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

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s