division ring

**xixi** · June 2nd 2010, 05:48 AM

Let $R$ be a ring and $Nil(R)=0$ . Suppose that $R=eRe \oplus fRf$ where $e$ and $f$ are idempotent elements and $f=1-e$ . Let $R_1=eRe$ and $R_2=fRf$ . If $R_1 \cong \mathbb{Z}_2$ and $|Max_l(R_2)|=1$ then prove that $R_2$ is a division ring .

**NonCommAlg** · June 2nd 2010, 06:34 PM

Originally Posted by xixi

Let $R$ be a ring and $Nil(R)=0$ . Suppose that $R=eRe \oplus fRf$ where $e$ and $f$ are idempotent elements and $f=1-e$ . Let $R_1=eRe$ and $R_2=fRf$ . If $R_1 \cong \mathbb{Z}_2$ and $|Max_l(R_2)|=1$ then prove that $R_2$ is a division ring .

first you need to clarify two thing for us:

1) for noncommutative rings, there are more than one nilradical. so you need to give us the definition of $Nil(R)$ in here.

2) is $|Max_l(R_2)|=1$ supposed to mean that $R_2$ has only one maximal left ideral? (this is the first time i see this notation!)

for now, just see that $e,f$ are central, i.e. they are in the center of $R$ and thus, since $R_1 \cong \mathbb{Z}_2,$ we have $R_1=\{0,e\}, \ R_2=fRf=Rf.$

**xixi** · June 3rd 2010, 05:19 AM

Originally Posted by NonCommAlg

first you need to clarify two thing for us:

1) for noncommutative rings, there are more than one nilradical. so you need to give us the definition of $Nil(R)$ in here.

2) is $|Max_l(R_2)|=1$ supposed to mean that $R_2$ has only one maximal left ideral? (this is the first time i see this notation!)

for now, just see that $e,f$ are central, i.e. they are in the center of $R$ and thus, since $R_1 \cong \mathbb{Z}_2,$ we have $R_1=\{0,e\}, \ R_2=fRf=Rf.$

1. By $Nil(R)=0$ I meant that R doesn't have any nonzero nilpotent element . Maybe I shouldn't have used this notation .
2. Yes , $|Max_l(R_2)|=1$ means that $R_2$ has only one maximal left ideal .

$e,f$ are central and $R_1=\{0,e\}, \ R_2=fRf=Rf$ . Now how do you prove that $R_2$ is a division ring ?

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