$mathbb Z[sqrt{-7}]$ is not a UFD












0












$begingroup$


I can prove this for $sqrt{-5}$. What changes for the above question? Is it a similar proof? What would be the irreducible elements? Are they $1$ and $7$ ?






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please include your thoughts and efforts on the problem.
    $endgroup$
    – Servaes
    Dec 6 '18 at 14:38










  • $begingroup$
    Hint: $3^2+7=4^2$.
    $endgroup$
    – egreg
    Dec 6 '18 at 14:55






  • 1




    $begingroup$
    Hint: $(1+sqrt{-7})(1-sqrt{-7})$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 6 '18 at 15:32






  • 1




    $begingroup$
    See also this closedly related question.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 20:24
















0












$begingroup$


I can prove this for $sqrt{-5}$. What changes for the above question? Is it a similar proof? What would be the irreducible elements? Are they $1$ and $7$ ?






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please include your thoughts and efforts on the problem.
    $endgroup$
    – Servaes
    Dec 6 '18 at 14:38










  • $begingroup$
    Hint: $3^2+7=4^2$.
    $endgroup$
    – egreg
    Dec 6 '18 at 14:55






  • 1




    $begingroup$
    Hint: $(1+sqrt{-7})(1-sqrt{-7})$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 6 '18 at 15:32






  • 1




    $begingroup$
    See also this closedly related question.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 20:24














0












0








0





$begingroup$


I can prove this for $sqrt{-5}$. What changes for the above question? Is it a similar proof? What would be the irreducible elements? Are they $1$ and $7$ ?






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$




I can prove this for $sqrt{-5}$. What changes for the above question? Is it a similar proof? What would be the irreducible elements? Are they $1$ and $7$ ?






abstract-algebra ring-theory




abstract-algebra ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 1:07









Andrews

3901317




3901317










asked Dec 6 '18 at 14:33









JacobKnightJacobKnight

163




163












  • $begingroup$
    Please include your thoughts and efforts on the problem.
    $endgroup$
    – Servaes
    Dec 6 '18 at 14:38










  • $begingroup$
    Hint: $3^2+7=4^2$.
    $endgroup$
    – egreg
    Dec 6 '18 at 14:55






  • 1




    $begingroup$
    Hint: $(1+sqrt{-7})(1-sqrt{-7})$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 6 '18 at 15:32






  • 1




    $begingroup$
    See also this closedly related question.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 20:24


















  • $begingroup$
    Please include your thoughts and efforts on the problem.
    $endgroup$
    – Servaes
    Dec 6 '18 at 14:38










  • $begingroup$
    Hint: $3^2+7=4^2$.
    $endgroup$
    – egreg
    Dec 6 '18 at 14:55






  • 1




    $begingroup$
    Hint: $(1+sqrt{-7})(1-sqrt{-7})$.
    $endgroup$
    – Jyrki Lahtonen
    Dec 6 '18 at 15:32






  • 1




    $begingroup$
    See also this closedly related question.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 20:24
















$begingroup$
Please include your thoughts and efforts on the problem.
$endgroup$
– Servaes
Dec 6 '18 at 14:38




$begingroup$
Please include your thoughts and efforts on the problem.
$endgroup$
– Servaes
Dec 6 '18 at 14:38












$begingroup$
Hint: $3^2+7=4^2$.
$endgroup$
– egreg
Dec 6 '18 at 14:55




$begingroup$
Hint: $3^2+7=4^2$.
$endgroup$
– egreg
Dec 6 '18 at 14:55




1




1




$begingroup$
Hint: $(1+sqrt{-7})(1-sqrt{-7})$.
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 15:32




$begingroup$
Hint: $(1+sqrt{-7})(1-sqrt{-7})$.
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 15:32




1




1




$begingroup$
See also this closedly related question.
$endgroup$
– Bill Dubuque
Dec 6 '18 at 20:24




$begingroup$
See also this closedly related question.
$endgroup$
– Bill Dubuque
Dec 6 '18 at 20:24










2 Answers
2






active

oldest

votes


















2












$begingroup$

Hint $ $ It's simpler: UFDs are integrally closed, but it is not since $,(1+sqrt{-7})/2 $ is a root of $,x^2-x+2.,$ Said more simply it fails RRT = Rational Root Test, which holds true in any UFD.



