# division ring means no non trivial left ideals

• Oct 17th 2011, 08:41 PM
abhishekkgp
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.
• Oct 17th 2011, 08:51 PM
Drexel28
Re: division ring means no non trivial left ideals
Quote:

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.
• Oct 17th 2011, 09:26 PM
abhishekkgp
Re: division ring means no non trivial left ideals
Quote:

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??
• Oct 17th 2011, 09:41 PM
Deveno
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)
• Oct 17th 2011, 09:55 PM
Deveno
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.

Quote:

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.
• Oct 17th 2011, 10:01 PM
abhishekkgp
Re: division ring means no non trivial left ideals
Quote:

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?
• Oct 17th 2011, 10:25 PM
Deveno
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.