 at 14:37






  • 1




    $begingroup$
    It becomes obvious viewed $bmod p,,$ namely: the product of two nonzero polynomials is nonzero because this holds true for their leading coef's, i,e. $p$ remains prime in $R[x]$ because $R[x]/p cong (R/p)[x]$ is a domain - see my answer.
    $endgroup$
    – Bill Dubuque
    Jan 11 at 15:09




















  • $begingroup$
    "yet all the remaining ones are (because either $i < r$ or $j < s$)".
    $endgroup$
    – darij grinberg
    Jan 11 at 14:37






  • 1




    $begingroup$
    It becomes obvious viewed $bmod p,,$ namely: the product of two nonzero polynomials is nonzero because this holds true for their leading coef's, i,e. $p$ remains prime in $R[x]$ because $R[x]/p cong (R/p)[x]$ is a domain - see my answer.
    $endgroup$
    – Bill Dubuque
    Jan 11 at 15:09


















$begingroup$
"yet all the remaining ones are (because either $i < r$ or $j < s$)".
$endgroup$
– darij grinberg
Jan 11 at 14:37




$begingroup$
"yet all the remaining ones are (because either $i < r$ or $j < s$)".
$endgroup$
– darij grinberg
Jan 11 at 14:37




1




1




$begingroup$
It becomes obvious viewed $bmod p,,$ namely: the product of two nonzero polynomials is nonzero because this holds true for their leading coef's, i,e. $p$ remains prime in $R[x]$ because $R[x]/p cong (R/p)[x]$ is a domain - see my answer.
$endgroup$
– Bill Dubuque
Jan 11 at 15:09






$begingroup$
It becomes obvious viewed $bmod p,,$ namely: the product of two nonzero polynomials is nonzero because this holds true for their leading coef's, i,e. $p$ remains prime in $R[x]$ because $R[x]/p cong (R/p)[x]$ is a domain - see my answer.
$endgroup$
– Bill Dubuque
Jan 11 at 15:09












3 Answers
3






active

oldest

votes


















1












$begingroup$

What are the other terms? They are $a_0b_{r+s},a_1b_{r+ss-1},ldots,a_{r+s-1}b_1,a_{r+s}b_0$. But $pmid a_0implies pmid a_0b_{r+s}$, $pmid a_1implies pmid a_1b_{r+s-1}$, and so on…






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot! I think I misstated the question: All the other elements can be divided by p, but why can't the sum of all the others and $a_r b_s$ be divided by p?
    $endgroup$
    – KingDingeling
    Jan 11 at 14:48






  • 1




    $begingroup$
    Because we have a sum of several numbers in which one of them is not a multiple of $p$, whereas all other numbers are multiples of $p$. Therefore, the sum cannot be a multiple of $p$.
    $endgroup$
    – José Carlos Santos
    Jan 11 at 14:54



















1












$begingroup$

By hypothesis, $p$ (prime) does not divide any of the coefficients of the polynomials $f=sum_r a_rx^r$ and $g=sum_s b_sx^s$ as both polynomials are primitive. If $p$ divides $a_rb_s$, then it divides one of the factors $a_r$ or $b_s$ by Euclid's lemma, which isn't possible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your help!
    $endgroup$
    – KingDingeling
    Jan 11 at 15:07



















1












$begingroup$

It's clearer when viewed this way: $,pnmid F,G,Rightarrow,pnmid FG.,$ Since $,pnmid F,G,$ when reduced mod $p$ both have lead coefs $,color{#0a0}{a,bnotequiv 0},$ so $FG$ has lead coef $,color{#c00}{abnotequiv 0},$ (by $p$ prime), hence $ABnotequiv 0,,$ i.e.



