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Let $D$ be a division ring and $M$ be an infinite dimensional vector space over $D$. Then $R=\text{End}_D(M)$ is a regular ring. Let $I=\{a\in R\mid \text{dim}_D(aM)<\infty\}$. Then, is the set $R\backslash I$ (the compliment of $I$ in $R$) an ideal of $R$? If not, how can I ascertain that? What about the set $R\backslash I+0$, that is, when I adjoin a zero.

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If $\kappa$ is the dimension of your vector space, then in fact there is an ideal corresponding to each cardinal less or equal to $\kappa$. Each one is give by the set of transformations which have images that are strictly lower dimension than the specified cardinal. Thus, they are linearly ordered.

Obviously the complement of $I$ does not contain $0$, so it can't be an ideal. Even if you throw in zero, it is still obviously not an ideal. Given any linear transformation, you can always compose with another transformation on the left to reduce the dimension of the image ( to whatever smaller dimension you want.) You should be able to confirm this on your own, now, but let me know if you are stuck.

By the way, you also said unit regular ring in the title . The type of endomorphism ring you are talking about is never unit regular. The dimension of $M$ is necessarily finite if the ring is to be unit regular.

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No, because $I$ is itself an ideal different from $\{0\}$ and $R$. The general statement is: If $I$ is an ideal of $R$ and $R\setminus I \cup \{0\}$ is also an ideal, then $I=\{0\}$ or $I=R$. Note that I assume that $R$ has an identity element.

To prove this, note that if $I\neq R$, then $R\setminus I \cup \{0\}$ contains $1_R$, so if that set was an ideal, we would have $R\setminus I \cup \{0\}= R$. This can only happen if $I=\{0\}$.

I should add that $R\setminus I$ is never an ideal, since it does not contain $0$.

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