Example of GCD in UFD that can't be expressed as linear combination
$begingroup$
I proved that any two elements in PID have GCD and it can be expressed as linear combination of those two elements.
I know that even in case of UFD GCD exists but it may not be expressed as linear combination. Can anyone give me an example for which GCD is not expressed as linear combination?
number-theory ring-theory greatest-common-divisor integral-domain unique-factorization-domains
edited Dec 15 '18 at 7:46
qwr
6,70342755
asked Apr 3 '16 at 6:03
user298543user298543
413
$endgroup$
add a comment |
$begingroup$
I proved that any two elements in PID have GCD and it can be expressed as linear combination of those two elements.
I know that even in case of UFD GCD exists but it may not be expressed as linear combination. Can anyone give me an example for which GCD is not expressed as linear combination?
number-theory ring-theory greatest-common-divisor integral-domain unique-factorization-domains
edited Dec 15 '18 at 7:46
qwr
6,70342755
asked Apr 3 '16 at 6:03
user298543user298543
413
$endgroup$
add a comment |
$begingroup$
I proved that any two elements in PID have GCD and it can be expressed as linear combination of those two elements.
I know that even in case of UFD GCD exists but it may not be expressed as linear combination. Can anyone give me an example for which GCD is not expressed as linear combination?
number-theory ring-theory greatest-common-divisor integral-domain unique-factorization-domains
edited Dec 15 '18 at 7:46
qwr
6,70342755
asked Apr 3 '16 at 6:03
user298543user298543
413
$endgroup$
I proved that any two elements in PID have GCD and it can be expressed as linear combination of those two elements.
I know that even in case of UFD GCD exists but it may not be expressed as linear combination. Can anyone give me an example for which GCD is not expressed as linear combination?
number-theory ring-theory greatest-common-divisor integral-domain unique-factorization-domains
number-theory ring-theory greatest-common-divisor integral-domain unique-factorization-domains
edited Dec 15 '18 at 7:46
qwr
6,70342755
asked Apr 3 '16 at 6:03
user298543user298543
413
edited Dec 15 '18 at 7:46
qwr
6,70342755
asked Apr 3 '16 at 6:03
user298543user298543
413
edited Dec 15 '18 at 7:46
qwr
6,70342755
edited Dec 15 '18 at 7:46
qwr
6,70342755
edited Dec 15 '18 at 7:46
qwr
6,70342755
6,70342755
asked Apr 3 '16 at 6:03
user298543user298543
413
asked Apr 3 '16 at 6:03
user298543user298543
413
asked Apr 3 '16 at 6:03
user298543user298543
413
413
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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It is true that in any UFD, any two elements have a GCD (this comes right away from the definition), but the Bezout property (i.e. such a GCD is a linear combination) does not hold in general. Counter-example: in the polynomial ring $mathbb{Z}[t]$, $2$ and $t$ are coprime, but the ideal $(2,t)$ is strictly contained in the whole ring, whereas it should be equal to it if Bezout were available.
edited Feb 28 '17 at 14:37
Andreas Caranti
56.6k34395
answered Apr 4 '16 at 8:34
nguyen quang donguyen quang do
8,9891724
$endgroup$
$begingroup$
To clarify: the GCD of $2$ and $t$ is $1$, and the ideal $(1)$ is the entire ring.
$endgroup$
– qwr
Dec 15 '18 at 7:49
add a comment |
$begingroup$
Let's consider $R[x,y]$. The GCD of $x^2y$ and $xy^2$ is $xy$ so we are looking for $a,bin R[x,y]$ such that $ax^2y+bxy^2=xyRightarrow ax+by=1$. But the constant term of both $ax$ and $by$ is $0$, so the constant term of their sum is also zero. Contradiction.
answered Apr 3 '16 at 6:19
Stella BidermanStella Biderman
26.7k63375
$endgroup$
add a comment |
$begingroup$
First we have the result that $mathbb Z$ is a UFD implies $mathbb Z[x, y, z]$ is a UFD.
$operatorname{gcd}(xy, xz) = x$ but $x$ cannot be written as a linear combination of $xy$ and $xz$.
answered Dec 5 '18 at 17:48
qwrqwr
6,70342755
$endgroup$
add a comment |
$begingroup$
In $ℚ[X,Y]$ we have $gcd (X,Y) = 1$, but $1 ≠ fX + gY$ for any $f, g ∈ ℚ[X,Y]$ and both of these assertions can be seen by looking at degrees in $X$ and $Y$.
answered Feb 13 at 7:59
k.stmk.stm
10.9k22250
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is true that in any UFD, any two elements have a GCD (this comes right away from the definition), but the Bezout property (i.e. such a GCD is a linear combination) does not hold in general. Counter-example: in the polynomial ring $mathbb{Z}[t]$, $2$ and $t$ are coprime, but the ideal $(2,t)$ is strictly contained in the whole ring, whereas it should be equal to it if Bezout were available.
edited Feb 28 '17 at 14:37
Andreas Caranti
56.6k34395
answered Apr 4 '16 at 8:34
nguyen quang donguyen quang do
8,9891724
$endgroup$
$begingroup$
To clarify: the GCD of $2$ and $t$ is $1$, and the ideal $(1)$ is the entire ring.
