Rings Without Nontrivial One-Sided Ideals

First, let us review the notion of an ideal generated by an element.

Let $R$ be a ring, and let $a \in R$.

1. If $R$ is an arbitrary ring, then the ideal generated by $a$ is
   $$
   (a)=\left\{ma+xa+ay+\sum_{i=1}^n x_iay_i \mid m\in\mathbb{Z},\ x,y,x_i,y_i\in R,\ n\in\mathbb{N}^*\right\}.
   $$
   The left ideal generated by $a$ is
   $$
   \{xa+na \mid x\in R,\ n\in\mathbb{Z}\}.
   $$

2. If $R$ is a ring with identity, then the ideal generated by $a$ is
   $$
   (a)=\left\{\sum_{i=1}^n x_iay_i \mid x_i,y_i\in R,\ n\in\mathbb{N}^*\right\}.
   $$
   The left ideal generated by $a$ is
   $$
   \{xa \mid x\in R\}=Ra.
   $$

3. If $R$ is a commutative ring, then the ideal generated by $a$ is
   $$
   (a)=\{ma+xa \mid m\in\mathbb{Z},\ x\in R\}.
   $$

4. If $R$ is a commutative ring with identity, then the ideal generated by $a$ is
   $$
   (a)=\{xa \mid x\in R\}=aR.
   $$

Theorem. A ring $R$ has no nontrivial left ideals if and only if either $R$ is a zero ring of prime order, or $R$ is a division ring.

Proof: We first prove the necessity.

Assume that $R$ has no nontrivial left ideals. Let $a \in R$ with $a \neq 0$. Then it is easy to verify that
$$
Ra=\{xa \mid x\in R\}
$$
is a left ideal of $R$. There are two possible cases:

1. There exists $a \in R \setminus \{0\}$ such that
   $$
   Ra=\{0\}.
   $$

2. For every $a \in R \setminus \{0\}$, we have
   $$
   Ra=R.
   $$

We first consider Case 1. Suppose that there exists $a \in R \setminus \{0\}$ such that
$$
Ra=\{0\}.
$$
Consider the left ideal generated by $a$:
$$
I=\{na+xa \mid n\in\mathbb{Z},\ x\in R\}.
$$
Since $a \in I$, we have
$$
I\neq \{0\}.
$$
By assumption, $R$ has no nontrivial left ideals, so necessarily
$$
I=R.
$$

Moreover, since $Ra=\{0\}$, we have
$$
xa=0,\qquad \forall x\in R.
$$
Hence
$$
I=\{na \mid n\in\mathbb{Z}\}.
$$
Therefore, for all $x,y\in R$, there exist $m,n\in\mathbb{Z}$ such that
$$
x=ma,\qquad y=na.
$$
Thus
$$
x\cdot y=(ma)(na)=mna^2=0.
$$
Indeed, this is because $mna\in R$ and $Ra=\{0\}$.

So $R$ is a zero ring. Furthermore, viewed as an additive group, $R$ has no nontrivial subgroups. Hence $R$ must be a zero ring of prime order.

Now consider Case 2. Assume that for every $a\in R\setminus\{0\}$, we have
$$
Ra=R.
$$

We first prove that $R$ has no zero divisors.

Let $a,b\in R$ with $a\neq 0$ and $b\neq 0$. Since
$$
Ra=R,
$$
we have
$$
Rab=(Ra)b=Rb.
$$
Because $b\neq 0$, we also have
$$
Rb=R.
$$
Hence
$$
Rab=R.
$$
Therefore
$$
ab\neq 0.
$$
So $R$ has no zero divisors.

Next we prove that $R$ has an identity element.

Since $Ra=R$, there exists $e\in R\setminus\{0\}$ such that
$$
ea=a.
$$
Then
$$
(ae-a)a=a^2-a^2=0.
$$
Since $R$ has no zero divisors and $a\neq 0$, it follows that
$$
ae=a.
$$

Now let $b\in R$ be arbitrary. Then
$$
a(eb-b)=ab-ab=0.
$$
Again, since $R$ has no zero divisors and $a\neq 0$, we obtain
$$
eb=b.
$$
Furthermore,
$$
(be-b)b=b^2-b^2=0.
$$
If $b\neq 0$, then the absence of zero divisors implies
$$
be=b.
$$
If $b=0$, this is obvious as well. Therefore
$$
be=b,\qquad \forall b\in R.
$$
Hence $e$ is the identity element of $R$.

Finally, we prove that every nonzero element of $R$ is invertible.

Let $a\neq 0$. Since $Ra=R$, there exists $b\in R$ such that
$$
ba=e.
$$
Then
$$
(ab-e)a=aba-ea=a-a=0.
$$
Since $R$ has no zero divisors and $a\neq 0$, it follows that
$$
ab=e.
$$
Thus $b$ is the inverse of $a$ in $R$.

Therefore $R$ is a division ring.

The sufficiency is immediate.

Proposition 1. A ring with identity $R$ has no nontrivial one-sided ideals if and only if $R$ is a division ring.

Proof: Since
$$
1\cdot 1=1\neq 0,
$$
the ring $R$ cannot be a zero ring. Therefore, by the theorem above, the conclusion follows.

Proposition 2. A commutative ring with identity is a field if and only if it has no nontrivial ideals.

Proof: This follows immediately from Proposition 1.

The cover image of this article was taken in Busan, South Korea.