Remark $ $ The RRT proof fails because some gcd it employs fails to exist in $,Bbb Z[sqrt{-7}].,$ Since gcds always exist in a UFD, we can use such gcd failure to prove rings are not UFDs. Let's use this to prove a more general case (for variety vs. traditional proofs).



Lemma $ Bbb Z[sqrt d]$ is not a UFD if $,color{#0a0}{dequiv 1pmod{!4}} $



Proof $ $ Assume it's a UFD. $ color{#0a0}{ww'}=(1!+!sqrt d)(1!-!sqrt d) = 1-d = color{#0a0}{4k},, color{#c00}{w} = 2-w',$



therefore the gcd product $ ,(2,color{#0a0}w)(2,color{#0a0}w') = (4,2w',2w,color{#0a0}{4k}) = 2(2,w',color{#c00}{w}) = 2(2,w')$



Cancelling $,(2,w'),Rightarrow, (2,w) = (2),Rightarrow, 2mid w,Rightarrow, w/2 = (1!+!sqrt d)/2in Bbb Z[sqrt d] Rightarrow!Leftarrow$



Further $ $ if $,w=sqrt{c^2d},$ then $,Bbb Z[w],$ UFD $,Rightarrow, cmid 1,,$ else $,(w/c)^2 = d,$ contra RRT (as above).

Alternatively by gcds $,(c,w)^2 = (c^2,cw,c^2d) = c(c,w),Rightarrow, (c,w) = (c),Rightarrow, cmid w,Rightarrow,cmid 1$



Combining yields a purely gcd-based proof of the following well-known fact.



Theorem $, Bbb Z[sqrt d], $ a UFD $,Rightarrow, d,$ squarefree and $,dnotequiv 1pmod{!4}$



Note $ $ The proof of the theorem also works if we instead assume the ring is a PID and read the tuples as ideals vs. gcds [since the proof uses only laws common to both: $ $ distributive, associative, commutative, and $(a,b,bc) = (a,b),$]. Then it yields $,(2,w),$ is noninvertible (so nonprincipal).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    And more generally this idea goes through in $mathbf Z[sqrt{d}]$ for nonsquare $d equiv 1 bmod 4$, not just d = -7$.
    $endgroup$
    – KCd
    Dec 6 '18 at 15:40










  • $begingroup$
    @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 17:25










  • $begingroup$
    +1, I think $Z[sqrt d]$ is integrally closed if and only if $d$ congurent to $1$ module $4$.
    $endgroup$
    – UserS
    Dec 6 '18 at 17:39












  • $begingroup$
    @UserS: it is integrally closed if and only if $d$ is squarefree and $d$ is not congruent to $1 bmod 4$.
    $endgroup$
    – KCd
    Dec 6 '18 at 17:56










  • $begingroup$
    @UserS KCd is correct. I added a proof (using only gcds for variety vs. traditional proofs).
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 19:33



















1












$begingroup$

$$(1+sqrt{-7})×(1-sqrt{-7})=8=2×2×2.$$We show $2,1+sqrt{-7},1-sqrt{-7}$ are irreducibles.



So let $2=(a+sqrt{-7}b)(c+sqrt{-7}d)$ where $a,b,c,d in Bbb Z$, then taking norms $4=(a^2+7b^2)(c^2+7d^2)implies b=d=0$ , since for $e,fin Bbb Z$ and $fnot =0implies e^2+7f^2geq 7$. Also either $a$ or $c$ is $+1,-1$ as $2^2=4=(ac)^2$ and $Bbb Z$ is a UFD i.e. either $a+sqrt{-7}b$ or $c+sqrt{-7}d$ unit.



