Archive for the ‘Quotient Rings’ Category

Ore domains (2)

Posted: January 4, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
Tags: , , , , ,
0

Here you can see part (1). In the following, Z \langle x,y \rangle is the algebra generated by non-commuting variables x,y and with coefficients in Z. So an element of  Z \langle x,y \rangle is a finite sum of terms in the form \alpha x_1x_2 \ldots x_n, where n is any positive integer, \alpha \in Z and each x_i is either 1 or x or y. For example f = \alpha_0 + \alpha_1 xyx + \alpha_2 yx^3y^2 where \alpha_i \in Z.

Lemma. Let Z be the center of a domain R. If R is not left (resp. right) Ore, then R contains a copy of Z \langle x,y \rangle.

Proof. Choose 0 \neq r_1,r_2 \in R such that Rr_1 \cap Rr_2 = (0). Then r_1,r_2 are left linearly independent over R and we’re done by Jategaonkar’s lemma. \Box

Corollary. Let A be an algebra over a field. If A is PI or has a finite GK-dimension, then A is both left and right Ore.

Proof. Otherwise, by the lemma, A would contain a free algebra and we know that such algebras are neither PI nor have a finite GK dimension. \Box

A left Ore domain which is not right Ore. Let D be a division ring and \sigma : D \longrightarrow D be a homomorphism which is not onto.  Since \sigma (1_D)=1_D \neq 0, we have \ker \sigma \neq D and thus \ker \sigma = (0), because D has no non-zero proper ideal. Therefore \sigma is injective and hence the left skew polynomial ring R=D[x,\sigma] is a domain.

Claim 1. R is not right Ore.

Proof. Suppose R is right Ore. Since we’ve assumed that \sigma is not onto, there exists d \in D - \sigma (D). Now we must have xR \cap dxR \neq (0). So, there exist non-zero elements g, h \in R such that xg=dxh. Clearly g, h must have the same degree, say n. Let d_1x^n, \ d_2x^n be the leading terms of g, h respectively. So the leading terms of xg , dxh, which have to be equal, are  \sigma (d_1)x^{n+1}, \ d \sigma (d_2) x^{n+1} respectively. Equating these leading terms gives us d=\sigma (d_1 d_2^{-1}) \in \sigma (D), which is a contradiction. \Box

Claim 2. Euclidean algorithm holds in R.

Proof. That means we need to prove for every f, g \in R, there exists r, s \in R such that f=sg + r, where either r = 0 or \deg r < \deg g. The proof is by induction over \deg f: if \deg f < \deg g, then we may choose s = 0 and r = f. Let d_1 x^n, \ d_2 x^m, \ n \geq m, be the leading terms of f, g respectively. Then f and d_1x^{n-m}d_2^{-1}g have the same leading terms and thus h = f - d_1x^{n-m}d_2^{-1}g has a degree < n. Thus, by induction, there exist r_1, s_1 \in R such that h = s_1g + r_1, where either r_1 = 0 or \deg r_1 < \deg g. Finally we have h=s_1g + r_1 = f - d_1 x^{n-m}d_2^{-1}g and so f = (s_1 + d_1x^{n-m}d_2^{-1})g + r_1. \ \Box

Claim 3. R is left Ore.

Proof. Let I be a non-zero left ideal of R and choose a non-zero g \in I with the smallest degree. For every f \in I, by Claim 2, there exist r, s \in R such that f = sg + r, where either r = 0 or \deg r < \deg g. But r = f - sg \in I and thus we cannot have \deg r < \deg g. Therefore r=0 and f \in Rg, i.e. I=Rg. So every left ideal of R is principal and, as a result, R is left Noetherian and therefore left Ore. \Box

Ore domains (1)

Posted: January 3, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
Tags: , , ,
1

Definition. Let R be a domain and S=R \setminus \{0\}. If S is left (resp. right) Ore, then R is called a left (resp. right) Ore domain.

Remark 1. Note that a domain R is a left Ore domain if and only if Rr_1 \cap Rr_2 \neq (0), for all non-zero elements r_1,r_2 \in R. Similarly R is a right Ore domain if and only if r_1R \cap r_2R \neq (0), for all non-zero elements r_1,r_2 \in R.

