Gaussian Integers form an Euclidean Ring












3














A Ring $R$ is called euclidean if a map $f:Rbackslash {0} rightarrow mathbb{N}$ exists with the following properties: For two elements $a,b in R$ with $bneq 0$ there exist $q,rin R$ with:



(i) $a=qb+r$ and



(ii) $r=0$ or $fleft(r right)<fleft(bright)$.



The Gaussian Integers are a Ring $G:= {a+bi:a,b inmathbb{Z}}$. I want to show that $G$ is an euclidean Ring. Because $G$ is obviously an Integral domain, I would be able to show that $G$ is a Principal ideal domain.



My problem is with the second part (ii). I saw that such a mapping can be defined with $a+bi mapsto a^2 + b^2$. Here is what I don't get: In our case the $b$ in (ii) is i, so I want to find a $f$ with $fleft(a right)<fleft(iright)$ for the Gaussian Integers right? How is $a+bi mapsto a^2 + b^2$ helping here? I am sorry if the notation is a bit odd. For the Gaussian integers I get (to put them into the frame like in (i)) that $a=q$, $b=i$ and $r=a$ right?



Appreciate your help, KingDingeling






abstract-algebra ring-theory ideals principal-ideal-domains euclidean-domain







share|cite|improve this question




















  • 1




    I get the impression that you're confusing two different things. The $a$ and $b$ in (i) are two complex numbers, and have nothing to do with the $a$ and $b$ in $a^2+b^2$ which are two real numbers.
    – saulspatz
    1 hour ago






  • 1




    You have to find $qinmathbf Z[i]$ such that the norm of $a-qb$ is less that the norm of $b$.
    – Bernard
    1 hour ago






  • 1




    I don't know if you want to do this as an exercise. If not I can also post a complete solution.
    – 0x539
    1 hour ago






  • 1




    You are using $b$ in two different ways. One is as an element of the ring in the first paragraph and one is as the imaginary component of an element in the second paragraph. In neither case is $b$ given to be $i$
    – Ross Millikan
    58 mins ago












  • Guys... thank you. I just got really confused. I will try it myself and share my ideas later. Thank you again!
    – KingDingeling
    58 mins ago
















3














A Ring $R$ is called euclidean if a map $f:Rbackslash {0} rightarrow mathbb{N}$ exists with the following properties: For two elements $a,b in R$ with $bneq 0$ there exist $q,rin R$ with:



(i) $a=qb+r$ and



(ii) $r=0$ or $fleft(r right)<fleft(bright)$.



The Gaussian Integers are a Ring $G:= {a+bi:a,b inmathbb{Z}}$. I want to show that $G$ is an euclidean Ring. Because $G$ is obviously an Integral domain, I would be able to show that $G$ is a Principal ideal domain.



My problem is with the second part (ii). I saw that such a mapping can be defined with $a+bi mapsto a^2 + b^2$. Here is what I don't get: In our case the $b$ in (ii) is i, so I want to find a $f$ with $fleft(a right)<fleft(iright)$ for the Gaussian Integers right? How is $a+bi mapsto a^2 + b^2$ helping here? I am sorry if the notation is a bit odd. For the Gaussian integers I get (to put them into the frame like in (i)) that $a=q$, $b=i$ and $r=a$ right?



Appreciate your help, KingDingeling






abstract-algebra ring-theory ideals principal-ideal-domains euclidean-domain







share|cite|improve this question




















  • 1




    I get the impression that you're confusing two different things. The $a$ and $b$ in (i) are two complex numbers, and have nothing to do with the $a$ and $b$ in $a^2+b^2$ which are two real numbers.
    – saulspatz
    1 hour ago






  • 1




    You have to find $qinmathbf Z[i]$ such that the norm of $a-qb$ is less that the norm of $b$.
    – Bernard
    1 hour ago






  • 1




    I don't know if you want to do this as an exercise. If not I can also post a complete solution.
    – 0x539
    1 hour ago






