Example of factorization in a polynomial ring which is not an 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 $mathbb{Z}[sqrt{-5}][x]$, 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 16 hours ago
Marta FornasierMarta Fornasier
314
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Marta Fornasier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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 $mathbb{Z}[sqrt{-5}][x]$, 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 16 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
16 hours ago
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
16 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 $mathbb{Z}[sqrt{-5}][x]$, 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 16 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 $mathbb{Z}[sqrt{-5}][x]$, 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 16 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 16 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 14 hours ago
Marta Fornasier
asked 16 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 16 hours ago
Marta FornasierMarta Fornasier
314
asked 16 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
16 hours ago
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
16 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
16 hours ago
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
16 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
16 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
16 hours ago
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
16 hours ago
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
16 hours ago
add a comment |
1 Answer
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oldest
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$begingroup$
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 15 hours ago
lhflhf
166k11172402
$endgroup$
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
15 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
14 hours ago
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 15 hours ago
lhflhf
166k11172402
$endgroup$
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
15 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
14 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 15 hours ago
lhflhf
166k11172402
$endgroup$
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
15 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
14 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 15 hours ago
lhflhf
166k11172402
$endgroup$
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 15 hours ago
lhflhf
166k11172402
answered 15 hours ago
lhflhf
166k11172402
answered 15 hours ago
lhflhf
166k11172402
answered 15 hours ago
lhflhf
166k11172402
166k11172402
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
15 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
14 hours ago
add a comment |
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
15 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
14 hours ago
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
15 hours ago
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
15 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
14 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
14 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
16 hours ago
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
16 hours ago