# Let $A$ be a finite-dimensional $k$-algebra and $V$ a completely reducible $A$-module. Show that the algebra $A_V$ is semisimple.

This is Problem 1.6 from I.M. Isaacs, Character Theory of Finite Groups:

Let $$A$$ be a finite-dimensional $$k$$-algebra and $$V$$ a completely reducible (left) $$A$$-module which is also finite-dimensional as a vector space over $$k$$. Show that the algebra $$A_V$$ is semisimple.

Note: $$A_V$$ denotes the image of the algebra homomorphism $$A \to \operatorname{End}_F(V)$$, $$a \mapsto a_V$$, where $$a_V$$ is the natural action by multiplicating $$a$$ on the left: $$a_V: V \to V, x \mapsto ax$$. Also, by semisimple I mean $$A_V$$ is completely reducible as a left module over itself.

I think it could be proved using Jacobson radical. What I want to ask is whether the proposition can be proved in a more direct way, without the help of Jacobson radical. Can anyone help?

My 'proof':

Lemma: Let $$M_1$$, $$M_2$$ be finite-dimensional simple modules over $$A$$, $$m_1\in M_1$$, $$m_2\in M_2$$, $$m_1, m_2 \neq 0$$. Either $$\operatorname{Ann}(m_1) = \operatorname{Ann}(m_2)$$, or $$\operatorname{Ann}(m_1) + \operatorname{Ann}(m_2) = A$$.

Proof of lemma: $$A/\operatorname{Ann}(m_i) \cong M_i$$ and $$M_i$$ simple imply that $$\operatorname{Ann}(m_i)$$'s are maximal ideals of $$A$$. Thus, if $$\operatorname{Ann}(m_1) \neq \operatorname{Ann}(m_2)$$, there must be $$\operatorname{Ann}(m_1) \subset \operatorname{Ann}(m_1) + \operatorname{Ann}(m_2)$$, hence $$\operatorname{Ann}(m_1) + \operatorname{Ann}(m_2) = A$$.

Suppose $$V=V_1\oplus\dots\oplus V_n$$, with $$V_i$$'s irreducible. Then $$\operatorname{Ann}(V) = \bigcap_{i=1}^n\operatorname{Ann}(V_i) = \bigcap_{i=1}^n \bigcap_{v\in V_i, v \neq 0}\operatorname{Ann}(v).$$

Now partition $$\bigcup_{i = 1} ^n V_i$$ with respect to equivalence relation ~, where $$v$$~$$w$$ iff $$\operatorname{Ann}(v) = \operatorname{Ann}(w)$$, and let $$\mathscr{V} = \{v_j: j \in J\}$$ be a representative set. Then $$\operatorname{Ann}(V) = \bigcap_{j\in J} \operatorname{Ann}(V_j)$$. If we had shown that $$J$$ is finite, by the lemma Chinese Remainder Theorem applies, showing that $$A_V \cong A/\operatorname{Ann}(V) \cong \bigoplus_{j\in J} (A/\operatorname{Ann}(v_j)),$$ completing the proof.

Note that the proof relies on the hypothesis that the representative set $$\mathscr{V}$$ is finite, which I believe is true but cannot prove. In particular, it suffices to prove that the set $$\{\operatorname{Ann}(m):m\in M\}$$ is finite for a simple $$A$$-module $$M$$.

edited under rschwieb's suggestion:

$$A/\operatorname{Ann}(v)$$'is generally not a ring, so CRT does not hold in the full sense that the mapping is onto, as is pointed out by rschwieb. Still, one may proceed using the injection $$\phi: A/\operatorname{Ann}(V) \to \bigoplus_{j\in J} (A/\operatorname{Ann}(v_j))$$ if $$J$$ can be chosen to be finite. $$\operatorname{Im} \pi_j \phi$$ is a submodule of $$A/\operatorname{Ann}(v_j)$$, hence must be either $$0$$ or a simple module ($$\pi_j$$ is projection of the sum into $$A/\operatorname{Ann}(v_j)$$). It follows that the sum is semisimple.

But to chose a finite $$J$$ one has to use a complicated device (see rschwieb's comment below), which makes this proof way more complicated than that using $$J(A)$$.

• @rschwieb I have edited the question to illustrate what $A_V$ is; I forgot it is not a standard notation because the book used it repeatedly. Also, by semisimple I mean the $A_V$ is completely reducible as a left module over itself. Thank you for pointing out the mistakes.
– zyy
Commented Oct 20, 2023 at 14:28
• @rschwieb I have mentioned in my last reply that, by semisimple I mean $A_V$ is completely reducible as a left module over itself.
– zyy
Commented Oct 20, 2023 at 16:52
• Yes, I meant for you to add it to the post as context. Thank you for doing that. Commented Oct 20, 2023 at 16:55

(Note: after this answer was given, the user inserted the "finite dimensional assumption.)

If semisimple means "is completely reducible as a left module over itself" then this claim is not correct as stated.

For example, Let $$A=\prod_{i=0}^\infty F$$ for a field $$F$$. It has a semisimple submodule $$V=\bigoplus_{i=0}^\infty F$$ which is faithful. Faithfulness means that the map $$A\to A_V$$ will be injective, hence an isomorphism. But $$A$$ is not Artinian, so it cannot be completely reducible as a left module over itself.

If semisimple were taken to mean "has trivial Jacobson radical" then the claim is correct: since $$A_V\cong A/ann(V)$$, we'd have that $$V$$ is a faithful semisimple $$A_V$$ module, which is the same thing as having trivial Jacobson radical.

Perhaps there is some assumption in the book that you've not included which would rectify the situation.

Added after finite dimensionality was introduced

If one assumes additionally that $$V$$ is finite dimensional, then also $$End_F(V)$$ and $$A_V$$ will also be finite dimensional. Thus $$A_V$$ would be Artinian with Jacobson radical zero, i.e. a semisimple ring.

I am thinking to see if one can do this without utilizing the radical but at this point it seems like the simplest route...

• Thank you for you answer. Please forgive me for my carelessness, because I do neglect two assumptions in the book. In the book (which is about character theory of finite groups), $A$ is assumed to be a finite-dimensional $k$-algebra, and $V$ is also finite-dimensional viewed as a vector space over the field $k$.
– zyy
Commented Oct 20, 2023 at 17:49
• @zyy In general it is very bad form to "move the goalposts" on a question after it has been answered. I will overlook it this time but in the future you should be very careful when posting not to omit hypotheses like this. Commented Oct 21, 2023 at 1:54
• Sorry for that. I have only come across topics like 'semisimplicity' and 'algebra' in the context of representation theory of finite groups, but have never learnt them in their full generality. I should be more careful in the future.
– zyy
Commented Oct 21, 2023 at 8:14
• @zyy I see: thank you for sharing the context. That is a good motivation. If I think of anything I'll expand the answer further. Commented Oct 22, 2023 at 15:47
• @zyy I see, I must have overlooked that. Part of the reason for doubt is that there isn't really a CRT for modules at least, not in the full sense that the mapping is onto. But yes, I agree the annihilators are comaximal right ideals when the points are taken from simple modules. Commented Oct 23, 2023 at 14:31