# Proof of Wedderburn's theorem - Dummit and Foote Section 18.2, Problem 6

I am trying to prove Wedderburn's theorem by solving problems 1-10 in Section 18.2 of Dummit and Foote. However, I find myself rather stuck on Problem 6.

Let R be a non-zero ring containing 1. Show that $R = \bigoplus_{i=1}^r R_i$ where $R_j$ is a 2-sided ideal and a simple ring (i.e: has no proper 2 sided ideals). Show each $R_j$ has an identity and satisfies the descending chain condition on left ideals (i.e: There is no infinitely long descending chain of left ideals).

(Originally the problem says $R = R_1 \times \ldots \times R_r$, however, I don't see how R can be equal to a cartesian product of its subsets, when viewed as a set. So I took the liberty of changing it into an (internal) direct sum)

From the previous problem I know that $R$ is the (internal) direct sum of its (finitely many) simple left R-modules (left ideals) and that $R$ satisfies the descending chain condition. The hint provided by the book is:

Use the preceding exercise to show that $R$ has a minimal 2-sided ideal $R_1$. As a left $R$-module $R = R_1 \bigoplus R'$ for some left ideal $R'$. Show that $R'$ is a right ideal and proceed inductively using D.C.C (descending chain condition)

I am stuck in the process of showing that $R'$ is a two sided ideal. Thus far I have only managed the following:

Say $J_1, ... , J_n$ are the simple left R-modules (left ideals) such that $\bigoplus_{i=1}^n J_i = R$. Without loss of generality, I have shown that there is some $k \in \{1, \ldots, n\}$ such that $R_1 = \bigoplus_{i=1}^k J_i$ and $R' = \bigoplus_{i=k+1}^n J_i$ [I assumed that $k < n$ since otherwise, $R_1 = R$ and in that case, what we're trying to prove is true trivially].

However, I don't see how this is useful or how to go about proving what is asked for. Any help is appreciated.

• @anon if it’s a proof of Wedderburn-Artin, semi-simplicity is implied. – David Hill Dec 9 '17 at 3:33
• Derp. ${}{}{}{}$ – anon Dec 9 '17 at 7:57

First prove that if $V$ is a simple $R$ module, and $I\subset R$ is a minimal left ideal such that $IV\neq 0$, then $I\cong V$ as a $R$-module (fix $v\in V$ nonzero and show that the map $I\to V$, $x\mapsto xv$ is an isomorphism).
Next, let $I=\{i\mid J_i\cong J_1\}$. Use the previous paragraph to show that $R_1=\bigoplus_{i\in I}J_i$ is a minimal two-sided ideal.
Write any $x\in R$ as $x=\sum_{i=1}^nx_i$ with $x_i\in J_i$. Now, if $y=\sum_{i\in I}y_i\in R_1$, then $yx=\sum_{i,j\in I}y_ix_j\in R_1$ showing $R_1$ is a 2-sided ideal. Conversely, if $R_1'\subset R_1$ is a 2-sided ideal, show that $R_1'=\bigoplus_{i\in I'}J_i$ for some subset $I'\subset I$. Now, act on the right with an element $J_j$ for some $j\in I\backslash I'$...
Finally, use the first paragraph to show that $R'=\bigoplus_{i\notin I}J_i$ is a two-sided ideal.
As above, take $x\in R$ and write $x=\sum_{i=1}^nx_i$ with $x_i\in J_i$. For $y\in R'$, write $y=\sum_{i\notin I}y_i$ and deduce that $yx=\sum_{i,j\notin I}y_ix_j \in R'$.