Next, let $1+sqrt{-7}=(a+sqrt{-7}b)(c+sqrt{-7}d)$ with $a,b,c,din Bbb Z$. Again taking norms $8=(a^2+7b^2)(c^2+7d^2)implies bd=0$ and both $b,d$ are not zero as $8$ has no integral square root. So without loss of generality assume $b=0$ and $dnot=0implies 8=a^2(c^2+7d^2)$ . Note that $d$ must be $+1,-1$ , otherwise $c^2+7d^2geq 28$. Also $cnot=0$ , as $7not |8$. Hence $c$ must be $+1,-1$ otherwise $c^2+7d^2geq 11$. Therefore $a=1,-1$ i.e. $a+sqrt{-7}b$ is unit.



Similarly $1-sqrt{-7}$ is irreducible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We only need $2$ is rred. Then $,2mid ww', 2nmid w,w',$ so $2$ is a non-prime irred so it's not a UFD.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 17:30










  • $begingroup$
    Yeah, that's right , but I tried to prove only using definition of UFD.
    $endgroup$
    – UserS
    Dec 6 '18 at 17:35











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2 Answers
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2 Answers
2






active

oldest

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active

oldest

votes






active

oldest

votes









2












$begingroup$

Hint $ $ It's simpler: UFDs are integrally closed, but it is not since $,(1+sqrt{-7})/2 $ is a root of $,x^2-x+2.,$ Said more simply it fails RRT = Rational Root Test, which holds true in any UFD.



Remark $ $ The RRT proof fails because some gcd it employs fails to exist in $,Bbb Z[sqrt{-7}].,$ Since gcds always exist in a UFD, we can use such gcd failure to prove rings are not UFDs. Let's use this to prove a more general case (for variety vs. traditional proofs).



Lemma $ Bbb Z[sqrt d]$ is not a UFD if $,color{#0a0}{dequiv 1pmod{!4}} $



Proof $ $ Assume it's a UFD. $ color{#0a0}{ww'}=(1!+!sqrt d)(1!-!sqrt d) = 1-d = color{#0a0}{4k},, color{#c00}{w} = 2-w',$



therefore the gcd product $ ,(2,color{#0a0}w)(2,color{#0a0}w') = (4,2w',2w,color{#0a0}{4k}) = 2(2,w',color{#c00}{w}) = 2(2,w')$



Cancelling $,(2,w'),Rightarrow, (2,w) = (2),Rightarrow, 2mid w,Rightarrow, w/2 = (1!+!sqrt d)/2in Bbb Z[sqrt d] Rightarrow!Leftarrow$



Further $ $ if $,w=sqrt{c^2d},$ then $,Bbb Z[w],$ UFD $,Rightarrow, cmid 1,,$ else $,(w/c)^2 = d,$ contra RRT (as above).

Alternatively by gcds $,(c,w)^2 = (c^2,cw,c^2d) = c(c,w),Rightarrow, (c,w) = (c),Rightarrow, cmid w,Rightarrow,cmid 1$



Combining yields a purely gcd-based proof of the following well-known fact.



Theorem $, Bbb Z[sqrt d], $ a UFD $,Rightarrow, d,$ squarefree and $,dnotequiv 1pmod{!4}$



Note $ $ The proof of the theorem also works if we instead assume the ring is a PID and read the tuples as ideals vs. gcds [since the proof uses only laws common to both: $ $ distributive, associative, commutative, and $(a,b,bc) = (a,b),$]. Then it yields $,(2,w),$ is noninvertible (so nonprincipal).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    And more generally this idea goes through in $mathbf Z[sqrt{d}]$ for nonsquare $d equiv 1 bmod 4$, not just d = -7$.
    $endgroup$
    – KCd
    Dec 6 '18 at 15:40










  • $begingroup$
    @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 17:25










  • $begingroup$
    +1, I think $Z[sqrt d]$ is integrally closed if and only if $d$ congurent to $1$ module $4$.
    $endgroup$
    – UserS
    Dec 6 '18 at 17:39












  • $begingroup$
    @UserS: it is integrally closed if and only if $d$ is squarefree and $d$ is not congruent to $1 bmod 4$.
    $endgroup$
    – KCd
    Dec 6 '18 at 17:56










  • $begingroup$
    @UserS KCd is correct. I added a proof (using only gcds for variety vs. traditional proofs).
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 19:33
















2












$begingroup$

Hint $ $ It's simpler: UFDs are integrally closed, but it is not since $,(1+sqrt{-7})/2 $ is a root of $,x^2-x+2.,$ Said more simply it fails RRT = Rational Root Test, which holds true in any UFD.



