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$4 b)$

(i) Any hints?

(ii) Well $R$ is not semi-simple since $|\mathbb{Z}/3|=3=0 \in F_3$ by the converse of Maschke's theorem.

(iii) The surjective $\mathbb{C}$-algebra map $\phi:R \to M_2(\mathbb{C})\times\mathbb{C}\times\mathbb{C}: (a_{i,j}) \mapsto \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix},a_{33},a_{44}$ has nilpotent kernel and semi-simple target.Hence the kernel is the Jacobson radical i.e. \begin{bmatrix} 0 & 0 & * & * \\ 0 & 0 & * & * \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}

Sample solution that may help with iii) enter image description here

Walter
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1 Answers1

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(i) If you know Maschke's theorem (as you hint in (ii)) then you already know the answer to this since the order of $D_8$ is $8$.

(ii) Yes. And it would not even be too hard to exhibit some nonzero nilpotent element to prove that the Jacobson radical is nonzero.

(iii) You can use exactly the same logic here at this similar question. I'm not sure about your attempt. A surjective homomorphism from $R\to M_2(\mathbb C)\times \mathbb C\times \mathbb C$ would have a six dimensional image and a $6$ dimensional kernel (not $4$ dimensional). But part of the logic at the other post confirms that you can find a homomorphism onto $M_2(\mathbb C)\times M_2(\mathbb C)$, which has and $8$-dimensional image and the $4$-dimensional kernel you describe. Since the quotient has Jacobson radical zero, that is excactly the Jacobson radical of $R$, then.

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rschwieb
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  • Why would it have a six dimensional image? – Walter Apr 10 '17 at 17:01
  • @Walter ... because $M_2(\mathbb C)$ is four dimensional and the two extra $\mathbb C$'s only add two more dimensions? If the kernel had four dimensions, you wouldn't be able to fit the remaining $8$ dimensions of the image into this ring. – rschwieb Apr 10 '17 at 17:02
  • Also, answer to (i) is $0$ since $R$ is semi-simple? – Walter Apr 10 '17 at 17:05
  • @Walter I think that you could write a solution for both questions in the style of that solution OR in the style that you want to do. It should work either way. You need to tell me about the homomorphism before I can help. (and YES about the answer of (i) you proposed in the last comment.) – rschwieb Apr 10 '17 at 17:05
  • @Walter Sorry, I don't click weird links. You can just tell me the map if you want. – rschwieb Apr 10 '17 at 17:07
  • Check the OP thanks – Walter Apr 10 '17 at 17:11
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    @Walter Sorry, the description of the homomorphism went right by me last time. Yes, your solution to the slightly-different problem works perfectly. The problem is that you've miscomputed the kernel in the original solution. The kernel is ${\left[\begin{smallmatrix}0&0&\ast&\ast \0&0&\ast&\ast\ 0&0&0&\ast \ 0&0&\ast&0\end{smallmatrix}\right]}$, and that isn't nilpotent. See? If you switch to the homomorphism that maps the $2\times 2$ diagonal blocks onto $M_2(\mathbb C)\times M_2(\mathbb C)$ then your idea works. – rschwieb Apr 10 '17 at 17:41
  • I see my kernel is wrong. What diagonal blocks? – Walter Apr 10 '17 at 17:54
  • @Walter do to the lower right hand $2\times 2$ what you did with the upper left hand $2\times 2$, of course – rschwieb Apr 10 '17 at 18:00
  • Should I make a new question to ask about bii)? I understand i) and iii) now but ii) comes up a lot but I don't understand it totally. – Walter Apr 15 '17 at 15:13
  • @Walter I don't think it deserves a whole new post. When the order of the group is not coprime with the characteristic, the sum of all elements of the group is a nilpotent element and it is also central, so the ideal generated by it is nilpotent too, and semisimple rings do not have such ideals. – rschwieb Apr 15 '17 at 19:40