Remark 2. The left (resp. right) quotient ring of a left (resp. right) Ore domain is a division ring. The reason is that every element of Q(R) is in the form x = s^{-1}r, where 0 \neq s \in R, \ r \in R. Now if x \neq 0, then r \neq 0 and then x^{-1}=r^{-1}s.

Theorem. Every left (resp. right) Noetherian domain is left (resp. right) Ore.

Proof. We will prove the theorem for left Noetherian domains only. The “right” version, as usual, has the same argument. Let r_1, r_2 \in R be non-zero. We need to show that Rr_1 \cap Rr_2 \neq (0). So suppose, on the contrary, that Rr_1 \cap Rr_2 = (0). We will show that the sum \sum_{n=0}^{\infty} R r_1r_2^n is direct and thus R cannot be left Noetherian because then we would have the non-stopping increasing chain of left ideals

Rr_1 \subset Rr_1 \oplus Rr_1r_2 \subset Rr_1 \oplus Rr_1r_2 \oplus Rr_1r_2^2 \subset \cdots.

To prove that the sum is direct, suppose that the sum is not direct and choose n to be the smallest postive integer for which there exist a_j \in R, \ j=0, \cdots , n, not all zero, such that \sum_{j=0}^n a_jr_1r_2^j = 0. Clearly n \geq 1. Thus

-a_0r_1=\left (\sum_{j=0}^{n-1} a_{j+1}r_1r_2^{j} \right)r_2 \in Rr_1 \cap Rr_2 = (0).

Therefore, since r_1, r_2 are non-zero and R is a domain, we must have a_0=0 and \sum_{j=0}^{n-1} a_{j+1}r_1r_2^j = 0, contradicting the minimality of n. \ \Box

Quotient rings; some facts (2)

Posted: January 2, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
Tags: , , , , , ,
0

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

Quotient rings; some facts (1)

Posted: January 1, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
Tags: , , , ,
0

Throughout R is a ring with unity, S \subset R is a regular submonoid which is left Ore. So Q, the left quotient ring of R with respect to S, exists and R is a subring of Q. The following facts also holds for the “right” version:

 1) If I \lhd_{\ell} Q, then I \cap R \lhd_{\ell} R and if J \lhd_{\ell} Q, then I \cap R = J \cap R \Longleftrightarrow I=J. Also I=Q(I \cap R).

Proof. The first part is trivial. Now suppose that s^{-1}a=x \in I. Then a = sx \in I \cap R = J \cap R and so x=s^{-1}a \in J. Hence I \subseteq J. Similarly J \subseteq I. For the last part, if x = s^{-1}a \in I, then a=sx \in I \cap R and so x=s^{-1}a \in Q(I \cap R).

2) If I \lhd_{\ell} R, then QI \lhd_{\ell} Q and QI=\{s^{-1}a : \ s \in S, \ a \in I \}. It is also true that QI \cap R = \{a \in R : \ sa \in I \ \text{for some} \ s \in S \}.

Proof.  Again, the first part is trivial. To prove the non-trivial side of the second part, let x = \sum_{i=1}^n x_i a_i \in QI, where x_i \in Q and a_i \in I. As we’ve showed before, there exist some s \in S and b_i \in R such that x_i = s^{-1}b_i  and thus x = s^{-1}a, where a=\sum_{i=1}^n b_ia_i \in I. For the last part, suppose that s^{-1}b \in QI \cap R, where s \in S, \ b \in I, by the second part. So we have s^{-1}b=a \in R and thus sa = b \in I.

3) For any I, J \lhd_{\ell} R we have I \cap J = (0) \Longrightarrow QI \cap QJ = (0).

Proof. Let x=s^{-1}a = t^{-1}b \in QI \cap QJ, where a \in I, \ b \in J. We have ts^{-1} \in Q and thus ts^{-1}=u^{-1}c, for some u \in S, \ c \in R. Then, since ts^{-1}a=b, we get ca=ub \in I \cap J = (0), which gives us b=0 because u is a unit in Q (or because u is regular in R). Thus x = 0.

4) If R is left Noetherian, then so is Q. If I \lhd R and Q is left Noetherian, then QI \lhd Q.

