How do I show a minimal left ideal of a finite dimensional simple algebra exists? What can you say about its endomorphism ring?

Question

How do I approach this ?

Let $k$ be an algebraically closed field and $R$ a $k$-algebra which is also a finite dimensional vector space over $k$. Suppose $R$ has no two-sided ideals other than $0$ and $R$. Show that $R$ has a simple left ideal $I$. What can you say about the ring $\mathrm{End}_R(I)$? Justify your answer. 6 Marks

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Also What does simple left ideal mean here ?

Answer 1

A simple left ideal is one which, considered as a left $R$ module, is simple (i.e. there are exactly two submodules: the zero submodule and the submodule itself.)

Every left ideal is patently a subspace of the finite dimensional space $R$. From all dimensions of nonzero left ideals there is necessarily one of minimal dimensionality, and that is clearly a simple left ideal.

The fact that the endomorphism ring of a simple module is a division ring is known as Schur's Lemma, and it is so well-known nobody asks for proofs of it here. Nevertheless, you can find one here. You can say even more about it, also on the same wiki page.