Question about proving $mathbbZ[i]$ is not UFDQuestion about UFDWhy is $mathbbZ[sqrt-n], nge 3$ not a UFD?An example of a non Noetherian UFDShow that $mathbbC[x,y]/(x^2+y^2-1)$ is a UFD.Determine if $mathbbZ[sqrt2i]$ is UFD.Proving $mathbbZ[sqrt 10]$ is not a UFDHow do I prove that $mathbbZ[sqrt19]$ is not a UFD?Is $mathbb Z_4$ a UFD (unique factorization domain)?Prove $ mathbbZ[sqrt -5] $ is not a UFD without factoring?$mathbb R left[ x _ 1 , ldots , x _ n right] / left( x _ 1 ^ 2 + cdots + x _ n ^ 2 - 1 right)$ is a UFD when $n>2$
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Question about proving $mathbbZ[i]$ is not UFD
Question about UFDWhy is $mathbbZ[sqrt-n], nge 3$ not a UFD?An example of a non Noetherian UFDShow that $mathbbC[x,y]/(x^2+y^2-1)$ is a UFD.Determine if $mathbbZ[sqrt2i]$ is UFD.Proving $mathbbZ[sqrt 10]$ is not a UFDHow do I prove that $mathbbZ[sqrt19]$ is not a UFD?Is $mathbb Z_4$ a UFD (unique factorization domain)?Prove $ mathbbZ[sqrt -5] $ is not a UFD without factoring?$mathbb R left[ x _ 1 , ldots , x _ n right] / left( x _ 1 ^ 2 + cdots + x _ n ^ 2 - 1 right)$ is a UFD when $n>2$
$begingroup$
I know is a silly question, but why $10=left(3+iright)left(3-iright)=2cdot5$ is not enough to prove that $mathbbZ[i]$ is not a UFD. Thanks in advance!
abstract-algebra ring-theory unique-factorization-domains
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$begingroup$
I know is a silly question, but why $10=left(3+iright)left(3-iright)=2cdot5$ is not enough to prove that $mathbbZ[i]$ is not a UFD. Thanks in advance!
abstract-algebra ring-theory unique-factorization-domains
New contributor
$endgroup$
add a comment |
$begingroup$
I know is a silly question, but why $10=left(3+iright)left(3-iright)=2cdot5$ is not enough to prove that $mathbbZ[i]$ is not a UFD. Thanks in advance!
abstract-algebra ring-theory unique-factorization-domains
New contributor
$endgroup$
I know is a silly question, but why $10=left(3+iright)left(3-iright)=2cdot5$ is not enough to prove that $mathbbZ[i]$ is not a UFD. Thanks in advance!
abstract-algebra ring-theory unique-factorization-domains
abstract-algebra ring-theory unique-factorization-domains
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New contributor
edited 2 days ago
José Carlos Santos
167k22132235
167k22132235
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asked 2 days ago
Jack TalionJack Talion
715
715
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4 Answers
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$begingroup$
Because $2=(1+i)(1-i)$ and $5=(2+i)(2-i)$, while $3+i=(1+i)(2-i)$ and $3-i=(1-i)(2+i)$. So really, you've just grouped the four prime factors of $10$ together in different ways.
$endgroup$
$begingroup$
Now I see, thank you!!
$endgroup$
– Jack Talion
2 days ago
add a comment |
$begingroup$
The ring $mathbb Z$ is a UFD, in spite of the fact that $12=2times6=4times3$. There is no contradiction, since these decompositions don't express $12$ as a product of irreducible integers.
$endgroup$
add a comment |
$begingroup$
$mathbbZ[i]$ is very much a UFD. When you allow complex factors, $4n+1$ "primes" are not primes anymore. They are sums of squares, which can then be rendered as products of complex conjugates; thus
$5=2^2+1^2=(2+i)(2-i)$
The number $2$ is similarly not prime anymore because it is also a sum of squares:
$2=1^2+1^2=(1+i)(1-i).$
Thereby
$10=2×5=(1+i)(1-i)(2+i)(2-i),$
which is unique except for unit factors (factors of $pm i$ or $-1$).
Here are some rules for identifying actual primes in $mathbbZ[i]$:
1) A positive whole number is prime only if it's an ordinary prime AND it cannot be the sum of two squares. So the ordinary prime has to be one less than a multiple of $4$. Other ordinary primes, with sum-of-two-squares representations, can be converted to complex conjugate products like those above.
2) A complex number $a+bi$ with $a$ and $b$ nonzero is prime only if $a^2+b^2$ is an ordinary prime. If this does not hold, you have factors whose corresponding sums of squares are factors of your $a^2+b^2$ value. Thus
$3+i=(1+i)(2-i)$
where
$3^2+1^2=10,1^2+1^2=2,2^2+1^2=5$.