Proof. For the first part, first note that, by 1) and 2), left ideals of Q are exactly in the form QI, where I is a left ideal of R. Let I \lhd_{\ell} R and I=\sum_{i=1}^n Ra_i. Then QI=\sum_{i=1}^n Qa_i. For the second part, let I be a two-sided ideal of R and s \in S. Then I \subseteq Is^{-1} \subseteq Is^{-2} \subseteq \cdots and thus QI \subseteq QIs^{-1} \subseteq QIs^{-2} \subseteq \cdots. So, since Q is left Noetherian, there exists some n \geq 0 such that QIs^{-n} = QIs^{-n-1}, and thus QI=QIs^{-1}. Therefore QIs^{-1}a = QIa \subseteq QI, for all s \in S, \ a \in R. Hence QIQ \subseteq QI and so QI is a two-sided ideal of Q. (In fact QIQ=QI.)

5) If R is prime, then so is Q.

Proof. Suppose that x_1Qx_2=(0), for some x_1=s_1^{-1}a_1, x_2=s_2^{-1}a_2 \in Q. Then s_1^{-1}a_1(as_2)s_2^{-1}a_2 = 0, for all a \in R. Thus s_1^{-1}a_1aa_2 = 0, which gives us a_1aa_2=0, i.e. a_1Ra_2=(0). Therefore either a_1=0 or a_2=0, which means either x_1=0 or x_2=0. This also proves that if R is semiprime, then so is Q.

Regular submonoids

Posted: December 28, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
Tags: , , , , ,
0

Throughout S is a regular submonoid of the ring R, i.e.S is multiplicatively closed, 1_R \in S and every element of S is regular.

Remarks. 1) S^{-1}R (resp. RS^{-1}) is defined if and only if S is left (resp. right) Ore. In this case, since every element of S is regular, the corresponding map \varphi : R \longrightarrow S^{-1}R (resp. \varphi : R \longrightarrow RS^{-1}) defined by  \varphi (r) = 1_R^{-1}r (resp. \varphi (r) = r1_R^{-1}) is one-to-one and hence S^{-1}R (resp. RS^{-1}) contains a copy of R.

2) If every element of S is a unit, then Rs = R, for all s \in S and thus Rs \cap Sr = Sr \neq \emptyset, for all s \in S, \ r \in R. So, S is left (and right) Ore and therefore S^{-1}R (and RS^{-1}) both exist and are equal to R. The reason that they are equal to R is that s^{-1} \in R, for all s \in S and so s^{-1}r \in R, for all s \in S, \ r \in R.

3) Let R be a left Artinian ring and s a regular element of R. Then the chain Rs \supseteq Rs^2 \supseteq \cdots must stop, i.e. there exists n \geq 1 such that Rs^n = Rs^{n+1}. Thus s^n = rs^{n+1}, for some r \in R. So (1-rs)s^n = 0, which gives rs = 1. Obviously the same result holds for right Artinian rings.

4) If S contains all the regular elements of R and it is left (resp. right) Ore, then the left (resp. right) quotient ring of R is denoted by Q(R).

5) If S is central, then it’s clearly both left and right Ore and thus S^{-1}R \cong RS^{-1}.

Lemma. Suppose that S is central and A=\{ s^{-1}r_1, \cdots , s^{-1}r_n \} \subset S^{-1}R. Let B=\{r_1, \cdots , r_n \}. Then A^m = (s^{-1})^m B^m, for all m \in \mathbb{N}.

Proof. Since S is central, we have s^{-1}x = xs^{-1} for all s \in S, \ x \in R. Therefore

s^m(s^{-1}x_1 s^{-1}x_2 \cdots s^{-1}x_m)=x_1x_2 \cdots x_m,

for all x_j \in R. \ \Box

We now use the above lemma to prove that the GK-dimension of an algebra is invariant under localization with respect to a central submonoid.

Theorem. If R is an algebra over a field F and S is central, then \text{GK}(S^{-1}R) = \text{GK}(R).

Proof. Let W=\sum_{j=1}^n Fq_j, where q_j \in S^{-1}R. As we’ve seen before, there exit s \in S, \ r_j \in R such that q_j = s^{-1}r_j, for all j. Let V=\sum_{j=1}^n F r_j. Then, by the lemma, s^m W^m = V^m, for all integers m \geq 0 (note that for m = 0 both sides are equal to F). Therefore W^m = (s^{-1})^m V^m and hence

\dim_F W^m = \dim_F (s^{-1})^m V^m \leq \dim_F V^m.

