division ring means no non trivial left ideals

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

I can prove that if $\displaystyle R$ is a division ring then every left ideal is trivial.

I cannot prove it in the other direction. Help needed on this.

Re: division ring means no non trivial left ideals

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Originally Posted by

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

I can prove that if $\displaystyle R$ 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.

Re: division ring means no non trivial left ideals

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Originally Posted by

**Drexel28** 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 $\displaystyle R$ is a ring with $\displaystyle 1 \neq 0$ with no zero divisors then each non-zero element of $\displaystyle R$ has a left inverse and a right inverse.

Proof: We claim that $\displaystyle aR=R,a \in R, a \neq 0$.

If this is not the case then $\displaystyle \exists r_1,r_2 \in R, r_1 \neq r_2, r_1 \neq 0, r_2 \neq 0$ such that $\displaystyle ar_1=ar_2 \Rightarrow a(r_1-r_2)=0$ which contradicts that $\displaystyle R$ has no zero divisors.

So $\displaystyle aR=R$.

Therefore there exists $\displaystyle r \in R$ with $\displaystyle ar=1$.

Similarly for left inverse.

Does the above also not prove that if $\displaystyle R$ is a ring with $\displaystyle 1 \neq 0$ with no zero divisors then every left ideal of $\displaystyle R$ is trivial??

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)

Re: division ring means no non trivial left ideals

Quote:

Originally Posted by

**abhishekkgp** I messed up. was using wrong definition of division ring.

Anyways, Can you please check whether the following is true:

If $\displaystyle R$ is a ring with $\displaystyle 1 \neq 0$ with no zero divisors then each non-zero element of $\displaystyle R$ has a left inverse and a right inverse.

Proof: We claim that $\displaystyle aR=R,a \in R, a \neq 0$.

If this is not the case then $\displaystyle \exists r_1,r_2 \in R, r_1 \neq r_2, r_1 \neq 0, r_2 \neq 0$ such that $\displaystyle ar_1=ar_2 \Rightarrow a(r_1-r_2)=0$ which contradicts that $\displaystyle R$ 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 $\displaystyle r_1$ and $\displaystyle r_2$ exist with $\displaystyle ar_1 = ar_2$, either.

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So $\displaystyle aR=R$.

Therefore there exists $\displaystyle r \in R$ with $\displaystyle ar=1$.

Similarly for left inverse.

Does the above also not prove that if $\displaystyle R$ is a ring with $\displaystyle 1 \neq 0$ with no zero divisors then every left ideal of $\displaystyle R$ is trivial??

your argument cannot possibly be right. $\displaystyle \mathbb{Z}$ is a ring with no zero divisors, and it definitely DOES have non-trivial ideals.

Re: division ring means no non trivial left ideals

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Originally Posted by

**Deveno** why are you using aR? this is not a left ideal, generally. i can't see how you can say that $\displaystyle r_1$ and $\displaystyle r_2$ exist with $\displaystyle ar_1 = ar_2$, either.

your argument cannot possibly be right. $\displaystyle \mathbb{Z}$ is a ring with no zero divisors, and it definitely DOES have non-trivial ideals.

By $\displaystyle aR$ i meant $\displaystyle \{ar: r \in R\}$.

I got my mistake. My argument works only if $\displaystyle R$ is finite. Do you agree?

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.