Example of factorization in a polynomial ring which is not an UFDDefining irreducible polynomials over polynomial ringsIs the coordinate ring of SL2 a UFD?Is this ring a UFDA non-UFD where we have different lengths of irreducible factorizations?Example of a Subgroup That Is Not NormalRegarding unique factorization in a polynomial ring and irreduciblesSimple question regarding proving some ring isn't a UFDShow that the ring $R$ of entire functions does not form a Unique Factorization Domainassuring factorization for R[x] when R is a UFDProblem with definition of unique factorization domain (UFD)Why a certain integral domain is not a UFD.
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Example of factorization in a polynomial ring which is not an UFD
Defining irreducible polynomials over polynomial ringsIs the coordinate ring of SL2 a UFD?Is this ring a UFDA non-UFD where we have different lengths of irreducible factorizations?Example of a Subgroup That Is Not NormalRegarding unique factorization in a polynomial ring and irreduciblesSimple question regarding proving some ring isn't a UFDShow that the ring $R$ of entire functions does not form a Unique Factorization Domainassuring factorization for R[x] when R is a UFDProblem with definition of unique factorization domain (UFD)Why a certain integral domain is not a UFD.
$begingroup$
I'm looking for a particular example of a polynomial ring $A[x]$, with $A$ integral domain, which is not an UFD. An easy example is $mathbbZ[sqrt-5][x]$, here $6 x^2 = (2x)(3x) = ((1+sqrt-5)x)((1-sqrt-5)x)$.
I would like to find a ring and a polynomial where the problem is not only in the coefficients. In other words a ring $A[x]$ where there exist a polynomial $f in A[x]$ with two factorizations that are not of the form $f = (a_1g)(b_1h) = (a_2 g)(b_2h)$, with $g,h in A[x]$, $a = a_1b_1 = a_2b_2 in A$ (with $a_1b_1, a_2b_2$ two factorization of $a$). Or not of a similar form with more factors.
abstract-algebra
asked 4 hours ago
Marta FornasierMarta Fornasier
314
New contributor
Marta Fornasier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm looking for a particular example of a polynomial ring $A[x]$, with $A$ integral domain, which is not an UFD. An easy example is $mathbbZ[sqrt-5][x]$, here $6 x^2 = (2x)(3x) = ((1+sqrt-5)x)((1-sqrt-5)x)$.
I would like to find a ring and a polynomial where the problem is not only in the coefficients. In other words a ring $A[x]$ where there exist a polynomial $f in A[x]$ with two factorizations that are not of the form $f = (a_1g)(b_1h) = (a_2 g)(b_2h)$, with $g,h in A[x]$, $a = a_1b_1 = a_2b_2 in A$ (with $a_1b_1, a_2b_2$ two factorization of $a$). Or not of a similar form with more factors.
abstract-algebra
asked 4 hours ago
Marta FornasierMarta Fornasier
314
New contributor
Marta Fornasier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
You're right. That example is a bit tame. We take $6x^2$, and then we take out the $x^2$, factor $6$ in two different ways, and then we redistribute the $x^2$. You can see that if you consider the full factorisation into irreducibles, which is $2cdot 3 cdot xcdot x$ versus $(1+sqrt-5)cdot (1-sqrt-5)cdot xcdot x$. So the way I see it, what you're asking about is basically an example showing the non-UFD-ness of $A[x]$ for some non-UFD $A$, which is not trivially reucible to an example from $A$.
$endgroup$
– Arthur
4 hours ago
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
4 hours ago
add a comment |
$begingroup$
I'm looking for a particular example of a polynomial ring $A[x]$, with $A$ integral domain, which is not an UFD. An easy example is $mathbbZ[sqrt-5][x]$, here $6 x^2 = (2x)(3x) = ((1+sqrt-5)x)((1-sqrt-5)x)$.
I would like to find a ring and a polynomial where the problem is not only in the coefficients. In other words a ring $A[x]$ where there exist a polynomial $f in A[x]$ with two factorizations that are not of the form $f = (a_1g)(b_1h) = (a_2 g)(b_2h)$, with $g,h in A[x]$, $a = a_1b_1 = a_2b_2 in A$ (with $a_1b_1, a_2b_2$ two factorization of $a$). Or not of a similar form with more factors.
abstract-algebra
asked 4 hours ago
Marta FornasierMarta Fornasier
314
New contributor
Marta Fornasier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm looking for a particular example of a polynomial ring $A[x]$, with $A$ integral domain, which is not an UFD. An easy example is $mathbbZ[sqrt-5][x]$, here $6 x^2 = (2x)(3x) = ((1+sqrt-5)x)((1-sqrt-5)x)$.