$endgroup$
– qwr
Dec 15 '18 at 7:49
add a comment |
$begingroup$
It is true that in any UFD, any two elements have a GCD (this comes right away from the definition), but the Bezout property (i.e. such a GCD is a linear combination) does not hold in general. Counter-example: in the polynomial ring $mathbb{Z}[t]$, $2$ and $t$ are coprime, but the ideal $(2,t)$ is strictly contained in the whole ring, whereas it should be equal to it if Bezout were available.
edited Feb 28 '17 at 14:37
Andreas Caranti
56.6k34395
answered Apr 4 '16 at 8:34
nguyen quang donguyen quang do
8,9891724
$endgroup$
$begingroup$
To clarify: the GCD of $2$ and $t$ is $1$, and the ideal $(1)$ is the entire ring.
$endgroup$
– qwr
Dec 15 '18 at 7:49
add a comment |
$begingroup$
It is true that in any UFD, any two elements have a GCD (this comes right away from the definition), but the Bezout property (i.e. such a GCD is a linear combination) does not hold in general. Counter-example: in the polynomial ring $mathbb{Z}[t]$, $2$ and $t$ are coprime, but the ideal $(2,t)$ is strictly contained in the whole ring, whereas it should be equal to it if Bezout were available.
edited Feb 28 '17 at 14:37
Andreas Caranti
56.6k34395
answered Apr 4 '16 at 8:34
nguyen quang donguyen quang do
8,9891724
$endgroup$
It is true that in any UFD, any two elements have a GCD (this comes right away from the definition), but the Bezout property (i.e. such a GCD is a linear combination) does not hold in general. Counter-example: in the polynomial ring $mathbb{Z}[t]$, $2$ and $t$ are coprime, but the ideal $(2,t)$ is strictly contained in the whole ring, whereas it should be equal to it if Bezout were available.
edited Feb 28 '17 at 14:37
Andreas Caranti
56.6k34395
answered Apr 4 '16 at 8:34
nguyen quang donguyen quang do
8,9891724
edited Feb 28 '17 at 14:37
Andreas Caranti
56.6k34395
edited Feb 28 '17 at 14:37
Andreas Caranti
56.6k34395
edited Feb 28 '17 at 14:37
Andreas Caranti
56.6k34395
56.6k34395
answered Apr 4 '16 at 8:34
nguyen quang donguyen quang do
8,9891724
answered Apr 4 '16 at 8:34
nguyen quang donguyen quang do
8,9891724
answered Apr 4 '16 at 8:34
nguyen quang donguyen quang do
8,9891724
8,9891724
$begingroup$
To clarify: the GCD of $2$ and $t$ is $1$, and the ideal $(1)$ is the entire ring.
$endgroup$
– qwr
Dec 15 '18 at 7:49
add a comment |
$begingroup$
To clarify: the GCD of $2$ and $t$ is $1$, and the ideal $(1)$ is the entire ring.
$endgroup$
– qwr
Dec 15 '18 at 7:49
$begingroup$
To clarify: the GCD of $2$ and $t$ is $1$, and the ideal $(1)$ is the entire ring.
$endgroup$
– qwr
Dec 15 '18 at 7:49
$begingroup$
To clarify: the GCD of $2$ and $t$ is $1$, and the ideal $(1)$ is the entire ring.
$endgroup$
– qwr
Dec 15 '18 at 7:49
add a comment |
$begingroup$
Let's consider $R[x,y]$. The GCD of $x^2y$ and $xy^2$ is $xy$ so we are looking for $a,bin R[x,y]$ such that $ax^2y+bxy^2=xyRightarrow ax+by=1$. But the constant term of both $ax$ and $by$ is $0$, so the constant term of their sum is also zero. Contradiction.
answered Apr 3 '16 at 6:19
Stella BidermanStella Biderman
26.7k63375
$endgroup$
add a comment |
$begingroup$
Let's consider $R[x,y]$. The GCD of $x^2y$ and $xy^2$ is $xy$ so we are looking for $a,bin R[x,y]$ such that $ax^2y+bxy^2=xyRightarrow ax+by=1$. But the constant term of both $ax$ and $by$ is $0$, so the constant term of their sum is also zero. Contradiction.