Remark $ $ The RRT proof fails because some gcd it employs fails to exist in $,Bbb Z[sqrt{-7}].,$ Since gcds always exist in a UFD, we can use such gcd failure to prove rings are not UFDs. Let's use this to prove a more general case (for variety vs. traditional proofs).



Lemma $ Bbb Z[sqrt d]$ is not a UFD if $,color{#0a0}{dequiv 1pmod{!4}} $



Proof $ $ Assume it's a UFD. $ color{#0a0}{ww'}=(1!+!sqrt d)(1!-!sqrt d) = 1-d = color{#0a0}{4k},, color{#c00}{w} = 2-w',$



therefore the gcd product $ ,(2,color{#0a0}w)(2,color{#0a0}w') = (4,2w',2w,color{#0a0}{4k}) = 2(2,w',color{#c00}{w}) = 2(2,w')$



Cancelling $,(2,w'),Rightarrow, (2,w) = (2),Rightarrow, 2mid w,Rightarrow, w/2 = (1!+!sqrt d)/2in Bbb Z[sqrt d] Rightarrow!Leftarrow$



Further $ $ if $,w=sqrt{c^2d},$ then $,Bbb Z[w],$ UFD $,Rightarrow, cmid 1,,$ else $,(w/c)^2 = d,$ contra RRT (as above).

Alternatively by gcds $,(c,w)^2 = (c^2,cw,c^2d) = c(c,w),Rightarrow, (c,w) = (c),Rightarrow, cmid w,Rightarrow,cmid 1$



Combining yields a purely gcd-based proof of the following well-known fact.



Theorem $, Bbb Z[sqrt d], $ a UFD $,Rightarrow, d,$ squarefree and $,dnotequiv 1pmod{!4}$



Note $ $ The proof of the theorem also works if we instead assume the ring is a PID and read the tuples as ideals vs. gcds [since the proof uses only laws common to both: $ $ distributive, associative, commutative, and $(a,b,bc) = (a,b),$]. Then it yields $,(2,w),$ is noninvertible (so nonprincipal).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    And more generally this idea goes through in $mathbf Z[sqrt{d}]$ for nonsquare $d equiv 1 bmod 4$, not just d = -7$.
    $endgroup$
    – KCd
    Dec 6 '18 at 15:40










  • $begingroup$
    @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 17:25










  • $begingroup$
    +1, I think $Z[sqrt d]$ is integrally closed if and only if $d$ congurent to $1$ module $4$.
    $endgroup$
    – UserS
    Dec 6 '18 at 17:39












  • $begingroup$
    @UserS: it is integrally closed if and only if $d$ is squarefree and $d$ is not congruent to $1 bmod 4$.
    $endgroup$
    – KCd
    Dec 6 '18 at 17:56










  • $begingroup$
    @UserS KCd is correct. I added a proof (using only gcds for variety vs. traditional proofs).
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 19:33














2












2








2





$begingroup$

Hint $ $ It's simpler: UFDs are integrally closed, but it is not since $,(1+sqrt{-7})/2 $ is a root of $,x^2-x+2.,$ Said more simply it fails RRT = Rational Root Test, which holds true in any UFD.



Remark $ $ The RRT proof fails because some gcd it employs fails to exist in $,Bbb Z[sqrt{-7}].,$ Since gcds always exist in a UFD, we can use such gcd failure to prove rings are not UFDs. Let's use this to prove a more general case (for variety vs. traditional proofs).



