# Question about an argument in Isaac's Algebra: Wedderburn-Artin Theorem

Theorem: Let $$R$$ be a Wedderburn ring, and let $$\mathcal{S}$$ be a representative set of the simple right $$R-$$modules. For each $$S \in \mathcal{S}$$, let $$I_S$$ be the sum of all those minimal right ideals of $$R$$ that are isomorphic to $$S$$. Then:

1. $$|\mathcal{S}|$$ is finite.

2. Each $$I_S$$ is a minimal (nonzero, two-sided) ideal of $$R$$.

3. $$R$$ is the direct sum of the ideals $$I_S$$.

4. Each ideal $$I_S$$ is a simple right artinian subring of $$R.$$

5. If $$S,T \in \mathcal{S}$$, with $$S \not \cong T$$, then $$T I_S = 0$$.

In particular, I have a question about the minimality argument of part (2).

We suppose that $$I$$ is an ideal of $$R$$ so that $$I < I_S$$. Since $$I$$ is proper, we can choose a minimal right ideal $$J \leq I_S$$ with $$J \cong S$$ and $$J \not \subseteq I$$. Then $$J \cap I$$ is a right ideal of $$R$$ properly contained in $$J$$, and hence $$J \cap I = 0$$. However, $$JI \subseteq J \cap I$$ since $$I$$ is a left ideal, so $$JI = 0$$.

Al this is fine with me. Next Isaacs writes: It follows that $$I_S I = 0$$ since $$I_S$$ is a sum of modules isomorphic to $$J$$, and each is annihilated by I.

But we have only shown that one of the modules isomorphic to $$J$$ is annihilated by $$I$$. Why does it follow that all of them are annihilated by $$I$$?

My thoughts: If $$J'$$ is another minimal right module isomorphic to $$J$$, then $$J J' \subseteq J$$. If $$J J' > 0$$, then $$J J' = J$$ since $$J$$ is right simple. So if $$I$$ contains any right module isomorphic to $$J$$, then it contains all of them - from which the result would follow. I am not sure how to argue that $$J J' > 0$$, or even if it is true...

Am I looking in the right direction?

The answer was very simple: $J$ and $J'$ are by definition both isomorphic as $R$-modules, so naturally if $I$ annihilates one then it annihilates the other. From this it follows that $I_S I = 0$, so $R I = 0$, implying that $I = 0$.