GRE 0568 #66: On whether or not exactly 2 right ideals give a non-commutative field (field except multiplication is not commutative)

Question

Algebra by Michael Ch11.3

Artin has different definitions of rings particularly that his rings are unital and commutative in both addition and multiplication. Therefore, ideals are neither right nor left, afaik. (*)

In the book, Artin proves that a commutative unital ring with exactly 2 ideals is a field. (*) For non-commutative unital rings, I would like to know what we get under certain circumstances eg a non-commutative field or a division ring. Edit: Apparently, non-commutative field isn't necessarily defined as a division ring or anything at all. What I mean by non-commutative field is that we have all the properties of fields except that multiplication is not commutative.

If a commutative unital ring with exactly 2 ideals is a field, then...

1. Is a non-commutative unital ring with exactly 2 right ideals a non-commutative field?

2. Is a non-commutative unital ring with exactly 2 left ideals a non-commutative field?

3. (Never mind this) Is a non-commutative unital ring with exactly 1 left ideal and exactly 1 right ideal a non-commutative field? (Never mind this)

4. Is a non-commutative unital ring with exactly 2 ideals a non-commutative field?

5. (Never mind this) Does a non-commutative unital ring with exactly 1 left ideal and exactly 1 right ideal have exactly 2 ideals? (Never mind this)

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This is based on a GRE question:

I mistakenly answered $D$ instead of $B$ because I forgot that Artin defines rings to be commutative.

It appears the answer to my first question is affirmative because of Ian Coley's solution (which I think forgets to say $r \ne 0$)

I don't see how the answer to my second question would be different because $0$ and $R$ I guess are still left ideals.

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(*)

Definition of ideals of a ring:

Prop 11.3.19(b) says a ring with exactly 2 ideals is a field:

Answer 1

Let $R$ be a (perhaps non-commutative) nonzero ring with a unit. Then the following are equivalent:

1. $R$ is a division ring, i.e. every nonzero element has an inverse.
2. $R$ has precisely two left ideals.
3. $R$ has precisely two right ideals.

Proof:

• ($1 \implies 2$) If $I \neq \{0\}$ is a left ideal, then let $a \in I$ be a nonzero element. By the division ring assumption, $a^{-1} \in R$ exists, and $1 = a^{-1} a \in RI \subseteq I$ by the fact that $I$ is a left ideal. So $1 \in I$, and $I = R$.
• ($2 \implies 1$) Since $\{0\} \neq R$ are always ideals, for any nonzero $a \in R$, the left ideal $aR = R$. Hence there exists some $b \in R$ such that $ab = 1$.
• The proof of $(1 \iff 3)$ is identical to $(1 \iff 2)$.

We also have the following implication: if $R$ is a division ring, then $R$ has precisely two two-sided ideals. This follows readily from the above. But the converse implication is not true: if $R$ has precisely two two-sided ideals, it may not be a division ring. For example, the matrix ring over a field has only two two-sided ideals.

Finally, if $R$ is commutative, a left ideal is a right ideal is a two-sided ideal, and so everything above is equivalent.