Wedderburn’s factorization theorem (2)

Posted: October 13, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , , ,

You can see part (1) in here.

Wedderburn’s Factorization Theorem. (Wedderburn, 1920) Let D be a division algebra with the center k. Suppose that a \in D is algebraic over k and let f(x) \in k[x] be the minimal polynomial of a over k. There exist non-zero elements c_1, c_2 , \ldots , c_n \in D such that f(x)=(x - c_1ac_1^{-1})(x - c_2ac_2^{-1}) \ldots (x - c_nac_n^{-1}).

Proof. By Remark 2 in part (1), there exists g(x) \in D[x] such that f(x)=g(x)(x-a). Now let m be the largest integer for which there exist non-zero elements c_1, c_2 , \ldots , c_m \in D and p(x) \in D[x] such that

f(x)=p(x)(x - c_1ac_1^{-1})(x - c_2ac_2^{-1}) \ldots (x - c_m a c_m^{-1}).

Let

h(x) = (x - c_1ac_1^{-1})(x - c_2ac_2^{-1}) \ldots (x - c_mac_m^{-1}).

Claim. h(cac^{-1})=0 for all 0 \neq c \in D.

Proof of the claim. Suppose, to the contrary, that there exists 0 \neq c \in D such that h(cac^{-1}) \neq 0. Then, since f(x)=p(x)h(x) and f(cac^{-1})=0, there exists 0 \neq b \in D such that p(bcac^{-1}b^{-1})=0, by Lemma 1 in part (1). So, by Remark 2 in part (1), there exists q(x) \in D[x] such that p(x)=q(x)(x - bcac^{-1}b^{-1}). Hence f(x)=p(x)h(x)=q(x)(x - bca (bc)^{-1})(x-c_1ac_1^{-1}) \ldots (x - c_m a c_m^{-1}), contradicting the maximality of m. \ \Box

Therefore, by Lemma 2 in part (1), \deg h(x) \geq \deg f(x) and so h(x)=f(x) because f(x)=p(x)h(x) and both f(x) and h(x) are monic. \Box

In the next post, I will use Wedderburn’s factorization theorem to find an expression for the reduced trace and the reduced norm of an element in a finite dimensional central division algebra.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s