Density Theorem

Motivation

We have a Ring homomorphism between $R$ and bicommutant $R^{\prime \prime }$. How big is the image of the ring homomorphism? Then answer is quite big through the Density Theorem.

Lemma(Key Argument)

Let $E$ be a Semi Simple module over $R$. Let $R^{\prime }=En{d}_R(E)$ and $R^{\prime \prime }=En{d}_{R^{\prime }}(E)$. Let $x\in R$. Then there exists an element $\alpha \in R$ such that $\alpha x=f(x)$, i.e. $f(x)=ev(r)(x)$.¹

Proof

Since $E$ is a semi simple module, $E=Rx\oplus F$ for some submodule $F$. Let $\pi :E\to Rx$ be a projection onto $Rx$. Then $\pi \in R^{\prime }$, so that $f\circ \pi =\pi \circ f$, which implies $f(x)=f(\pi (x))=\pi (f(x))=rx$ for some $r$.

Jacobson Density Theorem

Let $E$ be a semi simple module over $R$. Let $e_1,\cdots ,e_n\in E$, and $f\in R^{\prime \prime }$. Then $\exists r\in R$ such that $r{e}_i=f(e_i)$ for all $i$. Thus if $E$ is finitely generated over $R^{\prime }$, then $R\twoheadrightarrow R^{\prime \prime }$.

Proof

Special Case($E$ : simple)

Apply lemma with\underline{e} = (e_{1}, \cdots, e_{n}) \in E^{n}$with{f}^n:E^n\to E^n$ such that $f^n(y_1,\cdots ,y_n)=(f(y_1),\cdots ,f(y_n))$.

Since $f\in R^{\prime }(R^{\prime }(E))$, $f^n\in R^{\prime }(M_n(R))=R^{\prime }(R^{\prime }(E^n))$. Then by the key argument there exists $r\in R$ such that $r\underline{e}=f^n(\underline{e})$.

If $E$ is not simple, by the semi simple module (SS2), we can use the Block Decomposition.

If $E$ is finitely generated over $R^{\prime }$, an element $g\in En{d}_{R^{\prime }}(E)=R^{\prime \prime }$ is determined by its value on a finite number of elements of $E$. So we can see that surjectivity easily.

Application

Burnside’s Theorem

Footnotes

1. I am not sure but the form looks like a Riesz Representation theorem. ↩