Thus \text{GK}(S^{-1}A) \leq \text{GK}(A). The other direction of the inequality is trivial because A \subseteq S^{-1}A. \ \Box

Existence of quotient rings

Posted: December 27, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
Tags: , , , ,
0

We will assume throughout that R is a ring with unity and S \neq \emptyset a multiplicatively closed subset of R with 0 \notin S and 1_R \in S.

Definition 1. S is called a left (resp. right) denominator set if:

1) S is left (resp. right) Ore;

2) for r \in R and s \in S with rs = 0, there exists s' \in S such that s'r=0  (resp. for r \in R and s \in S with sr = 0, there exists s' \in S such that rs'=0).

Theorem. The left (resp. right) quotient ring of R with respect to S exists if and only if S is a left (resp. right) denominator set.

Proof. (Sketch) We’ve already proved that if the left (resp. right) quotient ring of R with respect to S exists, then S is a left (resp. right) denominator set. For the converse, assuming that S is a left denominator (the “right” version is the same), we’ll construct the left quotient ring of R by first defining a relation on S \times R: we say (s_1,r_1) \sim (s_2,r_2) if and only if there exist r_1',r_2' \in R such that r_1'r_1=r_2'r_2 and r_1's_1=r_2' s_2 \in S.

Claim. The relation \sim is an equivalence relation.

Proof. The reflexive property is proved by choosing r_1'=r_2'=1_R. Obviously \sim is symmetric. In order to prove that \sim is transitive suppose that (s_1,r_1) \sim (s_2,r_2) \sim (s_3,r_3). So there exist r_1',r_2'  such that r_1'r_1=r_2'r_2 and r_1's_1 = r_2's_2 \in S. There also exist r_2'',r_3'' \in R such that r_2''r_2=r_3''r_3 and r_2''s_2=r_3''s_3 \in S. On the other hand, since S is left Ore, we have Rr_2's_2 \cap S r_2''s_2 \neq\emptyset and thus there exist r \in R and s \in S such that rr_2's_2 = sr_2''s_2. So (rr_2' - sr_2'')s_2=0 and hence there exists t \in S such that t(rr_2' - sr_2'')=0, i.e. trr_2' = tsr_2''. Finally if we put x=trr_1', \ y=tsr_3'', then xr_1=yr_3, \ xs_1=ys_3 \in S, i.e. (r_1,s_1) \sim (r_3,s_3). \ \Box

Next, we will show the equivalence class of the element (s,r) \in S \times R by s^{-1}r and we let Q be the set of all s^{-1}r. We are going to put a ring structure on Q. Let \alpha = s_1^{-1}r_1, \ \beta = s_2^{-1}r_2 be two elements of Q. By left Ore condition, Rs_1 \cap Ss_2 \neq \emptyset and thus there exist r \in R and s \in S such that rs_1=ss_2=t \in S. Now define \alpha + \beta = t^{-1}(rr_1 + sr_2). Also, since Rs_2 \cap Sr_1 \neq \emptyset, there exist r' \in R and s' \in S such that r's_2=s'r_1. Let s's_1=t' \in S and define \alpha \beta = t'^{-1}r'r_2. It is proved that the addition and multiplication that we’ve defined are well-defined and satisfy all the conditions needed to make Q a ring. Let 1=1_R. It’s easy to see that 1^{-1}0 = 0_Q and 1^{-1}1 = 1_Q. Finally define \varphi : R \longrightarrow Q by \varphi (r)=1^{-1}r. See that \varphi is a ring homomorphism. Now r \in \ker \varphi  if and only if (1,r) \sim (1,0) if and only if there exist r_1,r_2 \in R such that r_1r=0 and r_1=r_2 = s \in S. Thus

\ker \varphi = \{r \in R : \ sr = 0, \text{for some} \ s \in S \}.

Therefore Q=S^{-1}R is the left quotient ring of R with respect to S. \ \Box

Remark. Every finite subset of S^{-1}R can be written as \{s^{-1}x_1, \cdots , s^{-1}x_n \}. This is easy to see: let \{s_1^{-1}y_1, \cdots s_n^{-1}y_n \} \subset S^{-1}R. By Remark 3 in this post, there exist r_1, \cdots , r_n \in R such that r_1s_1 = \cdots = r_ns_n = s \in S. Let x_j = r_jy_j, for every 1 \leq j \leq n, and see that (s_j , y_j) \sim (s , x_j).  So s_j^{-1}y_j = s^{-1}x_j, for all j.