3) Finally, any prime is linked to others by unit factors (factors of $pm i$ or $-1$). Thus $3$ is linked to $pm 3i$ and $-3$, all of which are in effect different images of a single prime. Similarly $1+i$ is linked to $1-i$ and $-1pm i$ through the unit factors.
$endgroup$
add a comment |
$begingroup$
Actually, there are a few more different-looking factorizations you could pile in there: $$(-1 + i)(1 + i)(-1 + 2i)(1 + 2i) = (1 - 3i)(1 + 3i) = 10.$$ However, that's no more distinct than saying $-2 times -5$ is a factorization of $10$ in $mathbbZ$ that's distinct from $5 times 2$.
For unique factorization, we ignore:
- ordering
- multiplication by units
In $mathbbZ[i]$, the units are, clockwise from "twelve": $i, 1, -i, -1$. So $3i, -3i, -3$ are all just as prime as $3$.
And of course you also have to check that you really have primes, and not composites with no obvious prime factors. For instance, in z, $1729 = 19 times 91$, but one of those is actually composite, though it sure looks like it could be prime.
$endgroup$
add a comment |
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4 Answers
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4 Answers
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$begingroup$
Because $2=(1+i)(1-i)$ and $5=(2+i)(2-i)$, while $3+i=(1+i)(2-i)$ and $3-i=(1-i)(2+i)$. So really, you've just grouped the four prime factors of $10$ together in different ways.
$endgroup$
$begingroup$
Now I see, thank you!!
$endgroup$
– Jack Talion
2 days ago
add a comment |
$begingroup$
Because $2=(1+i)(1-i)$ and $5=(2+i)(2-i)$, while $3+i=(1+i)(2-i)$ and $3-i=(1-i)(2+i)$. So really, you've just grouped the four prime factors of $10$ together in different ways.
$endgroup$
$begingroup$
Now I see, thank you!!
$endgroup$
– Jack Talion
2 days ago
add a comment |
$begingroup$
Because $2=(1+i)(1-i)$ and $5=(2+i)(2-i)$, while $3+i=(1+i)(2-i)$ and $3-i=(1-i)(2+i)$. So really, you've just grouped the four prime factors of $10$ together in different ways.
$endgroup$
Because $2=(1+i)(1-i)$ and $5=(2+i)(2-i)$, while $3+i=(1+i)(2-i)$ and $3-i=(1-i)(2+i)$. So really, you've just grouped the four prime factors of $10$ together in different ways.
answered 2 days ago
YiFanYiFan
4,6761727
4,6761727
$begingroup$
Now I see, thank you!!
$endgroup$
– Jack Talion
2 days ago
add a comment |
$begingroup$
Now I see, thank you!!
$endgroup$
– Jack Talion
2 days ago
$begingroup$
Now I see, thank you!!
$endgroup$
– Jack Talion
2 days ago
$begingroup$
Now I see, thank you!!
$endgroup$
– Jack Talion
2 days ago
add a comment |
$begingroup$
The ring $mathbb Z$ is a UFD, in spite of the fact that $12=2times6=4times3$. There is no contradiction, since these decompositions don't express $12$ as a product of irreducible integers.
$endgroup$
add a comment |
$begingroup$
The ring $mathbb Z$ is a UFD, in spite of the fact that $12=2times6=4times3$. There is no contradiction, since these decompositions don't express $12$ as a product of irreducible integers.
$endgroup$
add a comment |
$begingroup$
The ring $mathbb Z$ is a UFD, in spite of the fact that $12=2times6=4times3$. There is no contradiction, since these decompositions don't express $12$ as a product of irreducible integers.
$endgroup$
The ring $mathbb Z$ is a UFD, in spite of the fact that $12=2times6=4times3$. There is no contradiction, since these decompositions don't express $12$ as a product of irreducible integers.
edited 2 days ago
J. W. Tanner
3,1261320
3,1261320
answered 2 days ago
José Carlos SantosJosé Carlos Santos
167k22132235
167k22132235
add a comment |
add a comment |
$begingroup$
$mathbbZ[i]$ is very much a UFD. When you allow complex factors, $4n+1$ "primes" are not primes anymore. They are sums of squares, which can then be rendered as products of complex conjugates; thus
$5=2^2+1^2=(2+i)(2-i)$
The number $2$ is similarly not prime anymore because it is also a sum of squares:
$2=1^2+1^2=(1+i)(1-i).$
Thereby
$10=2×5=(1+i)(1-i)(2+i)(2-i),$
which is unique except for unit factors (factors of $pm i$ or $-1$).