answered Apr 3 '16 at 6:19
Stella BidermanStella Biderman
26.7k63375
$endgroup$
add a comment |
$begingroup$
Let's consider $R[x,y]$. The GCD of $x^2y$ and $xy^2$ is $xy$ so we are looking for $a,bin R[x,y]$ such that $ax^2y+bxy^2=xyRightarrow ax+by=1$. But the constant term of both $ax$ and $by$ is $0$, so the constant term of their sum is also zero. Contradiction.
answered Apr 3 '16 at 6:19
Stella BidermanStella Biderman
26.7k63375
$endgroup$
Let's consider $R[x,y]$. The GCD of $x^2y$ and $xy^2$ is $xy$ so we are looking for $a,bin R[x,y]$ such that $ax^2y+bxy^2=xyRightarrow ax+by=1$. But the constant term of both $ax$ and $by$ is $0$, so the constant term of their sum is also zero. Contradiction.
answered Apr 3 '16 at 6:19
Stella BidermanStella Biderman
26.7k63375
edited Feb 28 '17 at 14:31
answered Apr 3 '16 at 6:19
Stella BidermanStella Biderman
26.7k63375
answered Apr 3 '16 at 6:19
Stella BidermanStella Biderman
26.7k63375
answered Apr 3 '16 at 6:19
Stella BidermanStella Biderman
26.7k63375
26.7k63375
add a comment |
add a comment |
$begingroup$
First we have the result that $mathbb Z$ is a UFD implies $mathbb Z[x, y, z]$ is a UFD.
$operatorname{gcd}(xy, xz) = x$ but $x$ cannot be written as a linear combination of $xy$ and $xz$.
answered Dec 5 '18 at 17:48
qwrqwr
6,70342755
$endgroup$
add a comment |
$begingroup$
First we have the result that $mathbb Z$ is a UFD implies $mathbb Z[x, y, z]$ is a UFD.
$operatorname{gcd}(xy, xz) = x$ but $x$ cannot be written as a linear combination of $xy$ and $xz$.
answered Dec 5 '18 at 17:48
qwrqwr
6,70342755
$endgroup$
add a comment |
$begingroup$
First we have the result that $mathbb Z$ is a UFD implies $mathbb Z[x, y, z]$ is a UFD.
$operatorname{gcd}(xy, xz) = x$ but $x$ cannot be written as a linear combination of $xy$ and $xz$.
answered Dec 5 '18 at 17:48
qwrqwr
6,70342755
$endgroup$
First we have the result that $mathbb Z$ is a UFD implies $mathbb Z[x, y, z]$ is a UFD.
$operatorname{gcd}(xy, xz) = x$ but $x$ cannot be written as a linear combination of $xy$ and $xz$.
answered Dec 5 '18 at 17:48
qwrqwr
6,70342755
answered Dec 5 '18 at 17:48
qwrqwr
6,70342755
answered Dec 5 '18 at 17:48
qwrqwr
6,70342755
answered Dec 5 '18 at 17:48
qwrqwr
6,70342755
6,70342755
add a comment |
add a comment |
$begingroup$
In $ℚ[X,Y]$ we have $gcd (X,Y) = 1$, but $1 ≠ fX + gY$ for any $f, g ∈ ℚ[X,Y]$ and both of these assertions can be seen by looking at degrees in $X$ and $Y$.
answered Feb 13 at 7:59
k.stmk.stm
10.9k22250
$endgroup$
add a comment |
$begingroup$
In $ℚ[X,Y]$ we have $gcd (X,Y) = 1$, but $1 ≠ fX + gY$ for any $f, g ∈ ℚ[X,Y]$ and both of these assertions can be seen by looking at degrees in $X$ and $Y$.
answered Feb 13 at 7:59
k.stmk.stm
10.9k22250
$endgroup$
add a comment |
$begingroup$
In $ℚ[X,Y]$ we have $gcd (X,Y) = 1$, but $1 ≠ fX + gY$ for any $f, g ∈ ℚ[X,Y]$ and both of these assertions can be seen by looking at degrees in $X$ and $Y$.
answered Feb 13 at 7:59
k.stmk.stm
10.9k22250
$endgroup$
In $ℚ[X,Y]$ we have $gcd (X,Y) = 1$, but $1 ≠ fX + gY$ for any $f, g ∈ ℚ[X,Y]$ and both of these assertions can be seen by looking at degrees in $X$ and $Y$.
answered Feb 13 at 7:59
k.stmk.stm
10.9k22250
answered Feb 13 at 7:59
k.stmk.stm
10.9k22250
answered Feb 13 at 7:59
k.stmk.stm
10.9k22250
answered Feb 13 at 7:59
k.stmk.stm
10.9k22250
10.9k22250
add a comment |
add a comment |
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