division ring means no non trivial left ideals

**abhishekkgp** · October 17th 2011, 08:41 PM

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.

**Drexel28** · October 17th 2011, 08:51 PM

Originally Posted by abhishekkgp

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.

**abhishekkgp** · October 17th 2011, 09:26 PM

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 $R$ 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??

**Deveno** · October 17th 2011, 09:41 PM

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)

**Deveno** · October 17th 2011, 09:55 PM

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 $R$ 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.

**abhishekkgp** · October 17th 2011, 10:01 PM

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 $r_1$ 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?

**Deveno** · October 17th 2011, 10:25 PM

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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