$qquadqquad{rm mod} p!: begin{eqnarray}
&& 0 notequiv F equiv, color{#0a0}a, x^j! + :cdots,quad color{#0a0}{anotequiv 0}\
&& 0 notequiv G equiv, color{#0a0}b, x^k! + :cdots,quad color{#0a0}{bnotequiv 0}\
Rightarrow, &&0 notequiv FG equiv color{#c00}{ab} x^{j+k}! + :cdots,, color{#c00}{abnotequiv 0}end{eqnarray}$



i.e. primes $:pin R:$ remain prime in $:R[x]:$ because the prime divisor property $,color{#0a0}{pnmid a,b},Rightarrow color{#c00}{pnmid ab},$ persists when multiplying leading coefficients.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, thats also a nice way of showing it.
    $endgroup$
    – KingDingeling
    Jan 11 at 15:08






  • 1




    $begingroup$
    @KingDingeling Conceptually this is what the common proof in Wikipedia is really doing, but without explicitly using the quotient ring $R[x]/p cong (R/p)[x]$ of polynomials with coef's $bmod p$ (since this might not have been studied yet). The price one pays for using such "elementary" proofs is that it obfuscates some points (like that you mention) that are trivial in the quotient ring.
    $endgroup$
    – Bill Dubuque
    Jan 11 at 15:16














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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

What are the other terms? They are $a_0b_{r+s},a_1b_{r+ss-1},ldots,a_{r+s-1}b_1,a_{r+s}b_0$. But $pmid a_0implies pmid a_0b_{r+s}$, $pmid a_1implies pmid a_1b_{r+s-1}$, and so on…






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot! I think I misstated the question: All the other elements can be divided by p, but why can't the sum of all the others and $a_r b_s$ be divided by p?
    $endgroup$
    – KingDingeling
    Jan 11 at 14:48






  • 1




    $begingroup$
    Because we have a sum of several numbers in which one of them is not a multiple of $p$, whereas all other numbers are multiples of $p$. Therefore, the sum cannot be a multiple of $p$.
    $endgroup$
    – José Carlos Santos
    Jan 11 at 14:54
















1












$begingroup$

What are the other terms? They are $a_0b_{r+s},a_1b_{r+ss-1},ldots,a_{r+s-1}b_1,a_{r+s}b_0$. But $pmid a_0implies pmid a_0b_{r+s}$, $pmid a_1implies pmid a_1b_{r+s-1}$, and so on…






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot! I think I misstated the question: All the other elements can be divided by p, but why can't the sum of all the others and $a_r b_s$ be divided by p?
    $endgroup$
    – KingDingeling
    Jan 11 at 14:48






  • 1




    $begingroup$
    Because we have a sum of several numbers in which one of them is not a multiple of $p$, whereas all other numbers are multiples of $p$. Therefore, the sum cannot be a multiple of $p$.
    $endgroup$
    – José Carlos Santos
    Jan 11 at 14:54














1












1








1





$begingroup$

What are the other terms? They are $a_0b_{r+s},a_1b_{r+ss-1},ldots,a_{r+s-1}b_1,a_{r+s}b_0$. But $pmid a_0implies pmid a_0b_{r+s}$, $pmid a_1implies pmid a_1b_{r+s-1}$, and so on…






share|cite|improve this answer









$endgroup$



What are the other terms? They are $a_0b_{r+s},a_1b_{r+ss-1},ldots,a_{r+s-1}b_1,a_{r+s}b_0$. But $pmid a_0implies pmid a_0b_{r+s}$, $pmid a_1implies pmid a_1b_{r+s-1}$, and so on…







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 14:39









José Carlos SantosJosé Carlos Santos

174k23133242




174k23133242












  • $begingroup$
    Thanks a lot! I think I misstated the question: All the other elements can be divided by p, but why can't the sum of all the others and $a_r b_s$ be divided by p?
    $endgroup$
    – KingDingeling
    Jan 11 at 14:48






  • 1




    $begingroup$
    Because we have a sum of several numbers in which one of them is not a multiple of $p$, whereas all other numbers are multiples of $p$. Therefore, the sum cannot be a multiple of $p$.
    $endgroup$
    – José Carlos Santos
    Jan 11 at 14:54


















