Schur's Lemma for algebraically closed field . What is Schur's Lemma ? and why is it valid only for the algebraically closed field ? 
 A: A different version of Schur's lemma (which involves algebraically closed fields) is the following:

Let $(\rho,V)$ be a finite-dimensional, irreducible representation of a finite group $G$ (or some adequate substitute) over an algebraically closed field. Then, every $G$-equivariant map $f:V\to V$ is equal to a scalar multiplication.

The reason you want an algebraically closed field is that, as Benjamin Lim noted in his comment, the map $f$ has an eigenvalue $\lambda$. Then the map $f-\lambda I$ is a $G$-equivariant map, which is not invertible, and so (as noted in Julian's answer) it is the zero map. Therefore, we conclude that $f=\lambda I$.
A: Schur's Lemma says:

If $M$ and $N$ are two simple modules over a ring $R$, then any homomorphism $f: M \to N$ of $R$-modules is either invertible or zero. In particular, the endomorphism ring of a simple module is a division ring.

The difference for algebraically closed fields is that skew fields have to be the ground field (at least for finite dimensional modules).
