Factorization in prime elements of $mathbb{Z}left[sqrt{p}right]$ for a prime number $p$












3














I'm having troubles with the following problem:



Let $p$ be a prime number in $mathbb{Z}$, and $alphainmathbb{Z}left[sqrt{p}right]$ which is not a unit. Prove that $alpha$ have a factorization in irreducible elements of $mathbb{Z}left[sqrt{p}right]$.



At the beginning I though that that ring was a Euclidean Domain, but that fails for $mathbb{Z}left[sqrt{5}right]$. So, I don't know where to start now.



Thanks in advance.






abstract-algebra ring-theory integers







share|cite|improve this question





























    3














    I'm having troubles with the following problem:



    Let $p$ be a prime number in $mathbb{Z}$, and $alphainmathbb{Z}left[sqrt{p}right]$ which is not a unit. Prove that $alpha$ have a factorization in irreducible elements of $mathbb{Z}left[sqrt{p}right]$.



    At the beginning I though that that ring was a Euclidean Domain, but that fails for $mathbb{Z}left[sqrt{5}right]$. So, I don't know where to start now.



    Thanks in advance.






    abstract-algebra ring-theory integers







    share|cite|improve this question



























      3












      3








      3







      I'm having troubles with the following problem:



      Let $p$ be a prime number in $mathbb{Z}$, and $alphainmathbb{Z}left[sqrt{p}right]$ which is not a unit. Prove that $alpha$ have a factorization in irreducible elements of $mathbb{Z}left[sqrt{p}right]$.



      At the beginning I though that that ring was a Euclidean Domain, but that fails for $mathbb{Z}left[sqrt{5}right]$. So, I don't know where to start now.



      Thanks in advance.






      abstract-algebra ring-theory integers







      share|cite|improve this question















      I'm having troubles with the following problem:



      Let $p$ be a prime number in $mathbb{Z}$, and $alphainmathbb{Z}left[sqrt{p}right]$ which is not a unit. Prove that $alpha$ have a factorization in irreducible elements of $mathbb{Z}left[sqrt{p}right]$.



      At the beginning I though that that ring was a Euclidean Domain, but that fails for $mathbb{Z}left[sqrt{5}right]$. So, I don't know where to start now.



      Thanks in advance.






      abstract-algebra ring-theory integers




      abstract-algebra ring-theory integers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 14:37









      Eleven-Eleven

      5,39072659




      5,39072659










      asked Dec 10 '18 at 14:31









      José Alejandro Aburto Araneda

      756110




      756110






















          2 Answers
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          active

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          3














          Hint: Consider the norm $N(a+bsqrt{p})=|a^2-b^2p|$. Prove that if $delta$ divides $alpha$, then $N(delta) le N(alpha)$. Then use induction on $N(alpha)$.



          Or, more sophisticatedly, argue that $mathbb{Z}left[sqrt{p}right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.






          share|cite|improve this answer























          • I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
            – Matt Samuel
            Dec 10 '18 at 14:59






          • 1




            @MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
            – Bill Dubuque
            Dec 10 '18 at 21:43












          • @Bill Herstein and Artin undergraduate, Hungerford graduate.
            – Matt Samuel
            Dec 10 '18 at 23:29










          • @Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
            – Bill Dubuque
            Dec 10 '18 at 23:56










          • @Bill Then either we didn't cover them, or, heaven forbid...I forgot!
            – Matt Samuel
            Dec 10 '18 at 23:58



















          0














          Suppose $alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that
          $$alpha=a_1a_2$$
          If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so
          $$alpha = a_3a_4a_2$$
          In general there is an increasing chain of ideals
          $$(a_1)subseteq (a_3)subseteq cdots$$
          Since $mathbb{Z}[sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing,
          $$alpha = a_1a_2$$
          with $a_1$ irreducible. Thus you can iterate this to write
          $$alpha = a_1cdots a_n$$
          with $a_1,ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals
          $$(a_{n,1})subseteq (a_{n+1,2})subseteq cdots$$
          This must also stabilize, until you end up with a factorization into irreducible elements.






          share|cite|improve this answer





















          • At this level, one should justify the Noetherian claim.
            – Bill Dubuque
            Dec 10 '18 at 21:45











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          3














          Hint: Consider the norm $N(a+bsqrt{p})=|a^2-b^2p|$. Prove that if $delta$ divides $alpha$, then $N(delta) le N(alpha)$. Then use induction on $N(alpha)$.