  • 1




    You are using $b$ in two different ways. One is as an element of the ring in the first paragraph and one is as the imaginary component of an element in the second paragraph. In neither case is $b$ given to be $i$
    – Ross Millikan
    58 mins ago












  • Guys... thank you. I just got really confused. I will try it myself and share my ideas later. Thank you again!
    – KingDingeling
    58 mins ago














3












3








3


1





A Ring $R$ is called euclidean if a map $f:Rbackslash {0} rightarrow mathbb{N}$ exists with the following properties: For two elements $a,b in R$ with $bneq 0$ there exist $q,rin R$ with:



(i) $a=qb+r$ and



(ii) $r=0$ or $fleft(r right)<fleft(bright)$.



The Gaussian Integers are a Ring $G:= {a+bi:a,b inmathbb{Z}}$. I want to show that $G$ is an euclidean Ring. Because $G$ is obviously an Integral domain, I would be able to show that $G$ is a Principal ideal domain.



My problem is with the second part (ii). I saw that such a mapping can be defined with $a+bi mapsto a^2 + b^2$. Here is what I don't get: In our case the $b$ in (ii) is i, so I want to find a $f$ with $fleft(a right)<fleft(iright)$ for the Gaussian Integers right? How is $a+bi mapsto a^2 + b^2$ helping here? I am sorry if the notation is a bit odd. For the Gaussian integers I get (to put them into the frame like in (i)) that $a=q$, $b=i$ and $r=a$ right?



Appreciate your help, KingDingeling






abstract-algebra ring-theory ideals principal-ideal-domains euclidean-domain







share|cite|improve this question















A Ring $R$ is called euclidean if a map $f:Rbackslash {0} rightarrow mathbb{N}$ exists with the following properties: For two elements $a,b in R$ with $bneq 0$ there exist $q,rin R$ with:



(i) $a=qb+r$ and



(ii) $r=0$ or $fleft(r right)<fleft(bright)$.



The Gaussian Integers are a Ring $G:= {a+bi:a,b inmathbb{Z}}$. I want to show that $G$ is an euclidean Ring. Because $G$ is obviously an Integral domain, I would be able to show that $G$ is a Principal ideal domain.



My problem is with the second part (ii). I saw that such a mapping can be defined with $a+bi mapsto a^2 + b^2$. Here is what I don't get: In our case the $b$ in (ii) is i, so I want to find a $f$ with $fleft(a right)<fleft(iright)$ for the Gaussian Integers right? How is $a+bi mapsto a^2 + b^2$ helping here? I am sorry if the notation is a bit odd. For the Gaussian integers I get (to put them into the frame like in (i)) that $a=q$, $b=i$ and $r=a$ right?



Appreciate your help, KingDingeling






abstract-algebra ring-theory ideals principal-ideal-domains euclidean-domain




abstract-algebra ring-theory ideals principal-ideal-domains euclidean-domain






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited 1 hour ago

























asked 1 hour ago









KingDingeling

675




675








  • 1




    I get the impression that you're confusing two different things. The $a$ and $b$ in (i) are two complex numbers, and have nothing to do with the $a$ and $b$ in $a^2+b^2$ which are two real numbers.
    – saulspatz
    1 hour ago






  • 1




    You have to find $qinmathbf Z[i]$ such that the norm of $a-qb$ is less that the norm of $b$.
    – Bernard
    1 hour ago






  • 1




    I don't know if you want to do this as an exercise. If not I can also post a complete solution.
    – 0x539
    1 hour ago






  • 1




    You are using $b$ in two different ways. One is as an element of the ring in the first paragraph and one is as the imaginary component of an element in the second paragraph. In neither case is $b$ given to be $i$
    – Ross Millikan
    58 mins ago












  • Guys... thank you. I just got really confused. I will try it myself and share my ideas later. Thank you again!
    – KingDingeling
    58 mins ago














  • 1




    I get the impression that you're confusing two different things. The $a$ and $b$ in (i) are two complex numbers, and have nothing to do with the $a$ and $b$ in $a^2+b^2$ which are two real numbers.
    – saulspatz
    1 hour ago






