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R. HAZRAT AND A. R. WADSWORTH

the Wedderburn factorization theorem. We now carry over this theorem to the setting of graded division algebra. (This is used in proving Th. 3.3).

# Theorem A.1 (Wedderburn Factorization Theorem). Let E be a graded division ring with center T (with

## E

torsion-free abelian). Let a be a homogenous element of E which is algebraic over T with minimal

polynomial ha T [x]. of a such that ha = (x

# Then, ha splits completely in E. Furthermore, there exist n conjugates a1, . . . , an

an)(x

an

1) . . . (x

a1) in E[x].

Proof. The proof is similar to Wedderburn’s original proof for a division ring ([We], see also [L] for a nice a c c o u n t o f t h e p r o o f ) . W e s k e t c h t h e p r o o f f o r t h e c o n v e n i e n c e o f t h e r e a d e r . F o r f = P c i x i E [ x ] a n d a E , o u r c o n v e n t i o n i s t h a t f ( a ) m e a n s P c i a i . S i n c e E i s t o r s i o n - f r e e , w e h a v e E = E h \ { 0 } .

factorization f

= gk

some conjugate

a0 of a.

I:

Let

f

E[x] with

f (a)

=

g(a0)k(a), for

then

E satisfies k(a) T · E, any ring with T Z(E).)

in E[x].

If a

(Here E

could

be

P i b i x i . T h e n , f = P P bik(a)ai. But, k(a) = te, where t T and e E. b i k x Proof. Let g = P i , s o f ( a ) = P P 1 1)ite = g(eae 1)k(a). b Thus, f(a) = bi(eae i t e a b i e a te = = i e

# II: Let f ∈ E[x] be a non-zero polynomial. Then r ∈ E is a root of f if and only if x

r is a right

divisor of f in E[x]. (Here, E could be any ring.)

Proof. We have xi

ri = (xi

1

• +

xi

2r + . . . + ri 1

)(x

r) for any i 1. Hence,

f

f(r) = g · (x

r)

(A.1)

for some g E[x]. So, if f(r) = 0, then f = g · (x

r). Conversely, if x

r is a right divisor of f, then

equation (A.1) shows that x that f(r) = 0.

r is a right divisor of the constant f(r). Since x

r is monic, this implies

III: If a non-zero monic polynomial f E[x] vanishes identically on the conjugacy class A of a (i.e., f(b) = 0 for all b A), then deg(f) deg(ha).

P r o o f . C o n s i d e r f = x m + d 1 x m + . . . + dm E[x] such that f(A) = 0 and m < deg(ha) with m as small 1 a s p o s s i b l e . S u p p o s e a E γ , s o A E γ , a s t h e u n i t s o f w e m a y a s s u m e t h a t e a c h d i E i γ T [ x ] , s o m e d i of f(b) is 0 for each b A, T . m i n i m a l w i t h d j T , E are all homogeneous. Since the E-component a n d s o m e e E s u c h E, write c0 := ece Because f For any c Choose j 1. Thus t h =6 a t e d j . e. d j m1 d 0 j 6 = d j d 0 ` d E[x]. Now, for all b A, we have ` f 0 for ` d 0 but 1 + . . . + d 0 m xm Let j. + < x = = f 0 ( 1 b 0 ) = [ f ( b ) ] 0 = 0 0 = 0 . S i n c e e A e = A , t h i s s h o w s t h a t f 0 ( A ) = 0 . L e t g = f f 0 , w h i c h h a s

d e g r e e j < m w i t h l e a d i n g c o e ffi c i e n t d j

d 0 j . T h e n , g ( A ) = 0 . B u t , d j

d 0 j E j γ

\ {0} ⊆ E. Thus,

( d j

d 0 j )

1g is monic of degree j < m in E[x], and it vanishes on A. This contradicts the choice of f;

hence, m deg(ha).

We now prove the theorem. Since ha(a) = 0, by (II), ha E[x] · (x

a). Take a factorization

ha = g · (x

ar) . . . (x

a1) ,

where g E[x], a1, . . . , ar A and r is as large as possible. Let k = (x

ar) . . . (x

a1) E[x]. We claim

that

k(A) = 0,

where

A

is

the

conjugacy

class

of

a.

# For,

suppose

there

exists

bA

such

that

k ( b ) =6

0.

Since k(b) is homogenous, we have k(b) E. (b) = 0, (I) implies as b A; But, ha = gk, and h hence, a t h a t g ( b 0 ) = 0 f o r s o m e c o n j u g a t e b 0 o f b . W e c a n t h e n w r i t e g = g 1 · ( x b0), by (II). So ha has a right

factor (x

b0)k = (x

b0)(x

ar) . . . (x

a1), contradicting our choice of r.

Thus k(A) = 0, and using (III),

we have r deg(ha), which says that ha = (x

ar) . . . (x

a1).

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