22

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 h_{a }∈ T [x]. of a such that h_{a }= (x

# Then, h_{a }splits completely in E. Furthermore, there exist n conjugates a_{1}, . . . , a_{n }

a_{n})(x

a_{n }

_{1}) . . . (x

a_{1}) 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

a^{0 }of a.

I:

Let

f

∈ E[x] with

f (a)

=

g(a^{0})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 b_{i}k(a)a^{i}. 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})^{i}te = g(eae ^{1})k(a). b Thus, f(a) = b_{i}(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 x^{i }

r^{i }= (x^{i }

1

+

x

^{i }

^{2}r + . . . + r_{i }_{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(h_{a}).

P r o o f . C o n s i d e r f = x m + d 1 x m + . . . + d_{m }∈ E[x] such that f(A) = 0 and m < deg(h_{a}) 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_{mγ }-component a n d s o m e e ∈ E ∗ s u c h ∈ E, write c^{0 }:= 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 x^{m }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 )

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

hence, m ≥ deg(h_{a}).

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

a). Take a factorization

h_{a }= g · (x

a_{r}) . . . (x

a_{1}) ,

where g ∈ E[x], a_{1}, . . . , a_{r }∈ A and r is as large as possible. Let k = (x

a_{r}) . . . (x

a_{1}) ∈ E[x]. We claim

that

k(A) = 0,

where

A

is

the

conjugacy

class

of

a.

# For,

suppose

there

exists

b∈A

such

that

k ( b ) =6

0.

Since k(b) is homogenous, we have k(b) ∈ E^{∗}. (b) = 0, (I) implies as b ∈ A; But, h_{a }= 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 b^{0}), by (II). So h_{a }has a right

factor (x

b^{0})k = (x

b^{0})(x

a_{r}) . . . (x

a_{1}), contradicting our choice of r.

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

we have r ≥ deg(h_{a}), which says that h_{a }= (x

a_{r}) . . . (x

a_{1}).