Lemma $ Bbb Z[sqrt d]$ is not a UFD if $,color{#0a0}{dequiv 1pmod{!4}} $



Proof $ $ Assume it's a UFD. $ color{#0a0}{ww'}=(1!+!sqrt d)(1!-!sqrt d) = 1-d = color{#0a0}{4k},, color{#c00}{w} = 2-w',$



therefore the gcd product $ ,(2,color{#0a0}w)(2,color{#0a0}w') = (4,2w',2w,color{#0a0}{4k}) = 2(2,w',color{#c00}{w}) = 2(2,w')$



Cancelling $,(2,w'),Rightarrow, (2,w) = (2),Rightarrow, 2mid w,Rightarrow, w/2 = (1!+!sqrt d)/2in Bbb Z[sqrt d] Rightarrow!Leftarrow$



Further $ $ if $,w=sqrt{c^2d},$ then $,Bbb Z[w],$ UFD $,Rightarrow, cmid 1,,$ else $,(w/c)^2 = d,$ contra RRT (as above).

Alternatively by gcds $,(c,w)^2 = (c^2,cw,c^2d) = c(c,w),Rightarrow, (c,w) = (c),Rightarrow, cmid w,Rightarrow,cmid 1$



Combining yields a purely gcd-based proof of the following well-known fact.



Theorem $, Bbb Z[sqrt d], $ a UFD $,Rightarrow, d,$ squarefree and $,dnotequiv 1pmod{!4}$



Note $ $ The proof of the theorem also works if we instead assume the ring is a PID and read the tuples as ideals vs. gcds [since the proof uses only laws common to both: $ $ distributive, associative, commutative, and $(a,b,bc) = (a,b),$]. Then it yields $,(2,w),$ is noninvertible (so nonprincipal).






share|cite|improve this answer











$endgroup$



Hint $ $ It's simpler: UFDs are integrally closed, but it is not since $,(1+sqrt{-7})/2 $ is a root of $,x^2-x+2.,$ Said more simply it fails RRT = Rational Root Test, which holds true in any UFD.



Remark $ $ The RRT proof fails because some gcd it employs fails to exist in $,Bbb Z[sqrt{-7}].,$ Since gcds always exist in a UFD, we can use such gcd failure to prove rings are not UFDs. Let's use this to prove a more general case (for variety vs. traditional proofs).



Lemma $ Bbb Z[sqrt d]$ is not a UFD if $,color{#0a0}{dequiv 1pmod{!4}} $



Proof $ $ Assume it's a UFD. $ color{#0a0}{ww'}=(1!+!sqrt d)(1!-!sqrt d) = 1-d = color{#0a0}{4k},, color{#c00}{w} = 2-w',$



therefore the gcd product $ ,(2,color{#0a0}w)(2,color{#0a0}w') = (4,2w',2w,color{#0a0}{4k}) = 2(2,w',color{#c00}{w}) = 2(2,w')$



Cancelling $,(2,w'),Rightarrow, (2,w) = (2),Rightarrow, 2mid w,Rightarrow, w/2 = (1!+!sqrt d)/2in Bbb Z[sqrt d] Rightarrow!Leftarrow$



Further $ $ if $,w=sqrt{c^2d},$ then $,Bbb Z[w],$ UFD $,Rightarrow, cmid 1,,$ else $,(w/c)^2 = d,$ contra RRT (as above).

Alternatively by gcds $,(c,w)^2 = (c^2,cw,c^2d) = c(c,w),Rightarrow, (c,w) = (c),Rightarrow, cmid w,Rightarrow,cmid 1$



Combining yields a purely gcd-based proof of the following well-known fact.



