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What is Schur's Lemma ? and why is it valid only for the algebraically closed field ?

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Hint: You want every linear operator to have at least one eigenvalue. – user38268 Mar 25 '12 at 12:27
up vote 2 down vote accepted

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 $f=\lambda I$.

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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).

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The problem is that, there are (many) different versions of Schurs lemma. Hence, you should probably write "one version of Schurs lemma", especially since the version you are referring to does not address the necessity of the ground field to be algebraically closed. – M Turgeon Mar 25 '12 at 14:27
Although most versions are just a variation on the version you stated :) – M Turgeon Mar 25 '12 at 14:29

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