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 Mar 22 at 14:02
Marta FornasierMarta Fornasier
334
$endgroup$
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 Mar 22 at 14:02
Marta FornasierMarta Fornasier
334
$endgroup$
$begingroup$
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
Mar 22 at 14:08
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
Mar 22 at 14:19
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 Mar 22 at 14:02
Marta FornasierMarta Fornasier
334
$endgroup$
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 Mar 22 at 14:02
Marta FornasierMarta Fornasier
334
asked Mar 22 at 14:02
Marta FornasierMarta Fornasier
334
edited Mar 22 at 16:18
Marta Fornasier
asked Mar 22 at 14:02
Marta FornasierMarta Fornasier
334
asked Mar 22 at 14:02
Marta FornasierMarta Fornasier
334
asked Mar 22 at 14:02
Marta FornasierMarta Fornasier
334
334
$begingroup$
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
Mar 22 at 14:08
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
Mar 22 at 14:19
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
Mar 22 at 14:08
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
Mar 22 at 14:19
$begingroup$
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
Mar 22 at 14:08
$begingroup$
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
Mar 22 at 14:08
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
Mar 22 at 14:19
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
Mar 22 at 14:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Perhaps this is the sort of thing you are looking for: If $a,b,c,din A$ are irreducible non-associate elements such that $ab=cd$, then
$$
(ax+c)(dx-b) = adx^2 - bc = (ax-c)(dx+b).
$$
With $A = mathbbZ[sqrt-5]$, taking $(a,b,c,d) = (2,3,1+sqrt-5,1-sqrt-5)$, this
results in
$$
(2x+1+sqrt-5)((1-sqrt-5)x-3) = (2x-1-sqrt-5)((1-sqrt-5)x+3).
$$
However, you might still see this a kind of cheat, since $ax+c$ and $dx+b$ actually are scalar multiples; it's just that the scalar factor, $d/a = b/c$, isn't in $A$ but is instead in its field of fractions. But this is unavoidable: over the field of fractions, polynomials do have unique factorization; so seemingly distinct factorizations into linear factors over $A$ necessarily become identical aside from units over the field of fractions.
Another possibility is to use a higher-order polynomial that is irreducible over $A$, but not over the field of fractions. For example, if $ab=cd$ then
$$
(ax+c)(ax+d) = a(ax^2 + (c+d)x + b).
$$
In $mathbbZ[sqrt-5]$, this might be realized as
$$
(2x + 1 + sqrt-5)(2x + 1 - sqrt-5) = 2(2x^2 +2x + 3).
$$
This is, in essence, an example of the failure of Gauss's lemma for a non-UFD. I found the idea for this example in David E Speyer's answer to a MathOverflow question; the other responses may also be of interest.
answered Mar 23 at 12:38
FredHFredH
3,7251024
$endgroup$
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 Mar 22 at 15:19
lhflhf
168k11173406
$endgroup$
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
Mar 22 at 15:20
$begingroup$
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
Mar 22 at 16:17
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Perhaps this is the sort of thing you are looking for: If $a,b,c,din A$ are irreducible non-associate elements such that $ab=cd$, then
$$
(ax+c)(dx-b) = adx^2 - bc = (ax-c)(dx+b).
$$
With $A = mathbbZ[sqrt-5]$, taking $(a,b,c,d) = (2,3,1+sqrt-5,1-sqrt-5)$, this
results in
$$
(2x+1+sqrt-5)((1-sqrt-5)x-3) = (2x-1-sqrt-5)((1-sqrt-5)x+3).
$$
However, you might still see this a kind of cheat, since $ax+c$ and $dx+b$ actually are scalar multiples; it's just that the scalar factor, $d/a = b/c$, isn't in $A$ but is instead in its field of fractions. But this is unavoidable: over the field of fractions, polynomials do have unique factorization; so seemingly distinct factorizations into linear factors over $A$ necessarily become identical aside from units over the field of fractions.
Another possibility is to use a higher-order polynomial that is irreducible over $A$, but not over the field of fractions. For example, if $ab=cd$ then
$$
(ax+c)(ax+d) = a(ax^2 + (c+d)x + b).
$$
In $mathbbZ[sqrt-5]$, this might be realized as
$$
(2x + 1 + sqrt-5)(2x + 1 - sqrt-5) = 2(2x^2 +2x + 3).
$$
This is, in essence, an example of the failure of Gauss's lemma for a non-UFD. I found the idea for this example in David E Speyer's answer to a MathOverflow question; the other responses may also be of interest.
answered Mar 23 at 12:38
FredHFredH
3,7251024
$endgroup$
add a comment |
$begingroup$
Perhaps this is the sort of thing you are looking for: If $a,b,c,din A$ are irreducible non-associate elements such that $ab=cd$, then
$$
(ax+c)(dx-b) = adx^2 - bc = (ax-c)(dx+b).
$$
With $A = mathbbZ[sqrt-5]$, taking $(a,b,c,d) = (2,3,1+sqrt-5,1-sqrt-5)$, this
results in
$$
(2x+1+sqrt-5)((1-sqrt-5)x-3) = (2x-1-sqrt-5)((1-sqrt-5)x+3).
$$
However, you might still see this a kind of cheat, since $ax+c$ and $dx+b$ actually are scalar multiples; it's just that the scalar factor, $d/a = b/c$, isn't in $A$ but is instead in its field of fractions. But this is unavoidable: over the field of fractions, polynomials do have unique factorization; so seemingly distinct factorizations into linear factors over $A$ necessarily become identical aside from units over the field of fractions.
