Decomposition of $\mathbb{C}\mathbb{Z_5}$ into irreducible one dimensional modules using Artin-Wedderburn Theorem I am an undergraduate student taking a graduate class in Abstract Analysis and Representation Theory. For one of my HW's, I have a question where I have to decompose $\mathbb{C}$$\mathbb{Z_5}$ using Artin Wedderburn Theorem and then determine the isomorphism explicitly in terms of the natural basis on the right hand side and the basis {$e$, $g$, $g^2$,$g^3$, $g^4$} of $\mathbb{C}$$\mathbb{Z_5}$. I am not completely sure what the question is asking when it means determine the isomorphism explicitly but here is my attempt. Since the irreducible $\mathbb{C}$$\mathbb{Z_5}$ are 1-dimensional we can use Artin-Wedderburn to split them into $M_1$($\mathbb{C}$) $\times$$M_1$($\mathbb{C}$) $\times$$M_1$($\mathbb{C}$) $\times$$M_1$($\mathbb{C}$) $\times$$M_1$($\mathbb{C}$). The standard basis is obviously {(1,0,0,0,0),(0,1,0,0,0),(0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1)}={$e_1$, $e_2$, $e_3$,$e_4$, $e_5$}. I was thinking that maybe we can find an orthogonal basis using 5th root of unity and then represent it in therms of the natural basis. In particular, something along the lines of g maps to $e_1$ + $\omega$$e_2$ + $\omega^2$$e_3$ + $\omega^3$$e_4$ + $\omega^5$$e_5$. Does this seem correct?
 A: I understand the following part of the question:

Decompose $\mathbb{C} \mathbb{Z}_5$ using Artin Wedderburn Theorem and then determine the isomorphism explicitly in terms of the natural basis on the right hand side.

In this case, the Artin Wedderburn theorem allows us to conclude that there is an isomorphism $\Phi:\Bbb C \Bbb Z_5 \to [M_1(\Bbb C)]^5$, and that the diagonal entries in the image corresponds to the irreducible representations $\phi:\Bbb Z_5 \to \Bbb C^\times$.  
Every irreducible representation of $\Bbb Z_5$ has the form $\phi(g^k) = \omega^k$ for some $\omega$ satisfying $\omega^5 = 1$.  These solutions $\omega$ can be written in the form $\omega = e^{2 \pi ji/5}$ for $j = 0,1,2,3,4$.  With that, we note that one such isomorphism can be written explicitly as 
$$
\Phi(g^k) = \pmatrix{1\\&e^{2 \pi ki/5} \\ && \ddots \\ &&& e^{4(2 \pi k)i/5}}, \quad k = 0,\dots,4.
$$
This isomorphism is expressed in terms of the "natural basis on the right hand side".  That is, we have written the elements of $\Phi(G) =  [M_1(\Bbb C)]^5 \subset M_5(\Bbb C)$ in the way that they would "naturally" be written. This is consistent with the notation and definitions of Dummit and Foote.
Depending on how you or your professor prefer to to think about $[M_1(\Bbb C)]^5$, we could also write this in the form
$$
\Phi(g^k) = (1,e^{2 \pi ki/5} , \cdots , e^{4(2 \pi k)i/5}), \quad k = 0,\dots,4.
$$
In this case, the set corresponding to the product $[M_1(\Bbb C)]^5$ of algebras is taken to be a Cartesian product $M_1(\Bbb C) \times \cdots \times M_1(\Bbb C)$ rather than a "block-diagonal" subset of $M_5(\Bbb C)$.

I am not totally sure what your instructor means by "determine the isomorphism explicitly in terms of the basis {...} of $\Bbb C \Bbb Z_5$".  
My best guess it that for this part of the problem, we're meant to interpret each element of $\Bbb C \Bbb Z_5$ as a linear operator over the vector space $\Bbb C \Bbb Z_5$ and obtain the matrix $\Phi(g^k)$ by writing the matrix of that linear operator relative to the given basis.  
If that is the case, then we obtain an isomorphism with a subset of $M_5(\Bbb C)$ which is no longer a standard presentation of $[M_1(\Bbb C)]^5$.  Also, this construction doesn't have anything to do with the Artin-Wederburn theorem on its own.
