Product of gcd and lcm for multivariate polynomials
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This maybe trivial but I don't know how to conclude the proof. Consider the ring of multivariate polynomials with field coefficients $K[X_1,dots,X_n]$. Take two nonzero polynomials $F$ and $G$ and prove:
$$FG=gcd(F,G)lcm(F,G)$$
Since such ring is not Bezout I don't know how to prove this. I managed to prove that their gcd and their lcm both divide their product but I don't know how to do the converse relation.
polynomials commutative-algebra greatest-common-divisor least-common-multiple
asked Dec 6 '18 at 15:35
Renato FaraoneRenato Faraone
2,33111627
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add a comment |
$begingroup$
This maybe trivial but I don't know how to conclude the proof. Consider the ring of multivariate polynomials with field coefficients $K[X_1,dots,X_n]$. Take two nonzero polynomials $F$ and $G$ and prove:
$$FG=gcd(F,G)lcm(F,G)$$
Since such ring is not Bezout I don't know how to prove this. I managed to prove that their gcd and their lcm both divide their product but I don't know how to do the converse relation.
polynomials commutative-algebra greatest-common-divisor least-common-multiple
asked Dec 6 '18 at 15:35
Renato FaraoneRenato Faraone
2,33111627
$endgroup$
add a comment |
$begingroup$
This maybe trivial but I don't know how to conclude the proof. Consider the ring of multivariate polynomials with field coefficients $K[X_1,dots,X_n]$. Take two nonzero polynomials $F$ and $G$ and prove:
$$FG=gcd(F,G)lcm(F,G)$$
Since such ring is not Bezout I don't know how to prove this. I managed to prove that their gcd and their lcm both divide their product but I don't know how to do the converse relation.
polynomials commutative-algebra greatest-common-divisor least-common-multiple
asked Dec 6 '18 at 15:35
Renato FaraoneRenato Faraone
2,33111627
$endgroup$
This maybe trivial but I don't know how to conclude the proof. Consider the ring of multivariate polynomials with field coefficients $K[X_1,dots,X_n]$. Take two nonzero polynomials $F$ and $G$ and prove:
$$FG=gcd(F,G)lcm(F,G)$$
Since such ring is not Bezout I don't know how to prove this. I managed to prove that their gcd and their lcm both divide their product but I don't know how to do the converse relation.
polynomials commutative-algebra greatest-common-divisor least-common-multiple
polynomials commutative-algebra greatest-common-divisor least-common-multiple
asked Dec 6 '18 at 15:35
Renato FaraoneRenato Faraone
2,33111627
asked Dec 6 '18 at 15:35
Renato FaraoneRenato Faraone
2,33111627
asked Dec 6 '18 at 15:35
Renato FaraoneRenato Faraone
2,33111627
asked Dec 6 '18 at 15:35
Renato FaraoneRenato Faraone
2,33111627
asked Dec 6 '18 at 15:35
Renato FaraoneRenato Faraone
2,33111627
2,33111627
add a comment |
add a comment |
1 Answer
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The ring is a UFD so a GCD domain. As proved here the identity holds true in any GCD domain.
It can also be proved via prime factorizations, but that is less general than said proof using gcd laws.
answered Dec 6 '18 at 15:41
Bill DubuqueBill Dubuque
209k29191639
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add a comment |
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1 Answer
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$begingroup$
The ring is a UFD so a GCD domain. As proved here the identity holds true in any GCD domain.
It can also be proved via prime factorizations, but that is less general than said proof using gcd laws.
answered Dec 6 '18 at 15:41
Bill DubuqueBill Dubuque
209k29191639
$endgroup$
add a comment |
$begingroup$
The ring is a UFD so a GCD domain. As proved here the identity holds true in any GCD domain.
It can also be proved via prime factorizations, but that is less general than said proof using gcd laws.
answered Dec 6 '18 at 15:41
Bill DubuqueBill Dubuque
209k29191639
$endgroup$
add a comment |
$begingroup$
The ring is a UFD so a GCD domain. As proved here the identity holds true in any GCD domain.
It can also be proved via prime factorizations, but that is less general than said proof using gcd laws.
answered Dec 6 '18 at 15:41
Bill DubuqueBill Dubuque
209k29191639
$endgroup$
The ring is a UFD so a GCD domain. As proved here the identity holds true in any GCD domain.
It can also be proved via prime factorizations, but that is less general than said proof using gcd laws.
answered Dec 6 '18 at 15:41
Bill DubuqueBill Dubuque
209k29191639
answered Dec 6 '18 at 15:41
Bill DubuqueBill Dubuque
209k29191639
answered Dec 6 '18 at 15:41
Bill DubuqueBill Dubuque
209k29191639
answered Dec 6 '18 at 15:41
Bill DubuqueBill Dubuque
209k29191639
209k29191639
add a comment |
add a comment |
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