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Let $A$ be a finite-dimensinoal associative algebra with unit $1$ over $K$. Then $A$ is a $A$-right module (by module I always mean right module) over itself by right multiplication (the so called right regular module), denote it by $A^{\circ}$. An $A$-right module $V$ is semisimple if it is the sum of irreducible $A$-submodules, and this is equivalent with the fact that for every $A$-submodule $U \le V$ there exists another $A$-submodule $W$ such that $V = U\oplus W$. An algebra $A$ is called semisimple if $A^{\circ}$ is semisimple as a module. With this we could show that if $A^{\circ}$ is semisimple, then every irreducible $A$-right module is isomorphic to a submodule of $A^{\circ}$.

Now I noticed the following, but see it nowhere mentioned. So I ask myself if my reasoning is correct. If $W$ is an irreducible $A$-module for a semisimple algebra $A$, then I guess $W$ must be cyclic as an $A$-module, i.e. generated by a single element. This follows because an $A$-isomorphism commutes with multiplication by elements from $A$, and every element from $A$ (or $A^{\circ}$) could be written as a multiple of $1 \in A$.

Let $A$ be a semisimple algebra and let $W$ be an irreducible $A$-right module, then $W$ is generated by a single element.

Proof: By the above there exists a submodule $W_0$ of $A^{\circ}$ such that $W_0 \cong W$. Let $\varphi \in \operatorname{Hom}_A(W_0, W)$ be such an isomorphism, then for $x \in W_0$, as also $x \in A$, we have $\varphi(x) = \varphi(x1) = \varphi(1x) = \varphi(1)x$, hence every element $y \in W$ has the form $\varphi(1)\cdot x = y$ where $x \in A$ is choosen such that $\varphi(x) = y$, i.e. we have $\varphi(1)A = W$. $\square$

Is this correct? Or have I overlooked something? And is this of use, or not; as I see it nowhere mentioned?

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    $\begingroup$ You should not call such module $1$-dimensional, as this sounds like they are free (which they need not be). Being generated by a single element is called being cyclic, and any simple module is cyclic, regardless of the algebra. $\endgroup$ – Tobias Kildetoft Feb 22 '16 at 11:20
  • $\begingroup$ Thanks for your comment. I am not that much into module theory and just used the terminology which is common for vector spaces. Maybe you want to make an answer out of your comment; maybe with some further remarks about cyclic modules, where they are used and so on... $\endgroup$ – StefanH Feb 22 '16 at 11:25
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You're overthinking! ;-)

Let $M$ be an irreducible left module over the (unital) ring $R$. By definition, $M\ne\{0\}$ and the only submodules of $M$ are $\{0\}$ and $M$.

Take $x\in M$, $x\ne0$; then $Rx$ is a nonzero submodule of $M$, so $Rx=M$.

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