  • $begingroup$
    Thanks a lot! I think I misstated the question: All the other elements can be divided by p, but why can't the sum of all the others and $a_r b_s$ be divided by p?
    $endgroup$
    – KingDingeling
    Jan 11 at 14:48






  • 1




    $begingroup$
    Because we have a sum of several numbers in which one of them is not a multiple of $p$, whereas all other numbers are multiples of $p$. Therefore, the sum cannot be a multiple of $p$.
    $endgroup$
    – José Carlos Santos
    Jan 11 at 14:54
















$begingroup$
Thanks a lot! I think I misstated the question: All the other elements can be divided by p, but why can't the sum of all the others and $a_r b_s$ be divided by p?
$endgroup$
– KingDingeling
Jan 11 at 14:48




$begingroup$
Thanks a lot! I think I misstated the question: All the other elements can be divided by p, but why can't the sum of all the others and $a_r b_s$ be divided by p?
$endgroup$
– KingDingeling
Jan 11 at 14:48




1




1




$begingroup$
Because we have a sum of several numbers in which one of them is not a multiple of $p$, whereas all other numbers are multiples of $p$. Therefore, the sum cannot be a multiple of $p$.
$endgroup$
– José Carlos Santos
Jan 11 at 14:54




$begingroup$
Because we have a sum of several numbers in which one of them is not a multiple of $p$, whereas all other numbers are multiples of $p$. Therefore, the sum cannot be a multiple of $p$.
$endgroup$
– José Carlos Santos
Jan 11 at 14:54











1












$begingroup$

By hypothesis, $p$ (prime) does not divide any of the coefficients of the polynomials $f=sum_r a_rx^r$ and $g=sum_s b_sx^s$ as both polynomials are primitive. If $p$ divides $a_rb_s$, then it divides one of the factors $a_r$ or $b_s$ by Euclid's lemma, which isn't possible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your help!
    $endgroup$
    – KingDingeling
    Jan 11 at 15:07
















1












$begingroup$

By hypothesis, $p$ (prime) does not divide any of the coefficients of the polynomials $f=sum_r a_rx^r$ and $g=sum_s b_sx^s$ as both polynomials are primitive. If $p$ divides $a_rb_s$, then it divides one of the factors $a_r$ or $b_s$ by Euclid's lemma, which isn't possible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your help!
    $endgroup$
    – KingDingeling
    Jan 11 at 15:07














1












1








1





$begingroup$

By hypothesis, $p$ (prime) does not divide any of the coefficients of the polynomials $f=sum_r a_rx^r$ and $g=sum_s b_sx^s$ as both polynomials are primitive. If $p$ divides $a_rb_s$, then it divides one of the factors $a_r$ or $b_s$ by Euclid's lemma, which isn't possible.






share|cite|improve this answer









$endgroup$



By hypothesis, $p$ (prime) does not divide any of the coefficients of the polynomials $f=sum_r a_rx^r$ and $g=sum_s b_sx^s$ as both polynomials are primitive. If $p$ divides $a_rb_s$, then it divides one of the factors $a_r$ or $b_s$ by Euclid's lemma, which isn't possible.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 14:39









WuestenfuxWuestenfux

5,5131513




5,5131513












  • $begingroup$
    Thank you for your help!
    $endgroup$
    – KingDingeling
    Jan 11 at 15:07


















  • $begingroup$
    Thank you for your help!
    $endgroup$
    – KingDingeling
    Jan 11 at 15:07
















$begingroup$
Thank you for your help!
$endgroup$
– KingDingeling
Jan 11 at 15:07




$begingroup$
Thank you for your help!
$endgroup$
– KingDingeling
Jan 11 at 15:07











1












$begingroup$

It's clearer when viewed this way: $,pnmid F,G,Rightarrow,pnmid FG.,$ Since $,pnmid F,G,$ when reduced mod $p$ both have lead coefs $,color{#0a0}{a,bnotequiv 0},$ so $FG$ has lead coef $,color{#c00}{abnotequiv 0},$ (by $p$ prime), hence $ABnotequiv 0,,$ i.e.