Here are some rules for identifying actual primes in $mathbbZ[i]$:
1) A positive whole number is prime only if it's an ordinary prime AND it cannot be the sum of two squares. So the ordinary prime has to be one less than a multiple of $4$. Other ordinary primes, with sum-of-two-squares representations, can be converted to complex conjugate products like those above.
2) A complex number $a+bi$ with $a$ and $b$ nonzero is prime only if $a^2+b^2$ is an ordinary prime. If this does not hold, you have factors whose corresponding sums of squares are factors of your $a^2+b^2$ value. Thus
$3+i=(1+i)(2-i)$
where
$3^2+1^2=10,1^2+1^2=2,2^2+1^2=5$.
3) Finally, any prime is linked to others by unit factors (factors of $pm i$ or $-1$). Thus $3$ is linked to $pm 3i$ and $-3$, all of which are in effect different images of a single prime. Similarly $1+i$ is linked to $1-i$ and $-1pm i$ through the unit factors.
$endgroup$
add a comment |
$begingroup$
$mathbbZ[i]$ is very much a UFD. When you allow complex factors, $4n+1$ "primes" are not primes anymore. They are sums of squares, which can then be rendered as products of complex conjugates; thus
$5=2^2+1^2=(2+i)(2-i)$
The number $2$ is similarly not prime anymore because it is also a sum of squares:
$2=1^2+1^2=(1+i)(1-i).$
Thereby
$10=2×5=(1+i)(1-i)(2+i)(2-i),$
which is unique except for unit factors (factors of $pm i$ or $-1$).
Here are some rules for identifying actual primes in $mathbbZ[i]$:
1) A positive whole number is prime only if it's an ordinary prime AND it cannot be the sum of two squares. So the ordinary prime has to be one less than a multiple of $4$. Other ordinary primes, with sum-of-two-squares representations, can be converted to complex conjugate products like those above.
2) A complex number $a+bi$ with $a$ and $b$ nonzero is prime only if $a^2+b^2$ is an ordinary prime. If this does not hold, you have factors whose corresponding sums of squares are factors of your $a^2+b^2$ value. Thus
$3+i=(1+i)(2-i)$
where
$3^2+1^2=10,1^2+1^2=2,2^2+1^2=5$.
3) Finally, any prime is linked to others by unit factors (factors of $pm i$ or $-1$). Thus $3$ is linked to $pm 3i$ and $-3$, all of which are in effect different images of a single prime. Similarly $1+i$ is linked to $1-i$ and $-1pm i$ through the unit factors.
$endgroup$
add a comment |
$begingroup$
$mathbbZ[i]$ is very much a UFD. When you allow complex factors, $4n+1$ "primes" are not primes anymore. They are sums of squares, which can then be rendered as products of complex conjugates; thus
$5=2^2+1^2=(2+i)(2-i)$
The number $2$ is similarly not prime anymore because it is also a sum of squares:
$2=1^2+1^2=(1+i)(1-i).$
Thereby
$10=2×5=(1+i)(1-i)(2+i)(2-i),$
which is unique except for unit factors (factors of $pm i$ or $-1$).
Here are some rules for identifying actual primes in $mathbbZ[i]$:
1) A positive whole number is prime only if it's an ordinary prime AND it cannot be the sum of two squares. So the ordinary prime has to be one less than a multiple of $4$. Other ordinary primes, with sum-of-two-squares representations, can be converted to complex conjugate products like those above.
2) A complex number $a+bi$ with $a$ and $b$ nonzero is prime only if $a^2+b^2$ is an ordinary prime. If this does not hold, you have factors whose corresponding sums of squares are factors of your $a^2+b^2$ value. Thus
$3+i=(1+i)(2-i)$
where
$3^2+1^2=10,1^2+1^2=2,2^2+1^2=5$.
3) Finally, any prime is linked to others by unit factors (factors of $pm i$ or $-1$). Thus $3$ is linked to $pm 3i$ and $-3$, all of which are in effect different images of a single prime. Similarly $1+i$ is linked to $1-i$ and $-1pm i$ through the unit factors.
$endgroup$
$mathbbZ[i]$ is very much a UFD. When you allow complex factors, $4n+1$ "primes" are not primes anymore. They are sums of squares, which can then be rendered as products of complex conjugates; thus
$5=2^2+1^2=(2+i)(2-i)$
The number $2$ is similarly not prime anymore because it is also a sum of squares:
$2=1^2+1^2=(1+i)(1-i).$
Thereby
$10=2×5=(1+i)(1-i)(2+i)(2-i),$
which is unique except for unit factors (factors of $pm i$ or $-1$).