I would like to find a ring and a polynomial where the problem is not only in the coefficients. In other words a ring $A[x]$ where there exist a polynomial $f in A[x]$ with two factorizations that are not of the form $f = (a_1g)(b_1h) = (a_2 g)(b_2h)$, with $g,h in A[x]$, $a = a_1b_1 = a_2b_2 in A$ (with $a_1b_1, a_2b_2$ two factorization of $a$). Or not of a similar form with more factors.
abstract-algebra
abstract-algebra
asked 4 hours ago
Marta FornasierMarta Fornasier
314
New contributor
Marta Fornasier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Marta FornasierMarta Fornasier
314
New contributor
Marta Fornasier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Marta Fornasier
asked 4 hours ago
Marta FornasierMarta Fornasier
314
New contributor
Marta Fornasier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Marta FornasierMarta Fornasier
314
asked 4 hours ago
Marta FornasierMarta Fornasier
314
314
New contributor
Marta Fornasier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Marta Fornasier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Marta Fornasier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
You're right. That example is a bit tame. We take $6x^2$, and then we take out the $x^2$, factor $6$ in two different ways, and then we redistribute the $x^2$. You can see that if you consider the full factorisation into irreducibles, which is $2cdot 3 cdot xcdot x$ versus $(1+sqrt-5)cdot (1-sqrt-5)cdot xcdot x$. So the way I see it, what you're asking about is basically an example showing the non-UFD-ness of $A[x]$ for some non-UFD $A$, which is not trivially reucible to an example from $A$.
$endgroup$
– Arthur
4 hours ago
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
4 hours ago
add a comment |
$begingroup$
You're right. That example is a bit tame. We take $6x^2$, and then we take out the $x^2$, factor $6$ in two different ways, and then we redistribute the $x^2$. You can see that if you consider the full factorisation into irreducibles, which is $2cdot 3 cdot xcdot x$ versus $(1+sqrt-5)cdot (1-sqrt-5)cdot xcdot x$. So the way I see it, what you're asking about is basically an example showing the non-UFD-ness of $A[x]$ for some non-UFD $A$, which is not trivially reucible to an example from $A$.
$endgroup$
– Arthur
4 hours ago
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
4 hours ago
$begingroup$
You're right. That example is a bit tame. We take $6x^2$, and then we take out the $x^2$, factor $6$ in two different ways, and then we redistribute the $x^2$. You can see that if you consider the full factorisation into irreducibles, which is $2cdot 3 cdot xcdot x$ versus $(1+sqrt-5)cdot (1-sqrt-5)cdot xcdot x$. So the way I see it, what you're asking about is basically an example showing the non-UFD-ness of $A[x]$ for some non-UFD $A$, which is not trivially reucible to an example from $A$.
$endgroup$
– Arthur
4 hours ago
$begingroup$
You're right. That example is a bit tame. We take $6x^2$, and then we take out the $x^2$, factor $6$ in two different ways, and then we redistribute the $x^2$. You can see that if you consider the full factorisation into irreducibles, which is $2cdot 3 cdot xcdot x$ versus $(1+sqrt-5)cdot (1-sqrt-5)cdot xcdot x$. So the way I see it, what you're asking about is basically an example showing the non-UFD-ness of $A[x]$ for some non-UFD $A$, which is not trivially reucible to an example from $A$.
$endgroup$
– Arthur
4 hours ago
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
4 hours ago
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In A Classical Introduction to Modern Number Theory, Ireland and Rosen give two examples of non-UFDs that answer this question. First, they point out that a "generic" example is $k[x,y,z,w]/(xy-zw)$, where $k$ is any field. Second, they offer the subtler example of $mathbbC[x,y,z]/(x^2+y^2+z^2-1)$, where $mathbbC$ is the field of complex numbers. Here unique factorization fails because $(x + iy)(x - iy) = (1 + z)(1 - z)$.
answered 4 hours ago
FredHFredH
2,4941020
$endgroup$
2
$begingroup$
Are these rings of the form A[x]?
