Division ring

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Let [itex]R[/itex] be a ring . Suppose that [itex]e[/itex] and [itex]f=1-e[/itex] are two idempotent elements of [itex]R[/itex] and we have [itex]R=eRe \oplus fRf[/itex] (direct sum ) and [itex]R[/itex] doesn't have any non-trivial nilpotent element . Set [itex]R_1=eRe[/itex] and [itex]R_2=fRf[/itex] . If [itex]R_1=\{0,e\}[/itex] and [itex]R_2[/itex] is a local ring , then prove that [itex]R_2[/itex] is a division ring . (note that [itex]e[/itex] and [itex]f[/itex] are central idempotents and therefore [itex]fRf=fR[/itex] )
 
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I.) ##ef=e(1-e)=e-e^2=e-e=0## and ##fe=(1-e)e=e-e^2=e-e=0##

II.) With ##R=\{\,0,e\,\}\oplus fRf## we have for elements ##p=frf## by (I) that ##ep=0=pe## and for elements ##p=e+frf## we get ##ep=e^2+efrf=e^2=e=(e+frf)e=pe##, hence ##e \in R## is central. But ##fr=(1-e)r=r-er=r-re=r(1-e)=rf## so ##f\in R## is also central. Thus ##R=\{\,0,e\,\}\oplus fR##.

III.) If ##f## is no unit in ##R_2## then ##1-f=e \in R_2## is a unit by locality. Now ##e\in R## is also a unit, but ##ef=0##. So ##f=0## and ##e=1##.
If ##f## is a unit, then by ##ef=0## we get ##e=0## and ##f=1##.
Thus the only possibilities are ##(e,f)\in\{\,(1,0)\, , \,(0,1)\,\}##.

IV.) If ##f=0##, then ##R=R_1\oplus R_2=\{\,0,1\,\}\oplus \{\,0\,\} = \mathbb{Z}_2## is a field, and ##R_2=\{\,0\,\}##. Hence we have ##f=1## and ##e=0##, i.e. ##R=R_1\oplus R_2=\{\,0\,\}\oplus R_2=R_2##.
The nilradical of ##R## is zero, so the intersection of all prime ideals of ##R## is zero.
The Jacobson radical ##J(R)##, the intersection of all maximal ideals of ##R##, is zero if ##R## is a division ring. For a local ring ##R## we have that ##R/J(R)## is a division ring. So all comes down to show that ##J(R)=\{\,0\,\}##.

However, as I see it, this needs additional information which we do not have, e.g. ##R## could be left-Artinian.
 

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