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Let $A = \mathbb{C}[G]$ be finite group algebra. Assume $I = Ax$ is a minimal left ideal (has no non-trivial left ideal properly contained in it) for some element $x \in A$. Is it necessarily true that $x^2 = \alpha x$ for some $\alpha \in \mathbb{C}$?

There is a theorem:

If $I$ is a minimal left ideal of a ring $A$, then either $I^2 = 0$ or $I^2 = I = Ae$ for some idempotent $e^2 = e$ (additionally $eAe$ is a division ring).

SimB4
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    You already know it's generated by an idempotent. You're asking if every generator $x$ has the property that $x^2=\alpha x$ for some $\alpha\in \mathbb C$? You should probably revise the title question in that case. And also, interchange $e$ and $x$ in your explanation, as most people use $e$ for idempotents. – rschwieb Apr 02 '24 at 15:42
  • @rschwieb thanks for suggestions, edited accordingly. I want to understand, If this random $x$ happens to produce this minimal left ideal $Ax$, is it has to be (semi-)idempotent. – SimB4 Apr 02 '24 at 15:57
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    So somehow we need to show that $x$ is scalar multiple of $e$. if that were so, then every pair of elements in $I$ would commute. $\mathbb C[G]$ will contain copies of matrix rings when $G$ is nonabelian, and it is not hard to work out simple examples of minimal right ideals for which this is not so. – rschwieb Apr 02 '24 at 15:58
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    IIRC minimal left ideals in $M_n(k)$ consist of rank-one matrices. Cf. this. – Amateur_Algebraist Apr 02 '24 at 16:06
  • @Amateur_Algebraist interesting. So if an ideal is contained in matrix ring, then this would imply the statement? – SimB4 Apr 03 '24 at 06:42
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    @SimB4 Yes. And a minimal ideal in $\Bbb C[G]$ is a minimal ideal in one of the matrix rings in its Wedderburn-Artin decomposition (as rschwieb noted). – Amateur_Algebraist Apr 03 '24 at 07:11
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    @Amateur_Algebraist Thank you! Seems I got it. Since Wedderburn-Artin decomposition splits into matrix rings, each corresponding to certain isomorphism class of minimal left ideals, hence $I$ is ideal of a matrix ring. And $x$ then must be rank $1$ matrix of that ring, cause otherwise we can find rank $1$ matrix $y$ s.t. $Ayx \subseteq I$ is again an ideal. – SimB4 Apr 03 '24 at 08:27

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Credits to users Amateur_Algebraist and rschwieb, who gave ideas of the proof in the comments.

I am new to the subject, so this might be overkilling explanation, seems like the it does not follow the standard order of theorem derivations. But this is enough to ensure the positive answer to the question.

Assume $x \neq 0$, then $I \neq 0$. By Wedderburn-Artin theorem $$ A = \bigoplus_{i=1}^r End(V_i) $$ where $V_1, \ldots, V_r$ is a complete set of non-isomorphic irreducible representations of $G$. If $n_i = \dim V_i$ then $End(V_i) = M_{n_i}(\mathbb{C})$ is a matrix ring. Moreover, one can decompose matrix ring into minimal left ideals $End(V_i) = \oplus_{j=1}^{n_i} L^{(i)}_{j}$ where $L^{(i)}_j$ are mutually isomporhic.

Then each minimal left ideal $I$ must lie entirely in one of the components, say $End(V_1)$. Since, if $0 \neq y \in I$ then ideal $Ay = I$ (due to minimality of $I$), so if $y$ has something out $End(V_1)$ in direct sum representation, then we may take idempotent $e_1$ that projects onto $End(V_1)$, then $1-e_1$ projects outside (also an idempotent) and we obtain $A(1 - e_1)y \subseteq I$ is an ideal contradicting minimality.

Hence $I = Ax$ is an ideal in a matrix ring $M_n(\mathbb{C})$ for $n = n_1$. The claim is, that $x$ is then rank $1$ matrix of this ring. Indeed, since $x \neq 0$ there is a vector $v \in \mathbb{C}^n$ with $x(v) = u \neq 0$. Take a projector $y$ to the line $\mathbb{C}u$. Then $Ayx \subseteq I$ is an ideal of rank $1$ matrices (since we narrowed to the line) and by minimality $Ayx = I$, hence $I$ consists of rank-$1$ matrices and $x^2 = \mathrm{tr}(x) x$, by this.

SimB4
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