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Let $V$ be a nonzero finite dimensional representation of an algebra $A$.

a) Show that it has an irreducible subrepresentation.

b) Show by example that this does not always hold for infinite dimensional representations.

I did not have any problems with part a), but I'm struggling to find an example for part b). Any help?

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4 Answers 4

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Let $A=\mathbb{C}[x]$ and consider the regular representation of $A$ acting on itself. Every nonzero element generates a subrepresentation isomorphic to $A$, so this definitely does not have any irreducible subrepresentations.

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  • $\begingroup$ (+1) LOL. Why did I restrict myself to group algebras? $\endgroup$ Aug 27, 2015 at 21:57
  • $\begingroup$ How is any principal ideal in $A$ isomorphic to $A$? $\endgroup$
    – JDZ
    Oct 16, 2018 at 1:44
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    $\begingroup$ If $I$ is an ideal generated by some element $p$ it is isomorphic to $A$ (as an $A$-module) via the map $p \cdot f \rightarrow f$. $\endgroup$
    – Nate
    Oct 17, 2018 at 22:47
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Try the following. $A$ needs to be infinite dimensional, so let's use $A=\Bbb{C}[C_\infty]$, the group algebra of the infinite cyclic group $C_\infty=\langle c\rangle$. Let $V=\Bbb{C}[T,T^{-1}]$ be the ring of Laurent polynomials, i.e. $$ V=\{a(T)=\sum_{i=m}^n a_iT^i\mid m,n\in\Bbb{Z}, m\le n, a_i\in\Bbb{C}\}. $$ We can turn $V$ into an $A$-module by letting the element $c^j, j\in\Bbb{Z}$, act by shifting $j$ positions. IOW, $$c^j\cdot a(T)=T^ja(T).$$

I then claim that $V$ has no irreducible submodules. This is seen as follows. Assume contrariwise that $W\neq\{0\}$ is an irreducible submodule. If $a(T)=\sum_{i=m}^na_iT^i\neq0$ has $a_m\neq0\neq a_n$, then let's call $(n-m+1)$ the width of $a(T)$. So all the non-zero elements of $V$ have a width that is a positive integer. Therefore any submodule of $V$ has elements of a minimal width. But if $W$ is an irreducible submodule then the space $(T-1)W=\{(T-1)a(T)\mid a(T)\in W\}$ is a subspace of $W$. Because the minimum width of elements of $(T-1)W$ is one higher than the minimum width of elements of $W$, it follows that $(T-1)W$ is a proper subspace of $W$.

But $(T-1)W$ is clearly also an $A$-submodule of $W$ contradicting the assumption that $W$ is irreducible. Therefore $V$ has no irreducible $A$-submodules.

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  • $\begingroup$ Hi, why is $(T-1)W$ a subspace of $W$? $\endgroup$
    – Javi
    Mar 1, 2019 at 17:32
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    $\begingroup$ @Javi If $x\in W$ then also $(T-1)x=(c-1)\cdot x\in W$ because $W$ was assumed to be an $A$-module. Therefore $x\mapsto (T-1)x$ is a linear transformation from $W$ to itself. The image of a linear transformation is always a subspace. $\endgroup$ Mar 2, 2019 at 6:24
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Here is my try at this problem:

Let $k$ be a field. $A:=k[x_1, x_2]$, which is Noetherian by Hilbert Basis Theorem. Hence, if $I \triangleleft k[x_1, x_2]$, $I= \langle f_1, \ldots, f_j \rangle$. Then $I' := \langle f_1 g, \ldots, f_j g \rangle \subsetneq I$ for some $g \not| f_1$. If we had an equality,$$f_1 = \sum \alpha_i f_ig \Rightarrow g | f_1 $$ Contradiction (where I assumed $f_i$s are nonzero).

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I'm not quite sure if this contraption is correct, but I'll give it a go, because I'm also interested in this question.


b) Consider an algebra $A = \mathbb{Z}$ and its infinite-dimensional representation $V\subset\mathbb{Z^\infty}$ defined by recurrence relation $c_{i+1} = c_{i} + 1$ together with a homomorphism of algebras $\rho: A \rightarrow EndV$ defined by an operator $T$ such that:

$$\rho(a) = T_a: (c_i,c_{i+1},c_{i+2},\dots) \rightarrow (c_{i+a},c_{i+a+1},c_{i+a+2},\dots)$$

Irreducibility of this representation follows immediately from the observation that every $\rho(a)$, $a \in A$ has an infinite image.

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  • $\begingroup$ What you've written down is a representation of the monoid $\mathbb{Z}_{\ge 0}$, not the group $\mathbb{Z}$ or the ring $\mathbb{Z}$. The ring you want to talk about is, say, $\mathbb{Z}[x]$ (the monoid ring of $\mathbb{Z}_{\ge 0}$). I don't understand what you're trying to claim: this representation very much fails to be irreducible, but that's not what's being asked for anyway. $\endgroup$ Aug 27, 2015 at 21:58
  • $\begingroup$ Hmm. There I mean that $a$ can take any integer (positive or negative) that defines a shift operation (left or right) by that number on a recurrent sequence. In that way, we'd get an irreducible representation of infinite dimension, because whatever subspace $W \subset V$ we'd try to take, there will always be some $a \in A$ to shift it by, so that the result of $T_a$ is in $V$ and hence $W=V$. $\endgroup$
    – neek
    Aug 27, 2015 at 22:43
  • $\begingroup$ Ah. That's still not irreducible, though: for example, there is a subrepresentation consisting of the sequences with finite support (so $c_i = 0$ for sufficiently large and sufficiently small $i$). And again, that's still not what the question asked for. $\endgroup$ Aug 27, 2015 at 22:47
  • $\begingroup$ The question b) is to give an example of a representation $V$ of algebra $A$ that doesn't have an irreducible subrepresentation, i.e. $V$ is itself irreducible, isn't it? Also, wouldn't finite support be in contradiction with the infinite dimension of V? Sorry for the dumb questions, I've just started learning this. $\endgroup$
    – neek
    Aug 27, 2015 at 23:02
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    $\begingroup$ It's not true that if $V$ has no irreducible subrepresentations then it is itself irreducible; see the other answers. And what is the contradiction? Among all sequences, which might have finite or infinite support, there are some that have finite support. I claim this is a subrepresentation (although not an irreducible one); moreover, it is also infinite-dimensional. $\endgroup$ Aug 27, 2015 at 23:13

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