Theorem $, Bbb Z[sqrt d], $ a UFD $,Rightarrow, d,$ squarefree and $,dnotequiv 1pmod{!4}$



Note $ $ The proof of the theorem also works if we instead assume the ring is a PID and read the tuples as ideals vs. gcds [since the proof uses only laws common to both: $ $ distributive, associative, commutative, and $(a,b,bc) = (a,b),$]. Then it yields $,(2,w),$ is noninvertible (so nonprincipal).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 22:31

























answered Dec 6 '18 at 14:47









Bill DubuqueBill Dubuque

210k29191639




210k29191639












  • $begingroup$
    And more generally this idea goes through in $mathbf Z[sqrt{d}]$ for nonsquare $d equiv 1 bmod 4$, not just d = -7$.
    $endgroup$
    – KCd
    Dec 6 '18 at 15:40










  • $begingroup$
    @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 17:25










  • $begingroup$
    +1, I think $Z[sqrt d]$ is integrally closed if and only if $d$ congurent to $1$ module $4$.
    $endgroup$
    – UserS
    Dec 6 '18 at 17:39












  • $begingroup$
    @UserS: it is integrally closed if and only if $d$ is squarefree and $d$ is not congruent to $1 bmod 4$.
    $endgroup$
    – KCd
    Dec 6 '18 at 17:56










  • $begingroup$
    @UserS KCd is correct. I added a proof (using only gcds for variety vs. traditional proofs).
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 19:33


















  • $begingroup$
    And more generally this idea goes through in $mathbf Z[sqrt{d}]$ for nonsquare $d equiv 1 bmod 4$, not just d = -7$.
    $endgroup$
    – KCd
    Dec 6 '18 at 15:40










  • $begingroup$
    @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 17:25










  • $begingroup$
    +1, I think $Z[sqrt d]$ is integrally closed if and only if $d$ congurent to $1$ module $4$.
    $endgroup$
    – UserS
    Dec 6 '18 at 17:39












  • $begingroup$
    @UserS: it is integrally closed if and only if $d$ is squarefree and $d$ is not congruent to $1 bmod 4$.
    $endgroup$
    – KCd
    Dec 6 '18 at 17:56










  • $begingroup$
    @UserS KCd is correct. I added a proof (using only gcds for variety vs. traditional proofs).
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 19:33
















$begingroup$
And more generally this idea goes through in $mathbf Z[sqrt{d}]$ for nonsquare $d equiv 1 bmod 4$, not just d = -7$.
$endgroup$
– KCd
Dec 6 '18 at 15:40




$begingroup$
And more generally this idea goes through in $mathbf Z[sqrt{d}]$ for nonsquare $d equiv 1 bmod 4$, not just d = -7$.
$endgroup$
– KCd
Dec 6 '18 at 15:40












$begingroup$
@Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
$endgroup$
– Bill Dubuque
Dec 6 '18 at 17:25




$begingroup$
@Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
$endgroup$
– Bill Dubuque
Dec 6 '18 at 17:25












$begingroup$
+1, I think $Z[sqrt d]$ is integrally closed if and only if $d$ congurent to $1$ module $4$.
$endgroup$
– UserS
Dec 6 '18 at 17:39






$begingroup$
+1, I think $Z[sqrt d]$ is integrally closed if and only if $d$ congurent to $1$ module $4$.
$endgroup$
– UserS
Dec 6 '18 at 17:39














$begingroup$
@UserS: it is integrally closed if and only if $d$ is squarefree and $d$ is not congruent to $1 bmod 4$.
$endgroup$
– KCd
Dec 6 '18 at 17:56




$begingroup$
@UserS: it is integrally closed if and only if $d$ is squarefree and $d$ is not congruent to $1 bmod 4$.
$endgroup$
– KCd
Dec 6 '18 at 17:56












$begingroup$
@UserS KCd is correct. I added a proof (using only gcds for variety vs. traditional proofs).
$endgroup$
– Bill Dubuque
Dec 6 '18 at 19:33




$begingroup$
@UserS KCd is correct. I added a proof (using only gcds for variety vs. traditional proofs).
$endgroup$
– Bill Dubuque
Dec 6 '18 at 19:33











1












$begingroup$

$$(1+sqrt{-7})×(1-sqrt{-7})=8=2×2×2.$$We show $2,1+sqrt{-7},1-sqrt{-7}$ are irreducibles.



So let $2=(a+sqrt{-7}b)(c+sqrt{-7}d)$ where $a,b,c,d in Bbb Z$, then taking norms $4=(a^2+7b^2)(c^2+7d^2)implies b=d=0$ , since for $e,fin Bbb Z$ and $fnot =0implies e^2+7f^2geq 7$. Also either $a$ or $c$ is $+1,-1$ as $2^2=4=(ac)^2$ and $Bbb Z$ is a UFD i.e. either $a+sqrt{-7}b$ or $c+sqrt{-7}d$ unit.



Next, let $1+sqrt{-7}=(a+sqrt{-7}b)(c+sqrt{-7}d)$ with $a,b,c,din Bbb Z$. Again taking norms $8=(a^2+7b^2)(c^2+7d^2)implies bd=0$ and both $b,d$ are not zero as $8$ has no integral square root. So without loss of generality assume $b=0$ and $dnot=0implies 8=a^2(c^2+7d^2)$ . Note that $d$ must be $+1,-1$ , otherwise $c^2+7d^2geq 28$. Also $cnot=0$ , as $7not |8$. Hence $c$ must be $+1,-1$ otherwise $c^2+7d^2geq 11$. Therefore $a=1,-1$ i.e. $a+sqrt{-7}b$ is unit.



Similarly $1-sqrt{-7}$ is irreducible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We only need $2$ is rred. Then $,2mid ww', 2nmid w,w',$ so $2$ is a non-prime irred so it's not a UFD.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 17:30










  • $begingroup$
    Yeah, that's right , but I tried to prove only using definition of UFD.
    $endgroup$
    – UserS
    Dec 6 '18 at 17:35
















1












$begingroup$

$$(1+sqrt{-7})×(1-sqrt{-7})=8=2×2×2.$$We show $2,1+sqrt{-7},1-sqrt{-7}$ are irreducibles.



So let $2=(a+sqrt{-7}b)(c+sqrt{-7}d)$ where $a,b,c,d in Bbb Z$, then taking norms $4=(a^2+7b^2)(c^2+7d^2)implies b=d=0$ , since for $e,fin Bbb Z$ and $fnot =0implies e^2+7f^2geq 7$. Also either $a$ or $c$ is $+1,-1$ as $2^2=4=(ac)^2$ and $Bbb Z$ is a UFD i.e. either $a+sqrt{-7}b$ or $c+sqrt{-7}d$ unit.



Next, let $1+sqrt{-7}=(a+sqrt{-7}b)(c+sqrt{-7}d)$ with $a,b,c,din Bbb Z$. Again taking norms $8=(a^2+7b^2)(c^2+7d^2)implies bd=0$ and both $b,d$ are not zero as $8$ has no integral square root. So without loss of generality assume $b=0$ and $dnot=0implies 8=a^2(c^2+7d^2)$ . Note that $d$ must be $+1,-1$ , otherwise $c^2+7d^2geq 28$. Also $cnot=0$ , as $7not |8$. Hence $c$ must be $+1,-1$ otherwise $c^2+7d^2geq 11$. Therefore $a=1,-1$ i.e. $a+sqrt{-7}b$ is unit.



Similarly $1-sqrt{-7}$ is irreducible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We only need $2$ is rred. Then $,2mid ww', 2nmid w,w',$ so $2$ is a non-prime irred so it's not a UFD.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 17:30










  • $begingroup$
    Yeah, that's right , but I tried to prove only using definition of UFD.
    $endgroup$
    – UserS
    Dec 6 '18 at 17:35














1












1








1





$begingroup$

$$(1+sqrt{-7})×(1-sqrt{-7})=8=2×2×2.$$We show $2,1+sqrt{-7},1-sqrt{-7}$ are irreducibles.



So let $2=(a+sqrt{-7}b)(c+sqrt{-7}d)$ where $a,b,c,d in Bbb Z$, then taking norms $4=(a^2+7b^2)(c^2+7d^2)implies b=d=0$ , since for $e,fin Bbb Z$ and $fnot =0implies e^2+7f^2geq 7$. Also either $a$ or $c$ is $+1,-1$ as $2^2=4=(ac)^2$ and $Bbb Z$ is a UFD i.e. either $a+sqrt{-7}b$ or $c+sqrt{-7}d$ unit.



Next, let $1+sqrt{-7}=(a+sqrt{-7}b)(c+sqrt{-7}d)$ with $a,b,c,din Bbb Z$. Again taking norms $8=(a^2+7b^2)(c^2+7d^2)implies bd=0$ and both $b,d$ are not zero as $8$ has no integral square root. So without loss of generality assume $b=0$ and $dnot=0implies 8=a^2(c^2+7d^2)$ . Note that $d$ must be $+1,-1$ , otherwise $c^2+7d^2geq 28$. Also $cnot=0$ , as $7not |8$. Hence $c$ must be $+1,-1$ otherwise $c^2+7d^2geq 11$. Therefore $a=1,-1$ i.e. $a+sqrt{-7}b$ is unit.



Similarly $1-sqrt{-7}$ is irreducible.






share|cite|improve this answer











$endgroup$



$$(1+sqrt{-7})×(1-sqrt{-7})=8=2×2×2.$$We show $2,1+sqrt{-7},1-sqrt{-7}$ are irreducibles.



So let $2=(a+sqrt{-7}b)(c+sqrt{-7}d)$ where $a,b,c,d in Bbb Z$, then taking norms $4=(a^2+7b^2)(c^2+7d^2)implies b=d=0$ , since for $e,fin Bbb Z$ and $fnot =0implies e^2+7f^2geq 7$. Also either $a$ or $c$ is $+1,-1$ as $2^2=4=(ac)^2$ and $Bbb Z$ is a UFD i.e. either $a+sqrt{-7}b$ or $c+sqrt{-7}d$ unit.



Next, let $1+sqrt{-7}=(a+sqrt{-7}b)(c+sqrt{-7}d)$ with $a,b,c,din Bbb Z$. Again taking norms $8=(a^2+7b^2)(c^2+7d^2)implies bd=0$ and both $b,d$ are not zero as $8$ has no integral square root. So without loss of generality assume $b=0$ and $dnot=0implies 8=a^2(c^2+7d^2)$ . Note that $d$ must be $+1,-1$ , otherwise $c^2+7d^2geq 28$. Also $cnot=0$ , as $7not |8$. Hence $c$ must be $+1,-1$ otherwise $c^2+7d^2geq 11$. Therefore $a=1,-1$ i.e. $a+sqrt{-7}b$ is unit.



Similarly $1-sqrt{-7}$ is irreducible.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 15:49

























answered Dec 6 '18 at 15:38









UserSUserS

1,5391112




1,5391112












  • $begingroup$
    We only need $2$ is rred. Then $,2mid ww', 2nmid w,w',$ so $2$ is a non-prime irred so it's not a UFD.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 17:30










  • $begingroup$
    Yeah, that's right , but I tried to prove only using definition of UFD.
    $endgroup$
    – UserS
    Dec 6 '18 at 17:35


















  • $begingroup$
    We only need $2$ is rred. Then $,2mid ww', 2nmid w,w',$ so $2$ is a non-prime irred so it's not a UFD.
    $endgroup$
    – Bill Dubuque
    Dec 6 '18 at 17:30










  • $begingroup$
    Yeah, that's right , but I tried to prove only using definition of UFD.
    $endgroup$
    – UserS
    Dec 6 '18 at 17:35
















$begingroup$
We only need $2$ is rred. Then $,2mid ww', 2nmid w,w',$ so $2$ is a non-prime irred so it's not a UFD.
$endgroup$
– Bill Dubuque
Dec 6 '18 at 17:30




$begingroup$
We only need $2$ is rred. Then $,2mid ww', 2nmid w,w',$ so $2$ is a non-prime irred so it's not a UFD.
$endgroup$
– Bill Dubuque
Dec 6 '18 at 17:30












$begingroup$
Yeah, that's right , but I tried to prove only using definition of UFD.
$endgroup$
– UserS
Dec 6 '18 at 17:35




$begingroup$
Yeah, that's right , but I tried to prove only using definition of UFD.
$endgroup$
– UserS
Dec 6 '18 at 17:35


















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