If $G$ is any finite abelian group and $K$ an algebraically closed field with $|G|\neq 0$ in $K$, then the group algebra $K[G]\cong M_{n_1}(K)\times\cdots\times M_{n_k}(K)$ by Maschke's and Artin-Wedderburn's theorem. Since $K[G]$ is commutative, we have $n_1=\cdots=n_k=1$ and $K[G]\cong\prod_{g\in G}K$.

Thus only the cardinality of $G$ matters. For example, if $C_n$ is the cyclic group of order $n$, then $\mathbb{C}[C_m\times C_n]\cong\mathbb{C}[C_{mn}]$ even though $C_m\times C_n\ncong C_{mn}$ when $\gcd(m,n)\neq1$. This means $\mathbb{C}[C_m\times C_n]\cong \mathbb{C}[C_m]\otimes\mathbb{C}[C_n]\cong \mathbb{C}[x\,|\,x^m-1]\otimes\mathbb{C}[y\,|\,y^n-1]\cong\mathbb{C}[x,y\,|\,x^m-1,y^n-1]$ is isomorphic to $\mathbb{C}[C_{mn}]\cong\mathbb{C}[z\,|\,z^{mn}-1]$. I'd like a concrete isomorphism for this.

I'm guessing $x\mapsto z^n$ and $y\mapsto z^m$, but how should I send $z\mapsto f(x,y)$?


First, it is worth considering how $\mathbb C[x]/\left<x^k-1\right>\cong \mathbb C^k$.

Namely, the isomorphism sends:

$$f(x)\to \left(f(1),f(\zeta_k),f(\zeta_k^2),\dots,f(\zeta_k^{k-1})\right)$$

where $\zeta_k$ is a primitive $k$th root of unity.

Now, $$\mathbb C[x,y]/\langle x^n-1,y^m-1\rangle\cong \mathbb C[x]/\langle x^n-1\rangle\otimes \mathbb C[y]/\left<y^m-1\right>$$

So the isomorphism to $C^{nm}$ has has $\langle i,j\rangle$th component equal to $f(\zeta_n^i,\zeta_m^j)$.

The inverse of these functions involves a messy Chinese Remainder Theorem stuff (or linear algebra,) and is heavily dependent on the order of the list of components. There is no obviously primary isomorphism.

When $m,n$ relatively prime, there is an obvious way to map the values $0,1,\dots,mn-1$ to the pairs $\langle i,j\rangle$ using Chinese Remainder Theorem. But there are lots of other ring isomorphisms even in this case.

If you look at the case $m=n=2$, you get that $z\to \frac{1+i}{2}\left(x-ixy\right)$ is an isomorphism, for example. That's hardly obvious.

Another approach is to find a set of orthogonal idempotents for each ring.

The idempotents for $\mathbb C[x]/\langle x^k-1\rangle$ are defined in terms of $P(X)=1+X+\dots+X^{k-1}$ and $\zeta_k$ a primitive $k$th root of unity. Then we get the following idempotents:

$$\frac{1}{k}P(\zeta_k^i x)$$ for $i=0,1,\dots,k-1$.

This makes the idempotents for $\mathbb C[x^2=1,y^2=1]$ to be $\frac{1}{4}(1\pm x)(1\pm y)$.

The idempotents for $\mathbb C[z^4=1]$ are $$\frac{1}{4}(1+z+z^2+z^3)\\\frac14(1+iz-z^2-iz^3)\\\frac14(1-z+z^2-z^3)\\\frac14(1-iz-z^2+iz^3)$$

In this latter ring, if you write $z$ as the linear combination of these idempotents, and you send the idempotents to any shuffling of the idempotents in the former ring, you can figure out where $z$ goes.

  • $\begingroup$ Strange, how elusive general explicit formulas are. Thank you! $\endgroup$ – Leon Jan 24 '14 at 4:23
  • $\begingroup$ Does this map work when $K$ is any alg. closed field with $|G|\neq0$ and $\zeta$ is any element of $K$ with $\zeta^k\!=\!1$ and $\zeta^i\!\neq\!1$ for $i\!<\!k$? Does such $\zeta$ always exist? $\endgroup$ – Leon Jan 24 '14 at 5:02
  • $\begingroup$ Such $\zeta$ exists, because $K$ is algebraically closed, and you can choose a root to the cyclotomic polynomial for $k$. $\endgroup$ – Thomas Andrews Jan 24 '14 at 5:10
  • $\begingroup$ Yes, but why is such a root a simple root of unity? And do we need a simple root of unity. If an inverse is so difficult to construct, how do you know it is bijective? $\endgroup$ – Leon Jan 24 '14 at 5:19
  • 1
    $\begingroup$ You can prove relatively easily that no root of $x^k-1$ is repeated, and the cyclotomic polynomials are the polynomials with the roots for $x^d-1$ with $d<k$ factored out. As for the latter, we know there is an inverse by linear algebra, or simply by Artin-Wedderburn. $\endgroup$ – Thomas Andrews Jan 24 '14 at 5:40

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