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I am reading the paper Algebras graded by a group of Knus. Immediately I run into problems, which I will now detail:

Let $G$ be a group and $K$ a field. A $G$-graded algebra $A$ is a finite-dimensional $K$-algebra $A$ with unit together with a direct sum decomposition as $A$-module $A=\bigoplus_{g\in G}A_g$ such that $A_gA_h\subset A_{gh}$. A subspace $I$ of $A$ is graded if it is the direct sum of the intersections $I\cap A_g$. An $G$-graded algebra $A$ is called simple if there are no proper graded two-sided ideals.

Then Knus proves the following analogue of Maschke's theorem:

Let $A$ be a simple graded algebra. If the characteristic of $K$ does not divide $\dim_K(A)$, then $A$ is semisimple.

I will now write down the proof as in the paper:

Proof: Let $x=\sum_{g\in G}x_g\in \text{Rad}(A)\setminus \left\{0\right\}$. Since $A$ is simple graded, we may assume that $x_0=1$. Hence $\text{Trace}(x)=\dim_K(A)$ is not zero and $x$ is not nilpotent. Therefore the radical must be zero.

I do not understand this proof. I get that we need to show that $\text{Rad}(A)=\left\{0\right\}$. We know that the radical is nilpotent so if we find a non-zero element in $\text{Rad}(A)$ which is not nilpotent we get a contradiction, that much is clear to me.

I do not understand why we may assume $x_0\neq 0$. I do not understand what $\text{Trace}(x)$ is, is it the trace of the multiplication by $x$? I suspect we use the condition on the characteristic of $K$ in the equality $\text{Trace}(x)=\dim_K(A)$, but I don't understand this equality and I do not see how this implies that $x$ is not nilpotent.

Any help would be much appreciated.

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    $\begingroup$ I agree with your confusion, although context might clear it up (is the paper available online?). But I know how to get from $\operatorname{Tr} x \neq 0$ to the non-nilpotency of $x$: Namely, if $x$ was nilpotent, then the $K$-linear map $A \to A, \ a \mapsto xa$ would also be nilpotent, and therefore its trace $\operatorname{Tr} x$ would have to be $0$ (since the trace of any nilpotent endomorphism of a finite-dimensional $K$-vector space is $0$). $\endgroup$ – darij grinberg May 4 '17 at 15:43
  • $\begingroup$ @darijgrinberg: The paper is not (legally) available online. The context won't make it much clearer, I write everything down up to that point. Indeed, a map being nilpotent implies that the trace is zero. I know how to prove it over algebraically closed fields using the existence of eigenvalues, but after some more thought I realized this holds over all fields as well. $\endgroup$ – Mathematician 42 May 5 '17 at 7:42
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First, note that since $x\neq 0$, $x_g\neq 0$ for some $g\in G$. Now, as $x_g$ is a homogeneous element, $Ax_g A$ is a 2-sided graded ideal. Since $A$ is simple, $1\in Ax_gA$. That is, there exists $a,a'\in A$ such that $a x_g a'=1$. Now, $axa'=y\in Rad(A)$ is nonzero and satisfies $y_1=1$. Without loss of generality, we may assume that $x=y$, in which case $x_1=1$.

As you say, we understand the trace of $x$ to be the trace of the operator $y\mapsto xy$. Note that $x=1+\sum_{g\neq 1}x_g$, so $$Tr(x)=Tr(1)+\sum_{g\neq 1}Tr(x_g).$$ However, for $g\neq 1$, $x_gA_h\subset A_{gh}\neq A_h$, so $Tr(x_g)=0$. It finally remains to note that $Tr(1)=\dim_K(A)$.

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  • $\begingroup$ By the way, the fact that $\mathrm{char}(K)$ does not divide $\dim_K(A)$ implies that $\dim_K(A)$ is nonzero in $K$. $\endgroup$ – David Hill May 4 '17 at 16:16
  • $\begingroup$ That's a nice explanation. I figured we had to look at the graded ideal generated by $x$, but it makes more sense to look at the graded ideal generated by $x_g$. Thanks! $\endgroup$ – Mathematician 42 May 5 '17 at 6:14

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