          Or, more sophisticatedly, argue that $mathbb{Z}left[sqrt{p}right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.






          share|cite|improve this answer























          • I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
            – Matt Samuel
            Dec 10 '18 at 14:59






          • 1




            @MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
            – Bill Dubuque
            Dec 10 '18 at 21:43












          • @Bill Herstein and Artin undergraduate, Hungerford graduate.
            – Matt Samuel
            Dec 10 '18 at 23:29










          • @Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
            – Bill Dubuque
            Dec 10 '18 at 23:56










          • @Bill Then either we didn't cover them, or, heaven forbid...I forgot!
            – Matt Samuel
            Dec 10 '18 at 23:58
















          3














          Hint: Consider the norm $N(a+bsqrt{p})=|a^2-b^2p|$. Prove that if $delta$ divides $alpha$, then $N(delta) le N(alpha)$. Then use induction on $N(alpha)$.



          Or, more sophisticatedly, argue that $mathbb{Z}left[sqrt{p}right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.






          share|cite|improve this answer























          • I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
            – Matt Samuel
            Dec 10 '18 at 14:59






          • 1




            @MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
            – Bill Dubuque
            Dec 10 '18 at 21:43












          • @Bill Herstein and Artin undergraduate, Hungerford graduate.
            – Matt Samuel
            Dec 10 '18 at 23:29










          • @Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
            – Bill Dubuque
            Dec 10 '18 at 23:56










          • @Bill Then either we didn't cover them, or, heaven forbid...I forgot!
            – Matt Samuel
            Dec 10 '18 at 23:58














          3












          3








          3






          Hint: Consider the norm $N(a+bsqrt{p})=|a^2-b^2p|$. Prove that if $delta$ divides $alpha$, then $N(delta) le N(alpha)$. Then use induction on $N(alpha)$.



          Or, more sophisticatedly, argue that $mathbb{Z}left[sqrt{p}right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.






          share|cite|improve this answer














          Hint: Consider the norm $N(a+bsqrt{p})=|a^2-b^2p|$. Prove that if $delta$ divides $alpha$, then $N(delta) le N(alpha)$. Then use induction on $N(alpha)$.



          Or, more sophisticatedly, argue that $mathbb{Z}left[sqrt{p}right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 15:52

























          answered Dec 10 '18 at 14:53









          lhf

          163k10167386




          163k10167386












          • I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
            – Matt Samuel
            Dec 10 '18 at 14:59






          • 1




            @MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
            – Bill Dubuque
            Dec 10 '18 at 21:43












          • @Bill Herstein and Artin undergraduate, Hungerford graduate.
            – Matt Samuel
            Dec 10 '18 at 23:29










          • @Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
            – Bill Dubuque
            Dec 10 '18 at 23:56










          • @Bill Then either we didn't cover them, or, heaven forbid...I forgot!
            – Matt Samuel
            Dec 10 '18 at 23:58


















          • I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
            – Matt Samuel
            Dec 10 '18 at 14:59






          • 1




            @MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
            – Bill Dubuque
            Dec 10 '18 at 21:43












          • @Bill Herstein and Artin undergraduate, Hungerford graduate.
            – Matt Samuel
            Dec 10 '18 at 23:29










          • @Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
            – Bill Dubuque
            Dec 10 '18 at 23:56










          • @Bill Then either we didn't cover them, or, heaven forbid...I forgot!
            – Matt Samuel
            Dec 10 '18 at 23:58
















          I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
          – Matt Samuel
          Dec 10 '18 at 14:59




          I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
          – Matt Samuel
          Dec 10 '18 at 14:59




          1




          1




          @MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
          – Bill Dubuque
          Dec 10 '18 at 21:43






          @MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
          – Bill Dubuque
          Dec 10 '18 at 21:43














