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Math Help - division ring means no non trivial left ideals

  1. #1
    Senior Member abhishekkgp's Avatar
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    division ring means no non trivial left ideals

    Prove that R is a division ring if and only if the only left ideals in R are 0 and R.(I am NOT assuming that R is commutative)

    I can prove that if R is a division ring then every left ideal is trivial.
    I cannot prove it in the other direction. Help needed on this.
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    MHF Contributor Drexel28's Avatar
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    Re: division ring means no non trivial left ideals

    Quote Originally Posted by abhishekkgp View Post
    Prove that R is a division ring if and only if the only left ideals in R are 0 and R.(I am NOT assuming that R is commutative)

    I can prove that if R is a division ring then every left ideal is trivial.
    I cannot prove it in the other direction. Help needed on this.
    The result should follow more easily if you know that if a monoid has left inverses for all its elements its actually a group. More information can be found on my blog, here.
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    Senior Member abhishekkgp's Avatar
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    Re: division ring means no non trivial left ideals

    Quote Originally Posted by Drexel28 View Post
    The result should follow more easily if you know that if a monoid has left inverses for all its elements its actually a group. More information can be found on my blog, here.
    I messed up. was using wrong definition of division ring.
    Anyways, Can you please check whether the following is true:


    If R is a ring with 1 \neq 0 with no zero divisors then each non-zero element of R has a left inverse and a right inverse.
    Proof: We claim that aR=R,a \in R, a \neq 0.
    If this is not the case then \exists r_1,r_2 \in R, r_1 \neq r_2, r_1 \neq 0, r_2 \neq 0 such that ar_1=ar_2 \Rightarrow a(r_1-r_2)=0 which contradicts that R has no zero divisors.
    So aR=R.
    Therefore there exists r \in R with ar=1.
    Similarly for left inverse.

    Does the above also not prove that if R is a ring with 1 \neq 0 with no zero divisors then every left ideal of R is trivial??
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    Re: division ring means no non trivial left ideals

    tl,dr; version: Ra is a left ideal for all a, so 1 = sa, for some s in R

    (there is a few gaps in this, of course)
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    Re: division ring means no non trivial left ideals

    Quote Originally Posted by abhishekkgp View Post
    I messed up. was using wrong definition of division ring.
    Anyways, Can you please check whether the following is true:


    If R is a ring with 1 \neq 0 with no zero divisors then each non-zero element of R has a left inverse and a right inverse.
    Proof: We claim that aR=R,a \in R, a \neq 0.
    If this is not the case then \exists r_1,r_2 \in R, r_1 \neq r_2, r_1 \neq 0, r_2 \neq 0 such that ar_1=ar_2 \Rightarrow a(r_1-r_2)=0 which contradicts that R has no zero divisors.
    why are you using aR? this is not a left ideal, generally. i can't see how you can say that r_1 and r_2 exist with ar_1 = ar_2, either.

    So aR=R.
    Therefore there exists r \in R with ar=1.
    Similarly for left inverse.

    Does the above also not prove that if R is a ring with 1 \neq 0 with no zero divisors then every left ideal of R is trivial??
    your argument cannot possibly be right. \mathbb{Z} is a ring with no zero divisors, and it definitely DOES have non-trivial ideals.
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    Senior Member abhishekkgp's Avatar
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    Re: division ring means no non trivial left ideals

    Quote Originally Posted by Deveno View Post
    why are you using aR? this is not a left ideal, generally. i can't see how you can say that r_1 and r_2 exist with ar_1 = ar_2, either.



    your argument cannot possibly be right. \mathbb{Z} is a ring with no zero divisors, and it definitely DOES have non-trivial ideals.
    By aR i meant \{ar: r \in R\}.
    I got my mistake. My argument works only if R is finite. Do you agree?
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    Re: division ring means no non trivial left ideals

    yes, because then we know that aR ≠ R means that the map r --> ar is not injective. with infinite rings, we can make no such claim.
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