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Let $R$ be a unital, but not necessarily commutative ring. Define its Jacobson radical $J$ to be the intersection of all maximal left ideals of $R$. Three questions:

  1. Let $R$ have unique two-sided maximal ideal $I$. Must $I=J$?
  2. Define $J'$ to be the intersection of all maximal two-sided ideals of $R$. Is $J' = J$?

3(a). Is every maximal left ideal a maximal ideal?

3(b). Is every maximal ideal a maximal left ideal?

My thoughts:

  1. My lecturer seemed to assert this, but I don't see why this should be the case.
  2. My guess is no. It does not even seem that we have an inclusion either way. Are there matrix rings that show $J \not\subset J'$ and $J' \not\subset J$?
  3. I think no to both. (If the answers were yes, then item 2 would become trivial.)

3(a). No. A counterexample is: let $R$ be a matrix ring over a division ring $D$. Then $R$ is a simple ring; however it has a nonzero maximal left ideal.

3(b). I guess no, because even though a maximal two-sided ideal $K$ is a left ideal, and maximal among two-sided ideals, I don't see a reason why it should be maximal among left ideals. UPDATE: Actually, I'd like to correct my guess to 'yes'. Using Zorn's Lemma, a maximal left ideal $K'$ containing $K$ is contained in a maximal ideal $L$. Then $K=L$ (since $K$ is a maximal two-sided ideal and $L$ is proper two-sided-sided), which forces $K=K'$. Is this argument correct?

Definitions: https://en.wikipedia.org/wiki/Maximal_ideal

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  1. No. Take $V$ to be a countable dimensional $F$-vector space, and consider the ring $End(V_F)$.

  2. No. Same example as the previous point. There is only one maximal two-sided ideal in the intersection.

  3. 3a) As you said "no": square matrix rings over fields have only one maximal ideal (the zero ideal) but lots of nonzero maximal left ideals.

  4. 3b) No. A square matrix ring over a field is still a counterexample to that assertion. The zero ideal is a maximal two-sided ideal, and it is contained in many nonzero maximal left ideals.

Using Zorn's Lemma, a maximal left ideal $K'$ containing $K$ is contained in a maximal ideal $L$. Then $K=L$ (since $K$ is a maximal two-sided ideal and $L$ is proper two-sided-sided), which forces $K=K'$. Is this argument correct?

The first sentence is non sequitur. There is no reason for a maximal left ideal to be contained in any proper two-sided ideal of a ring. The square matrix ring over a field demonstrates this.

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