Talk:Maximal Ideal iff Quotient Ring is Division Ring

This theorem is not correct and can't be proved.

From Introductory Lectures on Rings and Modules By John A. Beachy: Proposition 1.5.4 states: Let $R$ be a ring. Then every ideal of $M_n\paren{R}$ (the ring of matrices over $R$) has the form $M_n\paren{I}$, for some ideal of $R$. Since a field $F$ has no proper nontrivial ideals it follows that the ring $M_n\paren{F}$ has no proper nontrivial ideals. For $n > 1$, $M_n\paren{F}$ is non-commutative and $M_n\paren{F}/0 = M_n\paren{F}$ is not a division ring.

The correct theorem is: Maximal Left and Right Ideal iff Quotient Ring is Division Ring

I propose that this theorem be deleted.

The only theorem that is dependent on the theorem Maximal Ideal iff Quotient Ring is Division Ring is Basis of Free Module is No Greater than Generator. The proof of Basis of Free Module is No Greater than Generator can be made dependent on Maximal Ideal iff Quotient Ring is Field by changing the Ring in the premise to be commutative and invoking the theroem Maximal Ideal iff Quotient Ring is Field.

The only theorem dependent on Basis of Free Module is No Greater than Generator is the theorem Bases of Free Module have Equal Cardinality where the premise already assumes a commutative ring.

So it seems that the proposed changes would bring things into alignment.

--Leigh.Samphier (talk) 22:39, 19 October 2018 (EDT)


 * Take it away, maestro. $\stackrel{. \ .} {\smile}$ --prime mover (talk) 05:11, 20 October 2018 (EDT)