Ore condition

Posted: December 26, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
Tags: , , ,
0

In this post, R is a unitary ring, S \subset R is multiplicatively closed with 0 \notin S and 1_R \in S.

Definition. S is called a left Ore set if Rs \cap Sr \neq \emptyset, for all r \in R, \ s \in S. Similarly S is called a right Ore set if sR \cap rS \neq \emptyset, for all r \in R, \ s \in S.

Remark 1. If Q is the left (resp., right) quotient ring of R with respect to S, then  S is a left (resp., right) Ore set. The reason is that if \varphi : R \longrightarrow Q is the corresponding map, then for any r \in R, \ s \in S we must have \varphi (r) (\varphi (s))^{-1}=(\varphi (s'))^{-1} \varphi (r'), for some r' \in R, \ s' \in S. Thus \varphi (s'r)=\varphi (r's) and hence s'r - rs \in \ker \varphi. So s''(s'r - r's)=0, for some s'' \in S. Hence s''s'r=s''r's \in Rs \cap Sr.

Remark 2. If S is a left Ore set, then I=\{r \in R : \ sr = 0, \ \text{for some} \ s \in S \} is an ideal of R. (Recall that if R has the left quotient ring Q, then I is nothing but the kernel of the corresponding map from R to Q). To see this, we note that I is clearly closed under right multiplication by elements of R. also:

i) for every r,r' \in I: \ r+r' \in I. To prove this, we have sr=s'r'=0, for some s,s' \in S. By Remark 1, Rs \cap Ss' \neq \emptyset. Therefore r_1s=s''s', for some r_1 \in R, \ s'' \in S. So s''s'r=r_1sr=0, \ s''s'r'=0 and hence s''s'(r+r')=0, i.e. r+r' \in I.

ii) for every r \in I, \ r' \in R: \ r'r \in I. To prove this one, we have sr=0, for some s \in S. Now, by Remark 1: Rs \cap Sr' \neq \emptyset. So r''s=s'r', for some r'' \in R, \ s' \in S. Thus 0=r''sr=s'r'r, i.e. r'r \in I.

iii) if S is a right Ore set, then J=\{r \in R : \ rs = 0, \ \text{for some} \ s \in S \} \lhd R.

Remark 3. Suppose S is left Ore and s_1, \cdots , s_n \in S. There exist r_1, \cdots , r_n \in R such that r_1s_1 = \cdots = r_ns_n \in S. The proof is by induction over n: if n = 1, then we may choose r_1=1. Suppose n > 1 and the claim is true for n-1. Choose r_1', \cdots , r_{n-1}' so that r_1's_1 = \cdots = r_{n-1}' s_{n-1}=s \in S. Also there exist r_n \in R and t \in S such that r_ns_n = ts, since Rs_n \cap Ss \neq \emptyset. Let r_j = tr_j' for j=1, \cdots , n-1. Then for all 1 \leq j \leq n-1 we have r_j s_j = t r_j' s_j = ts = r_ns_n and ts \in S. Similarly, if S is right Ore and s_1, \cdots , s_n \in S, then there exist r_1, \cdots , r_n \in R such that s_1r_1 = \cdots = s_n r_n \in S.

An example of a quotient ring

Posted: December 26, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
Tags: ,
0

Let A be a ring and \alpha \in \text{Aut}(A). Let R=A[x,\alpha] and S=\{1,x,x^2, \cdots \}. Clearly S is a multiplicatively closed subset of R. Let Q=A[x,x^{-1}, \alpha]. The claim is that Q=S^{-1}R. To see this we define \varphi : R \longrightarrow Q to be the inclusion map, i.e. \varphi (f) = f, for all f \in R. We will show that the three conditions in the definition of a left quotient ring, mentioned in the previous post, are satisfied.

1) For any x^n \in S we have (\varphi (x^n))^{-1}=(x^n)^{-1}=x^{-n} \in Q.

