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Let $V$ be a right $D$ (a division ring) vector space of countably infinite dimension. Let $E=\mathrm{End}(V_D)$, then I is the ideal of E consisting of endomorphisms of finite rank. The claim is that $R=E/I$ is a simple ring.

So I take an ideal in $R$, say $U$, which properly contains $I$, I will show it is unital so improper. So let $f$ belong to $U$ but not $I$, so it is not finite dimensional, so I write $V=\ker(f)+W$

Now my linear algebra is not that strong which is hindering me further. ker(f) can be written as direct summand I know. But either of $\ker(f)$ or $W$ can have finite dim or both will have infinite? And T Y Lam used one more thing by letting $\{u_1,u_2,u_3\ldots\}$ be a basis for $W$ such that $\{f(u_1),f(u_2)\ldots\}$ will be linearly independent?? why?

rschwieb
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Bhaskar Vashishth
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1 Answers1

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As you remark $V= \ker f \oplus W$ this is obtained by taking a basis $A$ for $\ker f$ and then extending $A$ to a basis $A\cup B$ of $V$, $W$ is then the span of $B$. By assumption $f$ has infinite rank, and therefore both $W$ and $V$ have dimension $\aleph_0$. There now exists linear maps $g:V \rightarrow V$ and $h:V \rightarrow V$ such that $g$ maps $V$ onto $W$ with zero kernel and $h$ maps $im f$ onto $V$ such that $h$ is injective on $im f$.

Now consider $h\circ f\circ g \in U$ this sends $V$ injectively onto $W$, then $f$ sends $W$ injectively onto $im f$ and finally then $h$ sends $im f$ injectively onto $V$ so the resulting map is invertible, and $U$ contains the identity.

Rene Schipperus
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  • you mean g is the projection map on W and h is lifting map which takes w to (0,w). right? shouldn't be krg = kerf and ker h=0.

    letting u1,u2,u3..... as basis for W that f(u1),f(u2).... will be linearly independent?? why this is used in Lam?

     – Bhaskar Vashishth Jun 11 '14 at 07:49
  • No $g$ is not the projection map. $g$ has zero kernel. It is just some map from $V$ to $W$ with zero kernel it exists because the two spaces have the same dimension. – Rene Schipperus Jun 11 '14 at 07:56
  • sorry, but if h maps W onto V with kernel kerf, then won't W/kerf isomorphic to V by 1st isomorphism theorem – Bhaskar Vashishth Jun 11 '14 at 08:12
  • I made some notation mistakes writing it down $h$ goes from $\im f$ to $V$. I corrected it. – Rene Schipperus Jun 11 '14 at 12:13