I'm trying to solve this problem in Curtis's Representation Theory text.
The problem is: If a ring $A$ with unity, satisfying the minimum condition, has a faithful irreducible $A$-module $M$, then $A$ is simple.
In the section that has this problem, I just learned about Wedderburn theorem about simple ring $A$ being isomorphic to Hom$_D(M,M)$ for some division ring $D$.
The book's def for being $A$ being simple is that it satisfies the minimum condition and it has no proper nontrivial two sided ideal.
I assume $I \subseteq A$ is a nontrivial two sided ideal, then $IM$ is an $A$-submodule of $M$ so $M$ being irreducible implies $IM = 0$ or $IM = M$. At this point, I can't deduct anything useful from this.
I also try to use the fact that $M \simeq A/J$ where $J$ is a left maximal ideal in $A$ but it doesn't get me anywhere.
Can someone point me in the right direction?
Thank you very much.
Best Answer
This is a standard equivalence of the many definitions that can be given to simple rings. The Algebra of Bourbaki has a very good presentation of it, so I will merely attempt here to provide a sketch of the proof.
Assume that ring $A$ is left artinian and admits a faithful simple left $A$-module $S$. By hypothesis we are given that: $$\mathrm{Ann}_A(S)=\bigcap_{x \in S^{\times}}\mathrm{Ann}_A(x)=\{0_A\},$$ where I have taken the liberty to denote $S^{\times}=S \setminus \{0_S\}$.
By virtue of artinianity, any intersection of an arbitrary (nonempty) family of left ideals of $A$ can be expressed as the intersection of only finitely many of those ideals. In particular, in our case there exists a nonempty finite subset $H \subseteq S^{\times}$ such that $\displaystyle\bigcap_{x \in H}\mathrm{Ann}_A(x)=\{0_A\}$.
As $S$ is simple, it is generated by any element $x \in S^{\times}$; for any such nonzero element we consider the map: $$\begin{align} h_x \colon A &\to S\\ h_x(\lambda)&=\lambda x. \end{align}$$ This map is clearly seen to be $A$-linear and surjective and we have $\mathrm{Ker} h_x=\mathrm{Ann}_A(x)$ by definition (hence $S \approx A/\mathrm{Ann}_A(x)$ as $A$-modules and therefore the annihilator in question is a maximal left ideal).
Consider now the map: $$\begin{align} f \colon A &\to S^H\\ f(\lambda)&=(\lambda x)_{x \in H}, \end{align}$$ the so-called direct product of $h_x$, $x \in H$ in restricted sense. It is clearly an $A$-module morphism and has the property that: $$\mathrm{Ker}f=\bigcap_{x \in H}\mathrm{Ker}h_x=\{0_A\},$$ which means that it is injective. Thus, $A$ is isomorphic to an $A$-submodule of the isotypical module $S^H$. Since all submodules of a given isotypical module are isotypical of the same type, this means that $A$ itself is isotypical as an $A$-module.
In general, if $M$ is a semisimple left $A$-module then it remains semisimple when regarded as a (left, left) $(A, \mathrm{End}_{\operatorname{A-\mathbf{Mod}}}(M))$-bimodule, for any simple $A$-module $T$ the isotypical component $M_T$ being a simple $(A, \mathrm{End}_{\operatorname{A-\mathbf{Mod}}}(M))$-submodule of $M$. In particular, if $M$ is isotypical as a left $A$-module then it is simple as an $(A, \mathrm{End}_{\operatorname{A-\mathbf{Mod}}}(M))$-bimodule.
We apply this observation to our situation, by taking into account that:
we gather that $A$ is bilaterally simple, since the $(A, A)$-submodules of $A$ are precisely its bilateral ideals. More explicitly, this means that the only bilateral ideals of $A$ are $\{0_A\}$ and $A$ itself, which together with the assumption of left artinianity recovers your textbook definition of simplicity for $A$.
Simpler variant of the above proof
Let us resume the above train of thought at the point where we established the embedding $f \colon A \to S^H$. I will ask you for the continuation of this proof to familiarise yourself with the very simple result which essentially states the following:
$$N \approx \bigoplus_{i \in J}T_i \quad (\operatorname{A-\mathbf{Mod}}).$$
In plain speech, this means that any submodule of a semisimple module is itself semisimple and has the particular consequence that any submodule of an isotypical (i.e. a semisimple in which all the simple summands are isomorphic to each other) module is itself isotypical. It is an elementary result in the theory of semisimplicity, with easy to understand proofs that you can find in either Bourbaki or Serge Lang's algebra treatise.
Once you have accepted and assimilated this result, we can return to our setting and deduce from the above embedding the existence of an isomorphism $g \in \mathrm{Hom}_{\operatorname{A-\mathbf{Mod}}}\left(A, S^F\right)$, where $F \subseteq H$.
To make explicit my notation conventions in the subsequent portion, for arbitrary $A$-module $M$ and additive subgroups $G \leqslant A$, $T \leqslant M$ I will write $G.T=\langle GT \rangle$ for the "subgroup product", which is the additive subgroup generated by the subset product $GT=\{\lambda t\}_{\substack{\lambda \in G\\t \in T}}$. As you will be able to gather, the whole purpose of the "." syntax is to avoid confusion between subgroup products and subset products, both being equally important objects in the study of linear algebra.
Consider now an arbitrary nonzero bilateral ideal $\{0_A\} \neq I \leqslant_{\mathrm{b}} A$. Since $I$ is in particular a left ideal we have that $I.S \leqslant_A S$ is a submodule of $S$, therefore either $I.S=\{0_S\}$ or $I.S=S$ by virtue of simplicity. The first case would entail the contradiction $\{0_A\} \subset I \subseteq \mathrm{Ann}_A(S)=\{0_A\}$, so we are led to the conclusion that $I.S=S$. As a remark, the same conclusion could have been more generally reached for any nonzero left ideal.
Let us now exploit that $I$ is also a right ideal, by noticing that $I.A=\langle IA \rangle=\langle I \rangle=I$, on the one hand. On the other hand, since $g$ is $A$-linear we also have: $$g[I]=g[I.A]=I.g[A]=I.\left(S^F\right)=\left(I.S\right)^F=S^F=g[A]$$ and on the grounds of $g$ being injective we infer that $I=A$. This means that there is no other nonzero bilateral ideal than the improper one.