$endgroup$
– Lior B-S
3 hours ago
add a comment |
$begingroup$
When $A$ is not a domain, decomposing a polynomial in $A[x]$ as $f=gh$ does not necessarily simplify it, that is, does not always reduce its degree.
For instance:
$$
5x+1=(2x+1)(3x+1) bmod 6
$$
It is even possible to decompose a linear polynomial as a product of two quadratic polynomials:
$$
x+1=(2x^2+x+7)(4x^2+6x+7) bmod 8
$$
answered 3 hours ago
lhflhf
166k11171402
$endgroup$
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
3 hours ago
$begingroup$
Uh, I'm sorry, I didn't write it, but I was looking for $A[x]$ with $A$ domain. Now I edit the question.
$endgroup$
– Marta Fornasier
2 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In A Classical Introduction to Modern Number Theory, Ireland and Rosen give two examples of non-UFDs that answer this question. First, they point out that a "generic" example is $k[x,y,z,w]/(xy-zw)$, where $k$ is any field. Second, they offer the subtler example of $mathbbC[x,y,z]/(x^2+y^2+z^2-1)$, where $mathbbC$ is the field of complex numbers. Here unique factorization fails because $(x + iy)(x - iy) = (1 + z)(1 - z)$.
answered 4 hours ago
FredHFredH
2,4941020
$endgroup$
2
$begingroup$
Are these rings of the form A[x]?
$endgroup$
– Lior B-S
3 hours ago
add a comment |
$begingroup$
In A Classical Introduction to Modern Number Theory, Ireland and Rosen give two examples of non-UFDs that answer this question. First, they point out that a "generic" example is $k[x,y,z,w]/(xy-zw)$, where $k$ is any field. Second, they offer the subtler example of $mathbbC[x,y,z]/(x^2+y^2+z^2-1)$, where $mathbbC$ is the field of complex numbers. Here unique factorization fails because $(x + iy)(x - iy) = (1 + z)(1 - z)$.
answered 4 hours ago
FredHFredH
2,4941020
$endgroup$
2
$begingroup$
Are these rings of the form A[x]?
$endgroup$
– Lior B-S
3 hours ago
add a comment |
$begingroup$
In A Classical Introduction to Modern Number Theory, Ireland and Rosen give two examples of non-UFDs that answer this question. First, they point out that a "generic" example is $k[x,y,z,w]/(xy-zw)$, where $k$ is any field. Second, they offer the subtler example of $mathbbC[x,y,z]/(x^2+y^2+z^2-1)$, where $mathbbC$ is the field of complex numbers. Here unique factorization fails because $(x + iy)(x - iy) = (1 + z)(1 - z)$.
answered 4 hours ago
FredHFredH
2,4941020
$endgroup$
In A Classical Introduction to Modern Number Theory, Ireland and Rosen give two examples of non-UFDs that answer this question. First, they point out that a "generic" example is $k[x,y,z,w]/(xy-zw)$, where $k$ is any field. Second, they offer the subtler example of $mathbbC[x,y,z]/(x^2+y^2+z^2-1)$, where $mathbbC$ is the field of complex numbers. Here unique factorization fails because $(x + iy)(x - iy) = (1 + z)(1 - z)$.
answered 4 hours ago
FredHFredH
2,4941020
answered 4 hours ago
FredHFredH
2,4941020
answered 4 hours ago
FredHFredH
2,4941020
answered 4 hours ago
FredHFredH
2,4941020
2,4941020
2
$begingroup$
Are these rings of the form A[x]?
$endgroup$
– Lior B-S
3 hours ago
add a comment |
2
$begingroup$
Are these rings of the form A[x]?
$endgroup$
– Lior B-S
3 hours ago
2
2
$begingroup$
Are these rings of the form A[x]?
$endgroup$
– Lior B-S
3 hours ago
$begingroup$
Are these rings of the form A[x]?
$endgroup$
– Lior B-S
3 hours ago
add a comment |
$begingroup$
When $A$ is not a domain, decomposing a polynomial in $A[x]$ as $f=gh$ does not necessarily simplify it, that is, does not always reduce its degree.
