Prove U=Z where addition is regular addition in Z, but multiplication is defined by a*b=a IS NOT A RING












0












$begingroup$


Ok,so clearly




  1. a+b ∈ R


  2. a+(b+c)=(a+b)+c


  3. a+b = b+a


  4. a+0 = a = 0+a, 0 ∈ R a ∈ R


  5. a+x = 0, x = -a ∈R



now




  1. a ∈ R b ∈ R and ab = a ∈R



  2. a(bc) = a(b) = a



    (ab)c = (a)c = a




  3. a(b+c) = ab+ac = a+a



    (a+b)c = ac+bc = a+b




Why is this not a ring? It seems to me that it satisfies all the axioms






abstract-algebra







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    So according to how it is defined $a(b+c)=aneq ab+ac=a+a$.
    $endgroup$
    – Yadati Kiran
    Dec 9 '18 at 6:13












  • $begingroup$
    ohhhhhhhhhhhhh ok thank you
    $endgroup$
    – s_healy
    Dec 9 '18 at 6:44
















0












$begingroup$


Ok,so clearly




  1. a+b ∈ R


  2. a+(b+c)=(a+b)+c


  3. a+b = b+a


  4. a+0 = a = 0+a, 0 ∈ R a ∈ R


  5. a+x = 0, x = -a ∈R



now




  1. a ∈ R b ∈ R and ab = a ∈R



  2. a(bc) = a(b) = a



    (ab)c = (a)c = a




  3. a(b+c) = ab+ac = a+a



    (a+b)c = ac+bc = a+b




Why is this not a ring? It seems to me that it satisfies all the axioms






abstract-algebra







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    So according to how it is defined $a(b+c)=aneq ab+ac=a+a$.
    $endgroup$
    – Yadati Kiran
    Dec 9 '18 at 6:13












  • $begingroup$
    ohhhhhhhhhhhhh ok thank you
    $endgroup$
    – s_healy
    Dec 9 '18 at 6:44














0












0








0





$begingroup$


Ok,so clearly




  1. a+b ∈ R


  2. a+(b+c)=(a+b)+c


  3. a+b = b+a


  4. a+0 = a = 0+a, 0 ∈ R a ∈ R


  5. a+x = 0, x = -a ∈R



now




  1. a ∈ R b ∈ R and ab = a ∈R



  2. a(bc) = a(b) = a



    (ab)c = (a)c = a




  3. a(b+c) = ab+ac = a+a



    (a+b)c = ac+bc = a+b




Why is this not a ring? It seems to me that it satisfies all the axioms






abstract-algebra







share|cite|improve this question









$endgroup$




Ok,so clearly




  1. a+b ∈ R


  2. a+(b+c)=(a+b)+c


  3. a+b = b+a


  4. a+0 = a = 0+a, 0 ∈ R a ∈ R


  5. a+x = 0, x = -a ∈R



now




  1. a ∈ R b ∈ R and ab = a ∈R



  2. a(bc) = a(b) = a



    (ab)c = (a)c = a




  3. a(b+c) = ab+ac = a+a



    (a+b)c = ac+bc = a+b




Why is this not a ring? It seems to me that it satisfies all the axioms






abstract-algebra




abstract-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 6:08









s_healys_healy

205




205








  • 4




    $begingroup$
    So according to how it is defined $a(b+c)=aneq ab+ac=a+a$.
    $endgroup$
    – Yadati Kiran
    Dec 9 '18 at 6:13












  • $begingroup$
    ohhhhhhhhhhhhh ok thank you
    $endgroup$
    – s_healy
    Dec 9 '18 at 6:44














  • 4




    $begingroup$
    So according to how it is defined $a(b+c)=aneq ab+ac=a+a$.
    $endgroup$
    – Yadati Kiran
    Dec 9 '18 at 6:13












  • $begingroup$
    ohhhhhhhhhhhhh ok thank you
    $endgroup$
    – s_healy
    Dec 9 '18 at 6:44








4




4




$begingroup$
So according to how it is defined $a(b+c)=aneq ab+ac=a+a$.
$endgroup$
– Yadati Kiran
Dec 9 '18 at 6:13






$begingroup$
So according to how it is defined $a(b+c)=aneq ab+ac=a+a$.
$endgroup$
– Yadati Kiran
Dec 9 '18 at 6:13














$begingroup$
ohhhhhhhhhhhhh ok thank you
$endgroup$
– s_healy
Dec 9 '18 at 6:44




$begingroup$
ohhhhhhhhhhhhh ok thank you
$endgroup$
– s_healy
Dec 9 '18 at 6:44










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