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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)


This is based on a GRE question:

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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 $r \ne 0$)

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


(*)

Definition of ideals of a ring:

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

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    $\begingroup$ Clearly the question’s definition of a ring does not require it to be commutative, otherwise option (I) would be silly to include. And I would think that the solution you posted answers most of your questions. $\endgroup$ – Joppy Oct 23 '18 at 3:48
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    $\begingroup$ Ah this is all getting quite confusing. Suffice to say, the answer to this question does not change whether or not the ring is commutative! If the ring is commutative, then right ideals are the same as two-sided ideals, but this does not change the answer! (But also, since (I) is an option, it is clear the question does not assume the ring is commutative). Have you come across a problem which does have a different answer if you assume things like unital or commutative for rings? $\endgroup$ – Joppy Oct 23 '18 at 4:53
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    $\begingroup$ "Non-commutative field" is not a standard term. Presumably you mean "division ring" (aka "skew field"). For the answer to I, see the accepted answer at math.stackexchange.com/questions/208066/… $\endgroup$ – symplectomorphic Oct 23 '18 at 4:59
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    $\begingroup$ Possible duplicate of fields are characterized by the property of having exactly 2 ideals $\endgroup$ – symplectomorphic Oct 23 '18 at 5:00
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    $\begingroup$ @symplectomorphic What Bourbaki calls fields we would call division rings, so perhaps that terminology is still sticking around somewhere... $\endgroup$ – Joppy Oct 23 '18 at 5:14
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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.

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  • $\begingroup$ Thanks a million, Joppy! ^-^ To clarify, a unital division ring is a field except multiplication is not commutative? That's what I think it says on wikipedia. What specifically are the answers to questions 1-5 please? 1 and 2 I guess are yes by your equivalence proof. 4 I guess is no by 'converse implication is not true', and 3 and 5 are actually wrong questions because for a non-commutative unital ring with exactly 1 right ideal and exactly 1 left ideal, such ideals are actually equal? $\endgroup$ – BCLC Oct 23 '18 at 5:46
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    $\begingroup$ I'm not going to answer explicitly what 1-5 are, there is more than enough information in this post for you to figure them out. Also, keep in mind that if a ring has only one ideal, it must be the zero ring. And yes, a division ring is basically a field without the commutativity condition. Note that "unital division ring" and "division ring" must mean the same thing, since to define what an inverse is, you need to have a multiplicative unit. $\endgroup$ – Joppy Oct 23 '18 at 5:48
  • $\begingroup$ Thanks Joppy! Ok wrong question. I mean, is it meaningless to say 'exactly 1 right ideal and exactly 1 left ideal' hoping that the non-commutative unital ring would have more than 1 (right or left) ideal? $\endgroup$ – BCLC Oct 23 '18 at 5:55
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    $\begingroup$ A ring (perhaps non-unital, perhaps non-commutative) always has one left ideal, the zero ideal. Since $R$ itself is always a left ideal, if $R$ has only one left ideal, then $R = \{0\}$ is the zero ring. So if a ring has exactly one left ideal, it is the zero ring. $\endgroup$ – Joppy Oct 23 '18 at 6:00

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