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 $n>2$













3












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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!










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    3












    $begingroup$


    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!










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    New contributor




    Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      3












      3








      3





      $begingroup$


      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!










      share|cite|improve this question









      New contributor




      Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      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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      share|cite|improve this question








      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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          6












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Now I see, thank you!!
            $endgroup$
            – Jack Talion
            2 days ago


















          4












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$




















            4












            $begingroup$

            $mathbbZ[i]$ 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.






            share|cite|improve this answer











            $endgroup$




















              0












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$












                Your Answer





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                4 Answers
                4






                active

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                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                6












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$












                • $begingroup$
                  Now I see, thank you!!
                  $endgroup$
                  – Jack Talion
                  2 days ago















                6












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$












                • $begingroup$
                  Now I see, thank you!!
                  $endgroup$
                  – Jack Talion
                  2 days ago













                6












                6








                6





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                YiFanYiFan

                4,6761727




                4,6761727











                • $begingroup$
                  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




                $begingroup$
                Now I see, thank you!!
                $endgroup$
                – Jack Talion
                2 days ago











                4












                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$

















                  4












                  $begingroup$

                  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.






                  share|cite|improve this answer











                  $endgroup$















                    4












                    4








                    4





                    $begingroup$

                    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.






                    share|cite|improve this answer











                    $endgroup$



                    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.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 2 days ago









                    J. W. Tanner

                    3,1261320




                    3,1261320










                    answered 2 days ago









                    José Carlos SantosJosé Carlos Santos

                    167k22132235




                    167k22132235





















                        4












                        $begingroup$

                        $mathbbZ[i]$ 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.






                        share|cite|improve this answer











                        $endgroup$

















                          4












                          $begingroup$

                          $mathbbZ[i]$ 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.






                          share|cite|improve this answer











                          $endgroup$















                            4












                            4








                            4





                            $begingroup$

                            $mathbbZ[i]$ 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.






                            share|cite|improve this answer











                            $endgroup$



                            $mathbbZ[i]$ 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.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 2 days ago

























                            answered 2 days ago









                            Oscar LanziOscar Lanzi

                            13.1k12136




                            13.1k12136





















                                0












                                $begingroup$

                                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.






                                share|cite|improve this answer











                                $endgroup$

















                                  0












                                  $begingroup$

                                  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.






                                  share|cite|improve this answer











                                  $endgroup$















                                    0












                                    0








                                    0





                                    $begingroup$

                                    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.






                                    share|cite|improve this answer











                                    $endgroup$



                                    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.







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    answered 4 hours ago


























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                                        Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye