Can anyone make me understand in simple language why the second condition for being an Euclidean domain is...












1














Can anyone make me understand in simple language why the second condition for being an Euclidean domain is superfluous ?



Why $v(a) leq v(ab)$ is not needed? How we can deduce from the first one?






abstract-algebra ring-theory integral-domain euclidean-domain







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  • Euclidean domain.
    – MisterRiemann
    Dec 7 at 14:48






  • 1




    It would be good if you could elaborate exactly which part of the construction you find to be "not comprehensible".
    – MisterRiemann
    Dec 7 at 15:12










  • How the new function satisfies the division algorithm?@MisterRiemann
    – cmi
    Dec 7 at 15:13






  • 1




    I'm not really an algebraist, so I'm learning this as we speak. Here is another reference, which seems to a bit more in-depth about this. (In particular, see Theorem 2.1).
    – MisterRiemann
    Dec 7 at 15:20


















1














Can anyone make me understand in simple language why the second condition for being an Euclidean domain is superfluous ?



Why $v(a) leq v(ab)$ is not needed? How we can deduce from the first one?






abstract-algebra ring-theory integral-domain euclidean-domain







share|cite|improve this question






















  • Euclidean domain.
    – MisterRiemann
    Dec 7 at 14:48






  • 1




    It would be good if you could elaborate exactly which part of the construction you find to be "not comprehensible".
    – MisterRiemann
    Dec 7 at 15:12










  • How the new function satisfies the division algorithm?@MisterRiemann
    – cmi
    Dec 7 at 15:13






  • 1




    I'm not really an algebraist, so I'm learning this as we speak. Here is another reference, which seems to a bit more in-depth about this. (In particular, see Theorem 2.1).
    – MisterRiemann
    Dec 7 at 15:20
















1












1








1







Can anyone make me understand in simple language why the second condition for being an Euclidean domain is superfluous ?



Why $v(a) leq v(ab)$ is not needed? How we can deduce from the first one?






abstract-algebra ring-theory integral-domain euclidean-domain







share|cite|improve this question













Can anyone make me understand in simple language why the second condition for being an Euclidean domain is superfluous ?



Why $v(a) leq v(ab)$ is not needed? How we can deduce from the first one?






abstract-algebra ring-theory integral-domain euclidean-domain




abstract-algebra ring-theory integral-domain euclidean-domain






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 at 14:44









cmi

1,000212




1,000212












  • Euclidean domain.
    – MisterRiemann
    Dec 7 at 14:48






  • 1




    It would be good if you could elaborate exactly which part of the construction you find to be "not comprehensible".
    – MisterRiemann
    Dec 7 at 15:12










  • How the new function satisfies the division algorithm?@MisterRiemann
    – cmi
    Dec 7 at 15:13






  • 1




    I'm not really an algebraist, so I'm learning this as we speak. Here is another reference, which seems to a bit more in-depth about this. (In particular, see Theorem 2.1).
    – MisterRiemann
    Dec 7 at 15:20




















  • Euclidean domain.
    – MisterRiemann
    Dec 7 at 14:48






  • 1




    It would be good if you could elaborate exactly which part of the construction you find to be "not comprehensible".
    – MisterRiemann
    Dec 7 at 15:12










  • How the new function satisfies the division algorithm?@MisterRiemann
    – cmi
    Dec 7 at 15:13






  • 1




    I'm not really an algebraist, so I'm learning this as we speak. Here is another reference, which seems to a bit more in-depth about this. (In particular, see Theorem 2.1).
    – MisterRiemann
    Dec 7 at 15:20


















Euclidean domain.
– MisterRiemann
Dec 7 at 14:48




Euclidean domain.
– MisterRiemann
Dec 7 at 14:48




1




1




It would be good if you could elaborate exactly which part of the construction you find to be "not comprehensible".
– MisterRiemann
Dec 7 at 15:12




It would be good if you could elaborate exactly which part of the construction you find to be "not comprehensible".
– MisterRiemann
Dec 7 at 15:12












How the new function satisfies the division algorithm?@MisterRiemann
– cmi
Dec 7 at 15:13




How the new function satisfies the division algorithm?@MisterRiemann
– cmi
Dec 7 at 15:13




1




1




I'm not really an algebraist, so I'm learning this as we speak. Here is another reference, which seems to a bit more in-depth about this. (In particular, see Theorem 2.1).
– MisterRiemann
Dec 7 at 15:20






I'm not really an algebraist, so I'm learning this as we speak. Here is another reference, which seems to a bit more in-depth about this. (In particular, see Theorem 2.1).
– MisterRiemann
Dec 7 at 15:20












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Suppose $V$ only satisfies the first Euclidean property, i.e. for all $,a,bin D,$ if $,bneq 0,$ then there are $,q,rin D,$ such that $, a = qb +r,$ with $, V(r) < V(b),,$ where $V$ maps $D$ into (well-ordered) $,Bbb N$.



Derive $,v,$ from $,V,$ as follows



$$begin{align}
v(0) &= V(0)\
v(a) &= {rm min}{ V(b) : bin aDbackslash 0}
end{align}$$



Note $,v(a)le V(a),$ since it is clear if $,a = 0,,$ else it it follows by $, ain aDbackslash 0$



$v$ is also a Euclidean function: if $,a,bin D,$ and $,bneq 0,$ then $,v(b) = V(bc),$ for $,0neq cin D.,$ Since $,V,$ is a Euclidean function there exist $,q,rin D,$ such that $, a = qbc + r,$ and $,V(r) < V(bc) = v(b).,$ But by above we know $,v(r)le V(r),$ thus $,v(r) < v(b),,$ so $,v,$ is a Euclidean function.



Note $, v(a) le v(ab),$ if $,abneq 0,$ since $,aDbackslash 0supseteq abDbackslash 0$ $,Rightarrow,{rm min},V(aDbackslash 0) le {rm min}, V(abDbackslash 0)$






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    Suppose $V$ only satisfies the first Euclidean property, i.e. for all $,a,bin D,$ if $,bneq 0,$ then there are $,q,rin D,$ such that $, a = qb +r,$ with $, V(r) < V(b),,$ where $V$ maps $D$ into (well-ordered) $,Bbb N$.



    Derive $,v,$ from $,V,$ as follows



    $$begin{align}
    v(0) &= V(0)\
    v(a) &= {rm min}{ V(b) : bin aDbackslash 0}
    end{align}$$



    Note $,v(a)le V(a),$ since it is clear if $,a = 0,,$ else it it follows by $, ain aDbackslash 0$



    $v$ is also a Euclidean function: if $,a,bin D,$ and $,bneq 0,$ then $,v(b) = V(bc),$ for $,0neq cin D.,$ Since $,V,$ is a Euclidean function there exist $,q,rin D,$ such that $, a = qbc + r,$ and $,V(r) < V(bc) = v(b).,$ But by above we know $,v(r)le V(r),$ thus $,v(r) < v(b),,$ so $,v,$ is a Euclidean function.



    Note $, v(a) le v(ab),$ if $,abneq 0,$ since $,aDbackslash 0supseteq abDbackslash 0$ $,Rightarrow,{rm min},V(aDbackslash 0) le {rm min}, V(abDbackslash 0)$






    share|cite|improve this answer




























      1














      Suppose $V$ only satisfies the first Euclidean property, i.e. for all $,a,bin D,$ if $,bneq 0,$ then there are $,q,rin D,$ such that $, a = qb +r,$ with $, V(r) < V(b),,$ where $V$ maps $D$ into (well-ordered) $,Bbb N$.



      Derive $,v,$ from $,V,$ as follows



      $$begin{align}
      v(0) &= V(0)\
      v(a) &= {rm min}{ V(b) : bin aDbackslash 0}
      end{align}$$



      Note $,v(a)le V(a),$ since it is clear if $,a = 0,,$ else it it follows by $, ain aDbackslash 0$



      $v$ is also a Euclidean function: if $,a,bin D,$ and $,bneq 0,$ then $,v(b) = V(bc),$ for $,0neq cin D.,$ Since $,V,$ is a Euclidean function there exist $,q,rin D,$ such that $, a = qbc + r,$ and $,V(r) < V(bc) = v(b).,$ But by above we know $,v(r)le V(r),$ thus $,v(r) < v(b),,$ so $,v,$ is a Euclidean function.



      Note $, v(a) le v(ab),$ if $,abneq 0,$ since $,aDbackslash 0supseteq abDbackslash 0$ $,Rightarrow,{rm min},V(aDbackslash 0) le {rm min}, V(abDbackslash 0)$






      share|cite|improve this answer


























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        Suppose $V$ only satisfies the first Euclidean property, i.e. for all $,a,bin D,$ if $,bneq 0,$ then there are $,q,rin D,$ such that $, a = qb +r,$ with $, V(r) < V(b),,$ where $V$ maps $D$ into (well-ordered) $,Bbb N$.



        Derive $,v,$ from $,V,$ as follows



        $$begin{align}
        v(0) &= V(0)\
        v(a) &= {rm min}{ V(b) : bin aDbackslash 0}
        end{align}$$



        Note $,v(a)le V(a),$ since it is clear if $,a = 0,,$ else it it follows by $, ain aDbackslash 0$



        $v$ is also a Euclidean function: if $,a,bin D,$ and $,bneq 0,$ then $,v(b) = V(bc),$ for $,0neq cin D.,$ Since $,V,$ is a Euclidean function there exist $,q,rin D,$ such that $, a = qbc + r,$ and $,V(r) < V(bc) = v(b).,$ But by above we know $,v(r)le V(r),$ thus $,v(r) < v(b),,$ so $,v,$ is a Euclidean function.



        Note $, v(a) le v(ab),$ if $,abneq 0,$ since $,aDbackslash 0supseteq abDbackslash 0$ $,Rightarrow,{rm min},V(aDbackslash 0) le {rm min}, V(abDbackslash 0)$






        share|cite|improve this answer














        Suppose $V$ only satisfies the first Euclidean property, i.e. for all $,a,bin D,$ if $,bneq 0,$ then there are $,q,rin D,$ such that $, a = qb +r,$ with $, V(r) < V(b),,$ where $V$ maps $D$ into (well-ordered) $,Bbb N$.



        Derive $,v,$ from $,V,$ as follows



        $$begin{align}
        v(0) &= V(0)\
        v(a) &= {rm min}{ V(b) : bin aDbackslash 0}
        end{align}$$



        Note $,v(a)le V(a),$ since it is clear if $,a = 0,,$ else it it follows by $, ain aDbackslash 0$



        $v$ is also a Euclidean function: if $,a,bin D,$ and $,bneq 0,$ then $,v(b) = V(bc),$ for $,0neq cin D.,$ Since $,V,$ is a Euclidean function there exist $,q,rin D,$ such that $, a = qbc + r,$ and $,V(r) < V(bc) = v(b).,$ But by above we know $,v(r)le V(r),$ thus $,v(r) < v(b),,$ so $,v,$ is a Euclidean function.



        Note $, v(a) le v(ab),$ if $,abneq 0,$ since $,aDbackslash 0supseteq abDbackslash 0$ $,Rightarrow,{rm min},V(aDbackslash 0) le {rm min}, V(abDbackslash 0)$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 at 17:14

























        answered Dec 7 at 16:46









        Bill Dubuque

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        208k29190626






























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