## Quotient rings; some facts (2)

Posted: January 2, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
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For the first part see here.

6) If $R$ is simple, then $Z(R)=Z(Q).$

Proof. Let $x=s^{-1}a \in Z(Q).$ Then from $xs=sx$ we get $sa=as$ and thus $s^{-1}a=as^{-1}.$ Hence for every $b \in R$ we’ll have $s^{-1}ab=bs^{-1}a=bas^{-1},$ which gives us $abs=sba.$ Also, since $R$ is simple, $RsR=R,$ which means $\sum_{i=1}^n b_isc_i = 1,$ for some $b_i, \ c_i \in R.$ Thus $\sum_{i=1}^n sb_iac_i = \sum_{i=1}^n ab_isc_i = a=sx$ and so $x=\sum_{i=1}^n b_iac_i \in R.$ Therefore $x \in Z(R)$ which proves $Z(Q) \subseteq Z(R).$ Conversely, let $b \in Z(R)$ and $x=s^{-1}a \in Q$. Since $bs=sb,$ we have $s^{-1}b=bs^{-1}$ and thus $bx=bs^{-1}a=s^{-1}ba=s^{-1}ab=xb$ and so $b \in Z(Q). \Box$

7) The left uniform dimension of $R$ and $Q$ are equal.

Proof. We saw in the previous section that the left ideals of $Q$ are exactly in the form $QI,$ where $I$ is a left ideal of $R.$ Clearly $\sum QI_i$ is direct iff $\sum I_i$ is direct.

8) Let $N$ be a nilpotent ideal of $R$ and let $I$ be the right annihilator of $N$ in $R.$ Then $I$ is an essential left ideal of $R$ and hence $QI$ is an essential left ideal of $Q.$

Proof. Let $I$ be the right annihilator of $N$ in $R.$ For an essential left ideal $J$ of $R$ the left ideal $QJ$ of $Q$ is essential in $Q$ because for every non-zero left ideal $K$ of $R : (0) \neq \ Q(J \cap K) \subseteq QJ \cap QK.$ So we only need to prove the first part of the claim. Let $J$ be any non-zero left ideal of $R$ and put $n=\min \{k \geq 0 : \ N^k J \neq (0) \}.$ Then $(0) \neq N^n J \subseteq I \cap J.$

9) If $Q$ is semisimple, then $R$ is semiprime.

Proof. So we need to prove that $R$ has no non-zero nilpotent ideal. Suppose that $N$ is a nilpotent ideal of $R$ and let $I$ be the right annihilator of $N$ in $R.$ Since $Q$ is semisimple, $QI \oplus A = Q,$ for some left ideal $A$ of $Q.$ But, from the previous fact, we know that $QI$ is essential in $Q$ and thus $A=(0),$ i.e. $QI=Q.$ Thus $s^{-1}a=1,$ for some $a \in I=\text{r.ann}_R N.$ So $s=a$ and $Ns=Na=(0).$ Thus $N=Nss^{-1}=(0).$

We proved, in the previous section, that if $R$ is prime, then $Q$ is prime too.

10) If $Q$ is simple, then $R$ is prime.

Proof. Let $I,J$ be two non-zero ideals of $R.$ We need to show that $IJ \neq (0).$ We have $QIQ=Q,$ because $I \neq (0)$ and $Q$ is simple. Therefore $1=\sum_{i=1}^n x_ia_iy_i,$ for some $x_i,y_i \in Q$ and $a_i \in I.$ We can write $x_i = s^{-1}b_i,$ for some $b_i \in R.$ Then $s=\sum_{i=1}^n b_ia_iy_i \in IQ.$ So $IQ$ is a right ideal of $Q$ which contains a unit. Thus $IQ=Q.$ Similarly $JQ=Q$ and hence $IJQ=Q.$ As a result, $IJ \neq (0). \Box$