2) Let g(x) = \sum_{j=m}^n a_j x^j \in Q, where m,n \in \mathbb{Z} with m \leq n. If m \ge 0, then g(x) \in R and g(x) = (\varphi (1))^{-1} \varphi (g(x)). If m < 0,  let b_j = \alpha^{-m}(a_j) and f(x)=b_m + \cdots + b_n x^{n-m}. See that g(x) = (\varphi (x^{-m}))^{-1} \varphi (f(x)).

3) Since \varphi is the inclusion map, we have \ker \varphi = (0). On the other hand if f(x)=\sum_{j=0}^n c_jx^j \in R, then x^n f(x)=\sum_{j=0}^n \alpha^n (c_j) x^{n+j} and hence, since \alpha is an automorphism, we have x^n f(x) = 0 if and only if f (x)=0.

It’s even easier to prove that Q=RS^{-1}. So Q is both the left and the right quotient ring of R with respect to S.

Definition of a quotient ring

Posted: December 26, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
Tags: ,
0

Throughout this section R is a ring, unitary as usual, and S \neq \emptyset a multiplicatively closed subset of R such that 0 \notin S, \ 1_R \in S.

Definition 1.  A ring Q is said to be a left quotient ring of R with respect to S if there exists a ring homomorphism \varphi : R \longrightarrow Q such that the following conditions are satisfied

1) \varphi (s) is a unit in Q for all s \in S;

2) every element of Q is in the form (\varphi (s))^{-1} \varphi (r), for some r \in R, \ s \in S;

3) \ker \varphi = \{r \in R: \ sr = 0, \ \text{for some s} \in S \}.

Remark. If rs = 0, for some r \in R, \ s \in S, then s'r =0 for some s' \in S. This is because

0 = \varphi (rs) = \varphi (r) \varphi (s)

and thus \varphi (r)=0, since, by the condition 1), \varphi (s) is a unit of Q.

Definition 2. A ring Q is said to be a right quotient ring of R with respect to S if there exists a ring homomorphism \varphi : R \longrightarrow Q such that the following conditions are satisfied

1) \varphi (s) is a unit in Q for all s \in S;

2) every element of Q is in the form \varphi (r)(\varphi (s))^{-1}, for some r \in R, \ s \in S;

3) \ker \varphi = \{r \in R: \ rs = 0, \ \text{for some s} \in S \}.

Lemma. Suppose f : R \longrightarrow R_1 is a ring homomorphism and Q is a left (resp. right) quotient ring of R with respect to S. If f(s) is a unit in R_1, for every s \in S, then there exists a (unique) ring homomorphism g : Q \longrightarrow R_1 which extends f.

Poof. If Q is a left quotient ring of R and \varphi is the map in Definition 1, then for all r \in R, \ s \in S we  define

g((\varphi (s))^{-1} \varphi (r))=(f(s))^{-1}f(r).

Similarly, if Q is a right quotient ring, then we define  g(\varphi (r) (\varphi (s))^{-1}) = f(r) (f(s))^{-1}, for all r \in R and s \in S. Proving that g is a well-defined ring homomorphism is lengthy. I will only prove that g is well-defined:

Suppose (\varphi (s))^{-1} \varphi (r) = (\varphi (s'))^{-1} \varphi (r'), for some r,r' \in R and s,s' \in S. Then

\varphi (s') (\varphi (s))^{-1} \varphi (r)=\varphi (r'),

which gives us \varphi (r''r)=\varphi (tr'), for some t \in S and r'' \in R which satisfy

(\varphi (t))^{-1} \varphi (r'')= \varphi (s') (\varphi (s))^{-1}.

So r''r - tr' \in \ker \varphi and ts' - r'' s \in \ker \varphi. Hence ur''r=utr', \ vts' = vr''s, for some u, v \in S. Therefore

f(r'')f(r)=f(t)f(r') and f(t)f(s')=f(r'')f(s),

which will give us (f(s))^{-1}f(r)=(f(s'))^{-1}f(r'). \Box

Theorem. A left (resp. right) quotient ring R with respect to S, if it exists, is unique up to isomorphism. If R has a left quotient ring Q and a right quotient ring Q' (with respect to S), then Q \cong Q'.

Proof. An easy consequence of the lemma. \Box

Notation. “The” left (resp. right) quotient ring of R w.r.t. S, if it exists, is called the left (resp. right) localization of R at S and it is denoted by S^{-1}R (resp. RS^{-1}).