  • 1




    You have to find $qinmathbf Z[i]$ such that the norm of $a-qb$ is less that the norm of $b$.
    – Bernard
    1 hour ago






  • 1




    I don't know if you want to do this as an exercise. If not I can also post a complete solution.
    – 0x539
    1 hour ago






  • 1




    You are using $b$ in two different ways. One is as an element of the ring in the first paragraph and one is as the imaginary component of an element in the second paragraph. In neither case is $b$ given to be $i$
    – Ross Millikan
    58 mins ago












  • Guys... thank you. I just got really confused. I will try it myself and share my ideas later. Thank you again!
    – KingDingeling
    58 mins ago








1




1




I get the impression that you're confusing two different things. The $a$ and $b$ in (i) are two complex numbers, and have nothing to do with the $a$ and $b$ in $a^2+b^2$ which are two real numbers.
– saulspatz
1 hour ago




I get the impression that you're confusing two different things. The $a$ and $b$ in (i) are two complex numbers, and have nothing to do with the $a$ and $b$ in $a^2+b^2$ which are two real numbers.
– saulspatz
1 hour ago




1




1




You have to find $qinmathbf Z[i]$ such that the norm of $a-qb$ is less that the norm of $b$.
– Bernard
1 hour ago




You have to find $qinmathbf Z[i]$ such that the norm of $a-qb$ is less that the norm of $b$.
– Bernard
1 hour ago




1




1




I don't know if you want to do this as an exercise. If not I can also post a complete solution.
– 0x539
1 hour ago




I don't know if you want to do this as an exercise. If not I can also post a complete solution.
– 0x539
1 hour ago




1




1




You are using $b$ in two different ways. One is as an element of the ring in the first paragraph and one is as the imaginary component of an element in the second paragraph. In neither case is $b$ given to be $i$
– Ross Millikan
58 mins ago






You are using $b$ in two different ways. One is as an element of the ring in the first paragraph and one is as the imaginary component of an element in the second paragraph. In neither case is $b$ given to be $i$
– Ross Millikan
58 mins ago














Guys... thank you. I just got really confused. I will try it myself and share my ideas later. Thank you again!
– KingDingeling
58 mins ago




Guys... thank you. I just got really confused. I will try it myself and share my ideas later. Thank you again!
– KingDingeling
58 mins ago










2 Answers
2






active

oldest

votes


















3














You seem to be a bit confused. $b$ could be any element of $G$ except $0$. You have already found a good choice for $f$ so the only part left is, given arbitrary $a, b, in G$ with $b neq 0$, to find $q, r in G$ satisfying (i) and (ii).






share|cite|improve this answer





























    2














    It might help you to understand what can happen when a ring is not Euclidean. For example, consider $mathbb Z[sqrt{-5}]$, consisting of all numbers of the form $m + n sqrt{-5}$, with $m, n in mathbb Z$.



    The natural choice of Euclidean function is $$f(m + n sqrt{-5}) = m^2 + 5n^2.$$ Note that $f(m + n sqrt{-5})$ can never be negative, but it can be 0. Now try $gcd(2, 1 + sqrt{-5})$.



    Since $f(1 + sqrt{-5}) > f(2)$, we assign $a = 1 + sqrt{-5}$ and $b = 2$. Now we wish to express $a = qb + r$ with $f(r) < f(b)$, meaning $f(r) < 4$. Then the only possibilities for $r$ are $-1$, 0 and 1, as all other numbers in this ring have norms of at least 4.



    But then we see that none of the numbers $sqrt{-5}$, $1 + sqrt{-5}$, $2 + sqrt{-5}$ are divisible by 2. Therefore the Euclidean algorithm has failed in this ring, and therefore this domain is not Euclidean for the function $f$.



    Your task then is to prove that this can never happen in $G$, or $mathbb G$, or $mathbb Z[sqrt{-1}]$, or as it's more commonly notated, $mathbb Z[i]$, to prove that a suitable choice of $r$ can always be found. The Euclidean function is $f(m + ni) = m^2 + 1n^2$.



