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,

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