Another possibility is to use a higher-order polynomial that is irreducible over $A$, but not over the field of fractions. For example, if $ab=cd$ then
$$
(ax+c)(ax+d) = a(ax^2 + (c+d)x + b).
$$
In $mathbbZ[sqrt-5]$, this might be realized as
$$
(2x + 1 + sqrt-5)(2x + 1 - sqrt-5) = 2(2x^2 +2x + 3).
$$
This is, in essence, an example of the failure of Gauss's lemma for a non-UFD. I found the idea for this example in David E Speyer's answer to a MathOverflow question; the other responses may also be of interest.
answered Mar 23 at 12:38
FredHFredH
3,7251024
$endgroup$
add a comment |
$begingroup$
Perhaps this is the sort of thing you are looking for: If $a,b,c,din A$ are irreducible non-associate elements such that $ab=cd$, then
$$
(ax+c)(dx-b) = adx^2 - bc = (ax-c)(dx+b).
$$
With $A = mathbbZ[sqrt-5]$, taking $(a,b,c,d) = (2,3,1+sqrt-5,1-sqrt-5)$, this
results in
$$
(2x+1+sqrt-5)((1-sqrt-5)x-3) = (2x-1-sqrt-5)((1-sqrt-5)x+3).
$$
However, you might still see this a kind of cheat, since $ax+c$ and $dx+b$ actually are scalar multiples; it's just that the scalar factor, $d/a = b/c$, isn't in $A$ but is instead in its field of fractions. But this is unavoidable: over the field of fractions, polynomials do have unique factorization; so seemingly distinct factorizations into linear factors over $A$ necessarily become identical aside from units over the field of fractions.
Another possibility is to use a higher-order polynomial that is irreducible over $A$, but not over the field of fractions. For example, if $ab=cd$ then
$$
(ax+c)(ax+d) = a(ax^2 + (c+d)x + b).
$$
In $mathbbZ[sqrt-5]$, this might be realized as
$$
(2x + 1 + sqrt-5)(2x + 1 - sqrt-5) = 2(2x^2 +2x + 3).
$$
This is, in essence, an example of the failure of Gauss's lemma for a non-UFD. I found the idea for this example in David E Speyer's answer to a MathOverflow question; the other responses may also be of interest.
answered Mar 23 at 12:38
FredHFredH
3,7251024
$endgroup$
Perhaps this is the sort of thing you are looking for: If $a,b,c,din A$ are irreducible non-associate elements such that $ab=cd$, then
$$
(ax+c)(dx-b) = adx^2 - bc = (ax-c)(dx+b).
$$
With $A = mathbbZ[sqrt-5]$, taking $(a,b,c,d) = (2,3,1+sqrt-5,1-sqrt-5)$, this
results in
$$
(2x+1+sqrt-5)((1-sqrt-5)x-3) = (2x-1-sqrt-5)((1-sqrt-5)x+3).
$$
However, you might still see this a kind of cheat, since $ax+c$ and $dx+b$ actually are scalar multiples; it's just that the scalar factor, $d/a = b/c$, isn't in $A$ but is instead in its field of fractions. But this is unavoidable: over the field of fractions, polynomials do have unique factorization; so seemingly distinct factorizations into linear factors over $A$ necessarily become identical aside from units over the field of fractions.
Another possibility is to use a higher-order polynomial that is irreducible over $A$, but not over the field of fractions. For example, if $ab=cd$ then
$$
(ax+c)(ax+d) = a(ax^2 + (c+d)x + b).
$$
In $mathbbZ[sqrt-5]$, this might be realized as
$$
(2x + 1 + sqrt-5)(2x + 1 - sqrt-5) = 2(2x^2 +2x + 3).
$$
This is, in essence, an example of the failure of Gauss's lemma for a non-UFD. I found the idea for this example in David E Speyer's answer to a MathOverflow question; the other responses may also be of interest.
answered Mar 23 at 12:38
FredHFredH
3,7251024
answered Mar 23 at 12:38
FredHFredH
3,7251024
answered Mar 23 at 12:38
FredHFredH
3,7251024
answered Mar 23 at 12:38
FredHFredH
3,7251024
3,7251024
add a comment |
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 Mar 22 at 15:19
lhflhf
168k11173406
$endgroup$
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
Mar 22 at 15:20
$begingroup$
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
Mar 22 at 16:17
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 Mar 22 at 15:19
lhflhf
168k11173406
$endgroup$
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
Mar 22 at 15:20
$begingroup$
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
Mar 22 at 16:17
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 Mar 22 at 15:19
lhflhf
168k11173406
$endgroup$
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 Mar 22 at 15:19
lhflhf
168k11173406
answered Mar 22 at 15:19
lhflhf
168k11173406
answered Mar 22 at 15:19
lhflhf
168k11173406
answered Mar 22 at 15:19
lhflhf
168k11173406
168k11173406
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
Mar 22 at 15:20
$begingroup$
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
Mar 22 at 16:17
add a comment |
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
Mar 22 at 15:20
$begingroup$
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
Mar 22 at 16:17
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
Mar 22 at 15:20
$begingroup$
Adapted from math.stackexchange.com/a/2539517/589
$endgroup$
– lhf
Mar 22 at 15:20
$begingroup$
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
Mar 22 at 16:17
$begingroup$
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
Mar 22 at 16:17
add a comment |
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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
Mar 22 at 14:08
$begingroup$
@Arthur Yes, thank you very much, I didn't know how to write it in a simple way.
$endgroup$
– Marta Fornasier
Mar 22 at 14:19