    A complete solution is to be found in certain number theory books, such as An Introduction to the Theory of Numbers by Ivan Niven and Herb Zuckermann.






    share|cite|improve this answer





















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      2 Answers
      2






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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

      votes









      3














      You seem to be a bit confused. $b$ could be any element of $G$ except $0$. You have already found a good choice for $f$ so the only part left is, given arbitrary $a, b, in G$ with $b neq 0$, to find $q, r in G$ satisfying (i) and (ii).






      share|cite|improve this answer


























        3














        You seem to be a bit confused. $b$ could be any element of $G$ except $0$. You have already found a good choice for $f$ so the only part left is, given arbitrary $a, b, in G$ with $b neq 0$, to find $q, r in G$ satisfying (i) and (ii).






        share|cite|improve this answer
























          3












          3








          3






          You seem to be a bit confused. $b$ could be any element of $G$ except $0$. You have already found a good choice for $f$ so the only part left is, given arbitrary $a, b, in G$ with $b neq 0$, to find $q, r in G$ satisfying (i) and (ii).






          share|cite|improve this answer












          You seem to be a bit confused. $b$ could be any element of $G$ except $0$. You have already found a good choice for $f$ so the only part left is, given arbitrary $a, b, in G$ with $b neq 0$, to find $q, r in G$ satisfying (i) and (ii).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          0x539

          1,022316




          1,022316























              2














              It might help you to understand what can happen when a ring is not Euclidean. For example, consider $mathbb Z[sqrt{-5}]$, consisting of all numbers of the form $m + n sqrt{-5}$, with $m, n in mathbb Z$.



              The natural choice of Euclidean function is $$f(m + n sqrt{-5}) = m^2 + 5n^2.$$ Note that $f(m + n sqrt{-5})$ can never be negative, but it can be 0. Now try $gcd(2, 1 + sqrt{-5})$.



              Since $f(1 + sqrt{-5}) > f(2)$, we assign $a = 1 + sqrt{-5}$ and $b = 2$. Now we wish to express $a = qb + r$ with $f(r) < f(b)$, meaning $f(r) < 4$. Then the only possibilities for $r$ are $-1$, 0 and 1, as all other numbers in this ring have norms of at least 4.



              But then we see that none of the numbers $sqrt{-5}$, $1 + sqrt{-5}$, $2 + sqrt{-5}$ are divisible by 2. Therefore the Euclidean algorithm has failed in this ring, and therefore this domain is not Euclidean for the function $f$.



              Your task then is to prove that this can never happen in $G$, or $mathbb G$, or $mathbb Z[sqrt{-1}]$, or as it's more commonly notated, $mathbb Z[i]$, to prove that a suitable choice of $r$ can always be found. The Euclidean function is $f(m + ni) = m^2 + 1n^2$.



              A complete solution is to be found in certain number theory books, such as An Introduction to the Theory of Numbers by Ivan Niven and Herb Zuckermann.






              share|cite|improve this answer


























                2














                It might help you to understand what can happen when a ring is not Euclidean. For example, consider $mathbb Z[sqrt{-5}]$, consisting of all numbers of the form $m + n sqrt{-5}$, with $m, n in mathbb Z$.



                The natural choice of Euclidean function is $$f(m + n sqrt{-5}) = m^2 + 5n^2.$$ Note that $f(m + n sqrt{-5})$ can never be negative, but it can be 0. Now try $gcd(2, 1 + sqrt{-5})$.



                Since $f(1 + sqrt{-5}) > f(2)$, we assign $a = 1 + sqrt{-5}$ and $b = 2$. Now we wish to express $a = qb + r$ with $f(r) < f(b)$, meaning $f(r) < 4$. Then the only possibilities for $r$ are $-1$, 0 and 1, as all other numbers in this ring have norms of at least 4.