$qquadqquad{rm mod} p!: begin{eqnarray}
&& 0 notequiv F equiv, color{#0a0}a, x^j! + :cdots,quad color{#0a0}{anotequiv 0}\
&& 0 notequiv G equiv, color{#0a0}b, x^k! + :cdots,quad color{#0a0}{bnotequiv 0}\
Rightarrow, &&0 notequiv FG equiv color{#c00}{ab} x^{j+k}! + :cdots,, color{#c00}{abnotequiv 0}end{eqnarray}$



i.e. primes $:pin R:$ remain prime in $:R[x]:$ because the prime divisor property $,color{#0a0}{pnmid a,b},Rightarrow color{#c00}{pnmid ab},$ persists when multiplying leading coefficients.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, thats also a nice way of showing it.
    $endgroup$
    – KingDingeling
    Jan 11 at 15:08






  • 1




    $begingroup$
    @KingDingeling Conceptually this is what the common proof in Wikipedia is really doing, but without explicitly using the quotient ring $R[x]/p cong (R/p)[x]$ of polynomials with coef's $bmod p$ (since this might not have been studied yet). The price one pays for using such "elementary" proofs is that it obfuscates some points (like that you mention) that are trivial in the quotient ring.
    $endgroup$
    – Bill Dubuque
    Jan 11 at 15:16


















1












$begingroup$

It's clearer when viewed this way: $,pnmid F,G,Rightarrow,pnmid FG.,$ Since $,pnmid F,G,$ when reduced mod $p$ both have lead coefs $,color{#0a0}{a,bnotequiv 0},$ so $FG$ has lead coef $,color{#c00}{abnotequiv 0},$ (by $p$ prime), hence $ABnotequiv 0,,$ i.e.



$qquadqquad{rm mod} p!: begin{eqnarray}
&& 0 notequiv F equiv, color{#0a0}a, x^j! + :cdots,quad color{#0a0}{anotequiv 0}\
&& 0 notequiv G equiv, color{#0a0}b, x^k! + :cdots,quad color{#0a0}{bnotequiv 0}\
Rightarrow, &&0 notequiv FG equiv color{#c00}{ab} x^{j+k}! + :cdots,, color{#c00}{abnotequiv 0}end{eqnarray}$



i.e. primes $:pin R:$ remain prime in $:R[x]:$ because the prime divisor property $,color{#0a0}{pnmid a,b},Rightarrow color{#c00}{pnmid ab},$ persists when multiplying leading coefficients.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, thats also a nice way of showing it.
    $endgroup$
    – KingDingeling
    Jan 11 at 15:08






  • 1




    $begingroup$
    @KingDingeling Conceptually this is what the common proof in Wikipedia is really doing, but without explicitly using the quotient ring $R[x]/p cong (R/p)[x]$ of polynomials with coef's $bmod p$ (since this might not have been studied yet). The price one pays for using such "elementary" proofs is that it obfuscates some points (like that you mention) that are trivial in the quotient ring.
    $endgroup$
    – Bill Dubuque
    Jan 11 at 15:16
















1












1








1





$begingroup$

It's clearer when viewed this way: $,pnmid F,G,Rightarrow,pnmid FG.,$ Since $,pnmid F,G,$ when reduced mod $p$ both have lead coefs $,color{#0a0}{a,bnotequiv 0},$ so $FG$ has lead coef $,color{#c00}{abnotequiv 0},$ (by $p$ prime), hence $ABnotequiv 0,,$ i.e.



