Factor theorem & noncommutative division ring

Question

enter image description here Show that the factor theorem is false in $R[x]$ if $ R $ is a noncommutative division ring.

I started with $f = bx - ba $ where $ab \neq ba $ What should I do next?

What do you call "the factor theorem" to? – Timbuc Dec 01 '14 at 03:36 — Timbuc, Dec 01 '14 at 03:36
No problem! Do you know how to do the next step? – user196644 Dec 01 '14 at 03:40 — user196644, Dec 01 '14 at 03:40
Do you mean left ideal in $;f\in\langle x-a\rangle;$ ? – Timbuc Dec 01 '14 at 03:48 — Timbuc, Dec 01 '14 at 03:48
See this answer by Arturo Magidin for more discussion. – Jyrki Lahtonen Dec 01 '14 at 07:40 — Jyrki Lahtonen, Dec 01 '14 at 07:40

score 2 · Accepted Answer · answered Dec 01 '14 at 03:48

Your example works fine:

$$f(x)=bx-ba\implies f(a)=ba-ba=0$$

If we had that $\;f(x)=(x-a)q(x)\;$ , then it must be $\;q(x)=q\in R\;$ a constant (why?) , so

$$bx-ba=(x-a)q=xq-aq\stackrel{\text{def. of identical pol's}}\implies\begin{cases}b=q\\{}\\ba=aq\end{cases}\;\implies qa=aq$$

contradicting that $\;ba\neq ab\;$ .

1 Answers1