For instance:
$$
5x+1=(2x+1)(3x+1) bmod 6
$$
It is even possible to decompose a linear polynomial as a product of two quadratic polynomials:
$$
x+1=(2x^2+x+7)(4x^2+6x+7) bmod 8
$$
answered 3 hours ago
lhflhf
166k11171402
$endgroup$
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
3 hours ago
$begingroup$
Uh, I'm sorry, I didn't write it, but I was looking for $A[x]$ with $A$ domain. Now I edit the question.
$endgroup$
– Marta Fornasier
2 hours ago
add a comment |
$begingroup$
When $A$ is not a domain, decomposing a polynomial in $A[x]$ as $f=gh$ does not necessarily simplify it, that is, does not always reduce its degree.
For instance:
$$
5x+1=(2x+1)(3x+1) bmod 6
$$
It is even possible to decompose a linear polynomial as a product of two quadratic polynomials:
$$
x+1=(2x^2+x+7)(4x^2+6x+7) bmod 8
$$
answered 3 hours ago
lhflhf
166k11171402
$endgroup$
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
3 hours ago
$begingroup$
Uh, I'm sorry, I didn't write it, but I was looking for $A[x]$ with $A$ domain. Now I edit the question.
$endgroup$
– Marta Fornasier
2 hours ago
add a comment |
$begingroup$
When $A$ is not a domain, decomposing a polynomial in $A[x]$ as $f=gh$ does not necessarily simplify it, that is, does not always reduce its degree.
For instance:
$$
5x+1=(2x+1)(3x+1) bmod 6
$$
It is even possible to decompose a linear polynomial as a product of two quadratic polynomials:
$$
x+1=(2x^2+x+7)(4x^2+6x+7) bmod 8
$$
answered 3 hours ago
lhflhf
166k11171402
$endgroup$
When $A$ is not a domain, decomposing a polynomial in $A[x]$ as $f=gh$ does not necessarily simplify it, that is, does not always reduce its degree.
For instance:
$$
5x+1=(2x+1)(3x+1) bmod 6
$$
It is even possible to decompose a linear polynomial as a product of two quadratic polynomials:
$$
x+1=(2x^2+x+7)(4x^2+6x+7) bmod 8
$$
answered 3 hours ago
lhflhf
166k11171402
answered 3 hours ago
lhflhf
166k11171402
answered 3 hours ago
lhflhf
166k11171402
answered 3 hours ago
lhflhf
166k11171402
166k11171402
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
3 hours ago
$begingroup$
Uh, I'm sorry, I didn't write it, but I was looking for $A[x]$ with $A$ domain. Now I edit the question.
$endgroup$
– Marta Fornasier
2 hours ago
add a comment |
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
3 hours ago
$begingroup$
Uh, I'm sorry, I didn't write it, but I was looking for $A[x]$ with $A$ domain. Now I edit the question.
$endgroup$
– Marta Fornasier
2 hours ago
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
3 hours ago
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
3 hours ago
$begingroup$
Uh, I'm sorry, I didn't write it, but I was looking for $A[x]$ with $A$ domain. Now I edit the question.
$endgroup$
– Marta Fornasier
2 hours ago
$begingroup$
Uh, I'm sorry, I didn't write it, but I was looking for $A[x]$ with $A$ domain. Now I edit the question.
$endgroup$
– Marta Fornasier
2 hours ago
add a comment |
Marta Fornasier is a new contributor. Be nice, and check out our Code of Conduct.
Marta Fornasier is a new contributor. Be nice, and check out our Code of Conduct.
Marta Fornasier is a new contributor. Be nice, and check out our Code of Conduct.
Marta Fornasier is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
You're right. That example is a bit tame. We take $6x^2$, and then we take out the $x^2$, factor $6$ in two different ways, and then we redistribute the $x^2$. You can see that if you consider the full factorisation into irreducibles, which is $2cdot 3 cdot xcdot x$ versus $(1+sqrt-5)cdot (1-sqrt-5)cdot xcdot x$. So the way I see it, what you're asking about is basically an example showing the non-UFD-ness of $A[x]$ for some non-UFD $A$, which is not trivially reucible to an example from $A$.
$endgroup$
– Arthur
4 hours ago
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
4 hours ago