field are domains. [duplicate]

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  • Unital subrings of fields are all domains

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I want to prove that field are domain. Let $K$ a field. Suppose it's not a domain. Then there are $a,bin Ksetminus0$ s.t. $ab=0$. Let $a'$ and $b'$ s.t. $a'a=aa'=1$ and $b'b=bb'=1$. Then $$1=(a'a)(bb')=a'(ab)b'=0,$$
which is impossible. Does it work ?




abstract-algebra proof-verification field-theory




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  • Unital subrings of fields are all domains

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I want to prove that field are domain. Let $K$ a field. Suppose it's not a domain. Then there are $a,bin Ksetminus0$ s.t. $ab=0$. Let $a'$ and $b'$ s.t. $a'a=aa'=1$ and $b'b=bb'=1$. Then $$1=(a'a)(bb')=a'(ab)b'=0,$$
which is impossible. Does it work ?




abstract-algebra proof-verification field-theory




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  • This works well.
    – Suzet
    Aug 1 at 12:14












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up vote
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This question already has an answer here:



  • Unital subrings of fields are all domains

    1 answer



I want to prove that field are domain. Let $K$ a field. Suppose it's not a domain. Then there are $a,bin Ksetminus0$ s.t. $ab=0$. Let $a'$ and $b'$ s.t. $a'a=aa'=1$ and $b'b=bb'=1$. Then $$1=(a'a)(bb')=a'(ab)b'=0,$$
which is impossible. Does it work ?




abstract-algebra proof-verification field-theory




share|cite|improve this question














This question already has an answer here:



  • Unital subrings of fields are all domains

    1 answer



I want to prove that field are domain. Let $K$ a field. Suppose it's not a domain. Then there are $a,bin Ksetminus0$ s.t. $ab=0$. Let $a'$ and $b'$ s.t. $a'a=aa'=1$ and $b'b=bb'=1$. Then $$1=(a'a)(bb')=a'(ab)b'=0,$$
which is impossible. Does it work ?





This question already has an answer here:



  • Unital subrings of fields are all domains

    1 answer





abstract-algebra proof-verification field-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 1 at 12:16









José Carlos Santos

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112k1696172









asked Aug 1 at 12:12









MathBeginner

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695312




marked as duplicate by rschwieb abstract-algebra
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • This works well.
    – Suzet
    Aug 1 at 12:14
















  • This works well.
    – Suzet
    Aug 1 at 12:14















This works well.
– Suzet
Aug 1 at 12:14




This works well.
– Suzet
Aug 1 at 12:14










1 Answer
1






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1
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Yes, it's fine.



A shorter proof would be: if $a'$ is such that $a'a=1$, then$$0=a'(ab)=(a'a)b=b.$$






share|cite|improve this answer






























    1 Answer
    1






    active

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes








    up vote
    1
    down vote













    Yes, it's fine.



    A shorter proof would be: if $a'$ is such that $a'a=1$, then$$0=a'(ab)=(a'a)b=b.$$






    share|cite|improve this answer



























      up vote
      1
      down vote













      Yes, it's fine.



      A shorter proof would be: if $a'$ is such that $a'a=1$, then$$0=a'(ab)=(a'a)b=b.$$






      share|cite|improve this answer

























        up vote
        1
        down vote










        up vote
        1
        down vote









        Yes, it's fine.



        A shorter proof would be: if $a'$ is such that $a'a=1$, then$$0=a'(ab)=(a'a)b=b.$$






        share|cite|improve this answer















        Yes, it's fine.



        A shorter proof would be: if $a'$ is such that $a'a=1$, then$$0=a'(ab)=(a'a)b=b.$$







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 1 at 12:21









        Arthur

        98.3k793174




        98.3k793174











        answered Aug 1 at 12:15









        José Carlos Santos

        112k1696172




        112k1696172












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