Here are some rules for identifying actual primes in $mathbbZ[i]$:
1) A positive whole number is prime only if it's an ordinary prime AND it cannot be the sum of two squares. So the ordinary prime has to be one less than a multiple of $4$. Other ordinary primes, with sum-of-two-squares representations, can be converted to complex conjugate products like those above.
2) A complex number $a+bi$ with $a$ and $b$ nonzero is prime only if $a^2+b^2$ is an ordinary prime. If this does not hold, you have factors whose corresponding sums of squares are factors of your $a^2+b^2$ value. Thus
$3+i=(1+i)(2-i)$
where
$3^2+1^2=10,1^2+1^2=2,2^2+1^2=5$.
3) Finally, any prime is linked to others by unit factors (factors of $pm i$ or $-1$). Thus $3$ is linked to $pm 3i$ and $-3$, all of which are in effect different images of a single prime. Similarly $1+i$ is linked to $1-i$ and $-1pm i$ through the unit factors.
edited 2 days ago
answered 2 days ago
Oscar LanziOscar Lanzi
13.1k12136
13.1k12136
add a comment |
add a comment |
$begingroup$
Actually, there are a few more different-looking factorizations you could pile in there: $$(-1 + i)(1 + i)(-1 + 2i)(1 + 2i) = (1 - 3i)(1 + 3i) = 10.$$ However, that's no more distinct than saying $-2 times -5$ is a factorization of $10$ in $mathbbZ$ that's distinct from $5 times 2$.
For unique factorization, we ignore:
- ordering
- multiplication by units
In $mathbbZ[i]$, the units are, clockwise from "twelve": $i, 1, -i, -1$. So $3i, -3i, -3$ are all just as prime as $3$.
And of course you also have to check that you really have primes, and not composites with no obvious prime factors. For instance, in z, $1729 = 19 times 91$, but one of those is actually composite, though it sure looks like it could be prime.
$endgroup$
add a comment |
$begingroup$
Actually, there are a few more different-looking factorizations you could pile in there: $$(-1 + i)(1 + i)(-1 + 2i)(1 + 2i) = (1 - 3i)(1 + 3i) = 10.$$ However, that's no more distinct than saying $-2 times -5$ is a factorization of $10$ in $mathbbZ$ that's distinct from $5 times 2$.
For unique factorization, we ignore:
- ordering
- multiplication by units
In $mathbbZ[i]$, the units are, clockwise from "twelve": $i, 1, -i, -1$. So $3i, -3i, -3$ are all just as prime as $3$.
And of course you also have to check that you really have primes, and not composites with no obvious prime factors. For instance, in z, $1729 = 19 times 91$, but one of those is actually composite, though it sure looks like it could be prime.
$endgroup$
add a comment |
$begingroup$
Actually, there are a few more different-looking factorizations you could pile in there: $$(-1 + i)(1 + i)(-1 + 2i)(1 + 2i) = (1 - 3i)(1 + 3i) = 10.$$ However, that's no more distinct than saying $-2 times -5$ is a factorization of $10$ in $mathbbZ$ that's distinct from $5 times 2$.
For unique factorization, we ignore:
- ordering
- multiplication by units
In $mathbbZ[i]$, the units are, clockwise from "twelve": $i, 1, -i, -1$. So $3i, -3i, -3$ are all just as prime as $3$.
And of course you also have to check that you really have primes, and not composites with no obvious prime factors. For instance, in z, $1729 = 19 times 91$, but one of those is actually composite, though it sure looks like it could be prime.
$endgroup$
Actually, there are a few more different-looking factorizations you could pile in there: $$(-1 + i)(1 + i)(-1 + 2i)(1 + 2i) = (1 - 3i)(1 + 3i) = 10.$$ However, that's no more distinct than saying $-2 times -5$ is a factorization of $10$ in $mathbbZ$ that's distinct from $5 times 2$.
For unique factorization, we ignore:
- ordering
- multiplication by units
In $mathbbZ[i]$, the units are, clockwise from "twelve": $i, 1, -i, -1$. So $3i, -3i, -3$ are all just as prime as $3$.
And of course you also have to check that you really have primes, and not composites with no obvious prime factors. For instance, in z, $1729 = 19 times 91$, but one of those is actually composite, though it sure looks like it could be prime.
answered 4 hours ago
community wiki
Mr. Brooks
add a comment |
add a comment |
Jack Talion is a new contributor. Be nice, and check out our Code of Conduct.
Jack Talion is a new contributor. Be nice, and check out our Code of Conduct.
Jack Talion is a new contributor. Be nice, and check out our Code of Conduct.
Jack Talion is a new contributor. Be nice, and check out our Code of Conduct.
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