          @Bill Herstein and Artin undergraduate, Hungerford graduate.
          – Matt Samuel
          Dec 10 '18 at 23:29




          @Bill Herstein and Artin undergraduate, Hungerford graduate.
          – Matt Samuel
          Dec 10 '18 at 23:29












          @Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
          – Bill Dubuque
          Dec 10 '18 at 23:56




          @Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
          – Bill Dubuque
          Dec 10 '18 at 23:56












          @Bill Then either we didn't cover them, or, heaven forbid...I forgot!
          – Matt Samuel
          Dec 10 '18 at 23:58




          @Bill Then either we didn't cover them, or, heaven forbid...I forgot!
          – Matt Samuel
          Dec 10 '18 at 23:58











          0














          Suppose $alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that
          $$alpha=a_1a_2$$
          If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so
          $$alpha = a_3a_4a_2$$
          In general there is an increasing chain of ideals
          $$(a_1)subseteq (a_3)subseteq cdots$$
          Since $mathbb{Z}[sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing,
          $$alpha = a_1a_2$$
          with $a_1$ irreducible. Thus you can iterate this to write
          $$alpha = a_1cdots a_n$$
          with $a_1,ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals
          $$(a_{n,1})subseteq (a_{n+1,2})subseteq cdots$$
          This must also stabilize, until you end up with a factorization into irreducible elements.






          share|cite|improve this answer





















          • At this level, one should justify the Noetherian claim.
            – Bill Dubuque
            Dec 10 '18 at 21:45
















          0














          Suppose $alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that
          $$alpha=a_1a_2$$
          If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so
          $$alpha = a_3a_4a_2$$
          In general there is an increasing chain of ideals
          $$(a_1)subseteq (a_3)subseteq cdots$$
          Since $mathbb{Z}[sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing,
          $$alpha = a_1a_2$$
          with $a_1$ irreducible. Thus you can iterate this to write
          $$alpha = a_1cdots a_n$$
          with $a_1,ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals
          $$(a_{n,1})subseteq (a_{n+1,2})subseteq cdots$$
          This must also stabilize, until you end up with a factorization into irreducible elements.






          share|cite|improve this answer





















          • At this level, one should justify the Noetherian claim.
            – Bill Dubuque
            Dec 10 '18 at 21:45














          0












          0








          0






          Suppose $alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that
          $$alpha=a_1a_2$$
          If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so
          $$alpha = a_3a_4a_2$$
          In general there is an increasing chain of ideals
          $$(a_1)subseteq (a_3)subseteq cdots$$
          Since $mathbb{Z}[sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing,
          $$alpha = a_1a_2$$
          with $a_1$ irreducible. Thus you can iterate this to write
          $$alpha = a_1cdots a_n$$
          with $a_1,ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals
          $$(a_{n,1})subseteq (a_{n+1,2})subseteq cdots$$
          This must also stabilize, until you end up with a factorization into irreducible elements.






          share|cite|improve this answer












          Suppose $alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that
          $$alpha=a_1a_2$$
          If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so
          $$alpha = a_3a_4a_2$$
          In general there is an increasing chain of ideals
          $$(a_1)subseteq (a_3)subseteq cdots$$
          Since $mathbb{Z}[sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing,
          $$alpha = a_1a_2$$
          with $a_1$ irreducible. Thus you can iterate this to write
          $$alpha = a_1cdots a_n$$
          with $a_1,ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals
          $$(a_{n,1})subseteq (a_{n+1,2})subseteq cdots$$
          This must also stabilize, until you end up with a factorization into irreducible elements.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 14:58









          Matt Samuel

          37.3k63465




          37.3k63465












          • At this level, one should justify the Noetherian claim.
            – Bill Dubuque
            Dec 10 '18 at 21:45


















          • At this level, one should justify the Noetherian claim.
            – Bill Dubuque
            Dec 10 '18 at 21:45
















          At this level, one should justify the Noetherian claim.
          – Bill Dubuque
          Dec 10 '18 at 21:45




          At this level, one should justify the Noetherian claim.
          – Bill Dubuque
          Dec 10 '18 at 21:45


















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