How Do You Solve Group Ring Isomorphisms in Mathematics?

[catcherintherye]

I have about 5 questions all of a similar form ...

\mathbb{F}[C_2] \cong \mathbb{F} \times \mahbb{F}

if 1+1 \neq 0 in \mathbb{F}

\Re [C_3] \cong \Re \times C

\Re [C_4] \cong \Re \times \Re \times C

these were on the first sheet given out and I still don't know how to do them after 4 weeks, any help on how to do them??

 

[HallsofIvy]

"Do them"? Do you mean prove them? Start by writing out the definitions.
(I would have thought that \mathbb{F} was a field but apparently not.)

What are C₂, C₃, C₄?

 

[morphism]

I guess C_n is the cyclic group of order n. But still, what the OP posted doesn't make a lot of sense, e.g. what's the distinction between F and \mathbb{F}? Also, is \Re actually \mathbb{R}? (The former is the "real part" of a complex number, the latter is the set of real numbers.) And, in the last two, what is C? Finally, in what context are the direct products being taken?

 

[catcherintherye]

obviously i haven't made myself clear, i'll post the question in it's entirety...

Let A,B be algebras over a field F. We say that A and B are isomorphic over F written A\cong_F B when there exists a bijective ring homomorphism \varphi : A \rightarrow B which is also linear over F, i.e. satisfies \\\varphi(a\lambda) = \varphi(a) \lambda \mbox{ for all a} \in A \ \lambda \in F

show that i) F[C_2] \cong_F F \times F if 1+1 not equal 0 in F
ii) R[C_3] \cong_R R \times C

iii) \\ R[C_4] \cong_R R \times R \times C

iv) \\ R[C_6] \cong R \times R \times C \times C


here R = field of real numbers, C = field of complex numbers, F is an arbitrary field although in the question it is constrained by the given condition. Also all C_n's are cyclics and by F[C_n] we mean the group ring of C_n over F, that is the ring whose elements are linear combinations of the group elements with coefficients in F.

 

[morphism]

For some reason this problem popped into my head today, so I gave it another go and managed to solve it. There are two things we need:
(1) F[C_n] \cong_F F[x]/(x^n - 1). This can be easily proved using the evaluation homomorphism at a generator of C_n.
(2) The http://planetmath.org/encyclopedia/ChineseRemainderTheorem2.html . (The CRT is something that slipped my mind at the time.)

So now we have:
F[C_2] \cong F[x]/(x^2 - 1) \cong F[x]/(x+1) \times F[x]/(x-1) \cong F \times F

\mathbb{R}[C_3] \cong \mathbb{R}[x]/(x^3 - 1) \cong \mathbb{R}[x]/(x-1) \times \mathbb{R}[x]/(x^2 + x + 1) \cong \mathbb{R} \times \mathbb{C}

\mathbb{R}[C_4] \cong \mathbb{R}[x]/(x^4 - 1) \cong \mathbb{R}[x]/(x-1) \times \mathbb{R}[x]/(x+1) \times \mathbb{R}[x]/(x^2 + 1) \cong \mathbb{R} \times \mathbb{R} \times \mathbb{C}

\mathbb{R}[C_6] \cong \mathbb{R}[x]/(x^6 - 1) \cong \mathbb{R}[x]/(x-1) \times \mathbb{R}[x]/(x+1) \times \mathbb{R}[x]/(x^2+x+1) \times \mathbb{R}[x]/(x^2-x+1) \cong \mathbb{R} \times \mathbb{R} \times \mathbb{C} \times \mathbb{C}

Better late than never I suppose!