Lemma 11.3.1. Let $A$ be a possibly noncommutative ring with $1$ which contains no nontrivial two-sided ideal. Let $M$ be a nonzero right ideal in $A$, and view $M$ as a right $A$-module. Then $A$ coincides with the bicommutant of $M$.

Proof. Let $A' = \text{End}_ A(M)$, so $M$ is a left $A'$-module. Set $A'' = \text{End}_{A'}(M)$ (the bicommutant of $M$). We view $M$ as a right $A''$-module1. Let $R : A \to A''$ be the natural homomorphism such that $mR(a) = ma$. Then $R$ is injective, since $R(1) = \text{id}_ M$ and $A$ contains no nontrivial two-sided ideal. We claim that $R(M)$ is a right ideal in $A''$. Namely, $R(m)a'' = R(ma'')$ for $a'' \in A''$ and $m$ in $M$, because left multiplication of $M$ by any element $n$ of $M$ represents an element of $A'$, and so $(nm)a'' = n(ma'')$ for all $n$ in $M$. Finally, the product ideal $AM$ is a two-sided ideal, and so $A = AM$. Thus $R(A) = R(A)R(M)$, so that $R(A)$ is a right ideal in $A''$. But $R(A)$ contains the identity element of $A''$, and so $R(A) = A''$. $\square$

[1] This means that given $a'' \in A''$ and $m \in M$ we have a product $m a'' \in M$. In particular, the multiplication in $A''$ is the opposite of what you'd get if you wrote elements of $A''$ as endomorphisms acting on the left.

There are also:

• 6 comment(s) on Section 11.3: Wedderburn's theorem

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).