Proof of Wedderburn Theorem (Henderson's) confusion

Question

In this page 2 of this article, the author provides a proof of Wedderburn's Theorem.

If $R$ is a simple ring, with a minimal left ideal, then $R \cong M_n(D)$ for some $n \ge 1$, $D$ division ring.

What I do not understand is the argument here.

Let $K$ be minimal left ideal... If $\dim_D K$ is infinite , the set $$S = \{ \alpha \in End_D(K) : \alpha(K) \text{ is finite dimensional } \}$$ is a proper ideal of $End_D(K)$, contrary to simplicity of $R$.

My confusions:

(i) I don't see how the set is proper: I only see $0 \in S, Id \notin S$.

(ii) How does this contradict simplicity of $R$?

Answer 1

Suppose $K$ is infinite dimensional over $D$. Then the identity doesn't belong to $S$, so $S$ is a proper subset of $\operatorname{End}_D(K)$.

Now prove $S$ is an ideal of $\operatorname{End}_D(K)$.

1. $0\in S$ is obvious
2. If $\alpha,\beta\in S$, then $\alpha+\beta\in S$ because…
3. If $\alpha\in S$ and $\beta\in\operatorname{End}_D(K)$, then $\alpha\beta$ and $\beta\alpha\in S$ because…

The ideal $S$ is nonzero: take a one-dimensional $D$-subspace of $K$ and the projection on it is a nonzero element in $S$. With more details: let $(x_i)_{i\in I}$ be a basis for $K_D$; fix an index $i_0$ and consider the linear map $\alpha\colon K\to K$ defined on the basis by $\alpha(x_{i_0})=x_{i_0}$ and $\alpha(x_i)=0$ for $i\ne i_0$. Then $\alpha\ne0$ and $\alpha\in S$.

Since $R\cong\operatorname{End}_D(K)$, this is a contradiction to $R$ being simple.