Jacobson Density Theorem - Proof

Proof

In the Jacobson density theorem, the right R-module U is simultaneously viewed as a left D-module where D=End(UR) module in the natural way: the action g·u is defined to be g(u). It can be verified that this is indeed a left module structure on U. As noted before, Schur's lemma proves D is a division ring if U is simple, and so U is a vector space over D.

The proof also relies on the following theorem proven in (Isaacs 1993) p. 185:

Theorem

Let U be a simple right R-module and let D = End(UR) - the set of all R module endomorphisms of U. Let X be a finite subset of U and write I = annR(X) - the annihilator of X in R. Let u be in U with u·I = 0. Then u is in XD; the D-span of X.


Proof (of the Jacobson density theorem)

We proceed by mathematical induction on the number n of elements in X. If n=0 so that X is empty, then the theorem is vacuously true and the base case for induction is verified. Now we assume that X is non-empty with cardinality n. Let x be an element of X and write Y = X \ {x}. If A is any D-linear transformation on U, the induction hypothesis guarantees that there exists an s in R such that A(y) = y·s for all y in Y.
Write I = annR(Y). It is easily seen that x·I is a submodule of U. If it were the case that x·I = 0, then the previous theorem would indicate that x would be in the D-span of Y. This would contradict the linear independence of X, so it must be that x·I ≠ 0. So, by simplicity of U, the submodule x·I = U. Since A(x) - x·s is in U=x·I, there exists i in I such that x·i = A(x) - x·s.
After defining r = s + i, we compute that y·r = y·(s + i) = y·s + y·i = y·s = A(y) for all y in Y. Also, x·r = x·(s + i) = x·s + A(x) - x·s = A(x). Therefore, A(z) = z·r for all z in X, as desired. This completes the inductive step of the proof. It follows now from mathematical induction that the theorem is true for finite sets X of any size.

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