                But then we see that none of the numbers $sqrt{-5}$, $1 + sqrt{-5}$, $2 + sqrt{-5}$ are divisible by 2. Therefore the Euclidean algorithm has failed in this ring, and therefore this domain is not Euclidean for the function $f$.



                Your task then is to prove that this can never happen in $G$, or $mathbb G$, or $mathbb Z[sqrt{-1}]$, or as it's more commonly notated, $mathbb Z[i]$, to prove that a suitable choice of $r$ can always be found. The Euclidean function is $f(m + ni) = m^2 + 1n^2$.



                A complete solution is to be found in certain number theory books, such as An Introduction to the Theory of Numbers by Ivan Niven and Herb Zuckermann.






                share|cite|improve this answer
























                  2












                  2








                  2






                  It might help you to understand what can happen when a ring is not Euclidean. For example, consider $mathbb Z[sqrt{-5}]$, consisting of all numbers of the form $m + n sqrt{-5}$, with $m, n in mathbb Z$.



                  The natural choice of Euclidean function is $$f(m + n sqrt{-5}) = m^2 + 5n^2.$$ Note that $f(m + n sqrt{-5})$ can never be negative, but it can be 0. Now try $gcd(2, 1 + sqrt{-5})$.



                  Since $f(1 + sqrt{-5}) > f(2)$, we assign $a = 1 + sqrt{-5}$ and $b = 2$. Now we wish to express $a = qb + r$ with $f(r) < f(b)$, meaning $f(r) < 4$. Then the only possibilities for $r$ are $-1$, 0 and 1, as all other numbers in this ring have norms of at least 4.



                  But then we see that none of the numbers $sqrt{-5}$, $1 + sqrt{-5}$, $2 + sqrt{-5}$ are divisible by 2. Therefore the Euclidean algorithm has failed in this ring, and therefore this domain is not Euclidean for the function $f$.



                  Your task then is to prove that this can never happen in $G$, or $mathbb G$, or $mathbb Z[sqrt{-1}]$, or as it's more commonly notated, $mathbb Z[i]$, to prove that a suitable choice of $r$ can always be found. The Euclidean function is $f(m + ni) = m^2 + 1n^2$.



                  A complete solution is to be found in certain number theory books, such as An Introduction to the Theory of Numbers by Ivan Niven and Herb Zuckermann.






                  share|cite|improve this answer












                  It might help you to understand what can happen when a ring is not Euclidean. For example, consider $mathbb Z[sqrt{-5}]$, consisting of all numbers of the form $m + n sqrt{-5}$, with $m, n in mathbb Z$.



                  The natural choice of Euclidean function is $$f(m + n sqrt{-5}) = m^2 + 5n^2.$$ Note that $f(m + n sqrt{-5})$ can never be negative, but it can be 0. Now try $gcd(2, 1 + sqrt{-5})$.



                  Since $f(1 + sqrt{-5}) > f(2)$, we assign $a = 1 + sqrt{-5}$ and $b = 2$. Now we wish to express $a = qb + r$ with $f(r) < f(b)$, meaning $f(r) < 4$. Then the only possibilities for $r$ are $-1$, 0 and 1, as all other numbers in this ring have norms of at least 4.



                  But then we see that none of the numbers $sqrt{-5}$, $1 + sqrt{-5}$, $2 + sqrt{-5}$ are divisible by 2. Therefore the Euclidean algorithm has failed in this ring, and therefore this domain is not Euclidean for the function $f$.



                  Your task then is to prove that this can never happen in $G$, or $mathbb G$, or $mathbb Z[sqrt{-1}]$, or as it's more commonly notated, $mathbb Z[i]$, to prove that a suitable choice of $r$ can always be found. The Euclidean function is $f(m + ni) = m^2 + 1n^2$.



                  A complete solution is to be found in certain number theory books, such as An Introduction to the Theory of Numbers by Ivan Niven and Herb Zuckermann.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 20 mins ago









                  Robert Soupe

                  10.9k21950




                  10.9k21950






























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                      濃尾地震

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                      濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸...
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