# Why $End(M_{D})$ is not simple when $M_{D}$ is infinite dimensional?

I am reading Rings and Categories of Modules by Frank W.Anderson. In the book on page 161:

$End(M_{D})$ is primitive for every vector space $M_{D}$, but if $M_{D}$ is infinite dimensional, then $End(M_{D})$is not simple.

I want to know why $End(M_{D})$ is not simple when $M_{D}$ is infinite dimensional.

Actually, if $\kappa$ is the cardinality of the basis of $M_D$, for each infinite cardinal $\gamma$ less than $\kappa$, there is such an ideal of elements having dimensionality strictly less than $\gamma$, and those turn out to be all the ideals. Consequently the ideals are linearly ordered and have this nice description.
• @guojm OK. That is true if $\dim(M_D)$ is countable. The lemma you need to work on is this: if $f$ has an infinite dimensional image, then there exists $g$ such that $gf=1_M$. You simply define $g$ to invert $f$ on $Im(f)$, and define it to be zero elsewhere, and then $gf(x)=x$ for every $x\in M$. That means $(f)=End(M_D)$ for any such $f$. – rschwieb May 19 '17 at 13:04
• @guojm If you wanted to go further, you could adapt that to prove in general, if $\dim(Im(h))\leq \dim(Im(f))$, then there exists $g\in End(M_D)$ such that $gf=h$. (This would be useful when you let the dimension of $M$ be much larger.) – rschwieb May 19 '17 at 13:06