Wedderburn’s factorization theorem (1)

Posted: October 12, 2011 in Division Rings, Noncommutative Ring Theory Notes
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Let R be a ring and let R[x] be the ring of polynomials in the central indeterminate x. We will always write the coefficients of an element of R[x] on the left. Now, if R is commutative and f(x)=g(x)h(x) in R[x], then f(a)=g(a)h(a) for all a \in R. This is not true if R is noncommutative. For example, let g(x)=x-a and h(x)=x - b, where a,b \in R. Then f(x)=g(x)h(x)=x^2-(a+b)x+ab and hence f(a)=ab-ba. Thus if ab \neq ba, then f(a) \neq 0 but g(a)h(a) = (a-a)(a-b)=0.

Remark 1. Let R be a ring. If f(x)=g(x)h(x) in R[x] and if every coefficient of h(x) is in the center of R, then f(a)=g(a)h(a) for all a \in R.

Proof. Let g(x)=\sum_{i=0}^n b_ix^i and h(x)=\sum_{j=0}^m c_jx^j. Then, since a commutes with every c_j, we have

f(a)=\sum_{\ell=0}^{m+n}\sum_{i+j=\ell}b_ic_ja^{\ell}=\sum_{\ell=0}^{m+n}\sum_{i+j=\ell}b_ia^ic_ja^j=g(a)h(a).

Remark 2. Let R be a ring and f(x) \in R[x]. Then f(a)=0 if and only if f(x)=g(x)(x-a) for some g(x) \in R[x].

Proof. Let f(x)=\sum_{i=0}^n a_ix^i. If f(a)=0, then since x-a commutes with a, we have

f(x)=\sum_{i=0}^na_i(x-a +a)^i = \sum_{i=0}^n a_i \sum_{j=0}^i \binom{i}{j}a^{i-j}(x-a)^j =

g(x)(x-a) + f(a)=g(x)(x-a).

Conversely, if f(x)=g(x)(x-a) for some g(x) = \sum_{i=0}^m b_ix^i, then

f(x)=\sum_{i=0}^mb_ix^{i+1} - \sum_{i=0}^m b_iax^i

and thus f(a)=\sum_{i=0}^mb_ia^{i+1} - \sum_{i=0}^m b_ia^{i+1}=0. \ \Box

Lemma 1. Let D be a division algebra with the center k. Suppose that f(x)=g(x)h(x) in D[x]. If a \in D and b = h(a) \neq 0, then f(a) = g(bab^{-1})b. Thus if f(a)=0, then g(bab^{-1})=0.

Proof. Let g(x)=\sum_{i=0}^n b_ix^i and h(x)=\sum_{j=0}^m c_jx^j. Then

f(a)=\sum_{i,j} b_i c_j a^{i+j} = \sum_{i=0}^n b_i \left (\sum_{j=0}^m c_ja^j \right)a^i=\sum_{i=0}^n b_i h(a)a^i = \sum_{i=0}^n b_iba^i =

\sum_{i=0}^n b_i(bab^{-1})^ib=g(bab^{-1})b. \ \Box

Lemma 2. Let D be a division algebra with the center k. Suppose that a \in D is algebraic over k and f(x) \in k[x] is the minimal polynomial of a over k. If g(x) \in D[x] is non-zero and g(cac^{-1}) = 0, for all 0 \neq c \in D, then \deg g(x) \geq \deg f(x).

Proof. Let \deg f(x) = n and suppose that the lemma is false. Let m \geq 1 be the smallest integer for which there exists a non-zero polynomial g(x) \in D[x] of degree m such that m < n and g(cac^{-1})=0 for all 0 \neq c \in D. Note that since for every 0 \neq u \in D, the polynomial ug(x) has degree m and ug(cac^{-1})=0, for all 0 \neq c \in D, we may assume that g(x) is monic. Let g(x)=x^m + b_{m-1}x^{m-1} + \ldots + b_1x + b_0. To get a contradiction, we are going to find a non-zero polynomial h(x) \in D[x] such that \deg h(x) < m and h(cac^{-1})=0 for all 0 \neq c \in D. We first note that, since f(x) is the minimal polynomial of a over k, \ m < n and g(a)=0, we have g(x) \notin k[x]. So there exists 0 \leq j \leq m-1 such that b_j \notin k. Let b \in D be such that b_jb - bb_j \neq 0. So b \neq 0. Now we let

h(x) = g(x)b - bg(x).

Then h(x)= \sum_{i=0}^{m-1}(b_ib - bb_i)x^i. Hence \deg h(x) < m and h(x) \neq 0, because b_jb - bb_j \neq 0. Let 0 \neq c \in D. Then, by Lemma 1,

h(cac^{-1}) = g(bcac^{-1}b^{-1})b - bg(cac^{-1}) = g(bca(bc)^{-1})b = 0. \ \Box

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