$qquadqquad{rm mod} p!: begin{eqnarray}
&& 0 notequiv F equiv, color{#0a0}a, x^j! + :cdots,quad color{#0a0}{anotequiv 0}\
&& 0 notequiv G equiv, color{#0a0}b, x^k! + :cdots,quad color{#0a0}{bnotequiv 0}\
Rightarrow, &&0 notequiv FG equiv color{#c00}{ab} x^{j+k}! + :cdots,, color{#c00}{abnotequiv 0}end{eqnarray}$



i.e. primes $:pin R:$ remain prime in $:R[x]:$ because the prime divisor property $,color{#0a0}{pnmid a,b},Rightarrow color{#c00}{pnmid ab},$ persists when multiplying leading coefficients.






share|cite|improve this answer









$endgroup$



It's clearer when viewed this way: $,pnmid F,G,Rightarrow,pnmid FG.,$ Since $,pnmid F,G,$ when reduced mod $p$ both have lead coefs $,color{#0a0}{a,bnotequiv 0},$ so $FG$ has lead coef $,color{#c00}{abnotequiv 0},$ (by $p$ prime), hence $ABnotequiv 0,,$ i.e.



$qquadqquad{rm mod} p!: begin{eqnarray}
&& 0 notequiv F equiv, color{#0a0}a, x^j! + :cdots,quad color{#0a0}{anotequiv 0}\
&& 0 notequiv G equiv, color{#0a0}b, x^k! + :cdots,quad color{#0a0}{bnotequiv 0}\
Rightarrow, &&0 notequiv FG equiv color{#c00}{ab} x^{j+k}! + :cdots,, color{#c00}{abnotequiv 0}end{eqnarray}$



i.e. primes $:pin R:$ remain prime in $:R[x]:$ because the prime divisor property $,color{#0a0}{pnmid a,b},Rightarrow color{#c00}{pnmid ab},$ persists when multiplying leading coefficients.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 14:54









Bill DubuqueBill Dubuque

214k29197656




214k29197656












  • $begingroup$
    Thanks, thats also a nice way of showing it.
    $endgroup$
    – KingDingeling
    Jan 11 at 15:08






  • 1




    $begingroup$
    @KingDingeling Conceptually this is what the common proof in Wikipedia is really doing, but without explicitly using the quotient ring $R[x]/p cong (R/p)[x]$ of polynomials with coef's $bmod p$ (since this might not have been studied yet). The price one pays for using such "elementary" proofs is that it obfuscates some points (like that you mention) that are trivial in the quotient ring.
    $endgroup$
    – Bill Dubuque
    Jan 11 at 15:16




















  • $begingroup$
    Thanks, thats also a nice way of showing it.
    $endgroup$
    – KingDingeling
    Jan 11 at 15:08






  • 1




    $begingroup$
    @KingDingeling Conceptually this is what the common proof in Wikipedia is really doing, but without explicitly using the quotient ring $R[x]/p cong (R/p)[x]$ of polynomials with coef's $bmod p$ (since this might not have been studied yet). The price one pays for using such "elementary" proofs is that it obfuscates some points (like that you mention) that are trivial in the quotient ring.
    $endgroup$
    – Bill Dubuque
    Jan 11 at 15:16


















$begingroup$
Thanks, thats also a nice way of showing it.
$endgroup$
– KingDingeling
Jan 11 at 15:08




$begingroup$
Thanks, thats also a nice way of showing it.
$endgroup$
– KingDingeling
Jan 11 at 15:08




1




1




$begingroup$
@KingDingeling Conceptually this is what the common proof in Wikipedia is really doing, but without explicitly using the quotient ring $R[x]/p cong (R/p)[x]$ of polynomials with coef's $bmod p$ (since this might not have been studied yet). The price one pays for using such "elementary" proofs is that it obfuscates some points (like that you mention) that are trivial in the quotient ring.
$endgroup$
– Bill Dubuque
Jan 11 at 15:16






$begingroup$
@KingDingeling Conceptually this is what the common proof in Wikipedia is really doing, but without explicitly using the quotient ring $R[x]/p cong (R/p)[x]$ of polynomials with coef's $bmod p$ (since this might not have been studied yet). The price one pays for using such "elementary" proofs is that it obfuscates some points (like that you mention) that are trivial in the quotient ring.
$endgroup$
– Bill Dubuque
Jan 11 at 15:16




















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