Posts Tagged ‘quotient ring’

Examples of division rings

Posted: March 25, 2022 in Division Rings, Noncommutative Ring Theory Notes
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Division rings have a simple definition: a ring with identity is a division ring if every non-zero element of the ring is invertible. So every field is a division ring. Also, by the Wedderburn’s little theorem, every finite division ring is a field. So interesting division rings are non-commutative infinite ones. In this post, I’m going to give several examples of division rings.

Example 1. By Remark vii) in this post, the quaternion algebra \mathbb{H}(F):=(-1,-1)_F is a division ring for every field F \subseteq \mathbb{R}. The division rings \mathbb{H}:=\mathbb{H}(\mathbb{R}) and \mathbb{H}(\mathbb{Q}) are called the real and rational quaternions.

Example 2. Let p be a prime number such that p \equiv 3 \mod 4. Then the quaternion algebra A:=(-1,p)_{\mathbb{Q}} is a division ring.

Proof. By Proposition 1 in this post, we only need to show that -x^2+py^2 \ne 1 for all x,y \in \mathbb{Q}. So suppose, to the contrary, that there are rational numbers x,y such that -x^2+py^2=1. So x,y \ne 0 and if we write x=m/n, y=a/b, where m,n,a,b are non-zero integers, then -(m/n)^2+p(a/b)^2=1, which gives (bn)^2+(bm)^2=p(an)^2. Thus (bn)^2+(bm)^2=p^{2k+1}c^2 for some integers k \ge 0 and c \ge 1 with \gcd(c,p)=1. So the exponent of p in the prime factorization of (bn)^2+(bm)^2 is an odd number and that is a contradiction because we know from elementary number theory that in the prime factorization of sum of two squares, prime factors that are 3 \mod 4 occur to even powers. \ \Box

Example 3. If R is a ring with identity, and \mathfrak{m} is a maximal left ideal of R, then \text{End}_R(R/\mathfrak{m}), the ring of R-module homomorphisms R/\mathfrak{m} \to R/\mathfrak{m}, is a division ring.

Proof. That’s just Schur’s lemma because R/\mathfrak{m} is clearly a simple R-module. \ \Box

Example 4. The quotient ring of every (left or right) Noetherian domain is a division ring.

Proof. See Remark 2 and the Theorem in this post. \Box

Example 5. Let R be an algebra over a field, and suppose that R is a domain. If R is PI or has a finite GK-dimension, then the quotient ring of R is a division ring.

Proof. See the Corollary in this post, and Remark 2 in this post. \ \Box

Now that we have some decent examples of division rings, we can use them to build new examples.

Example 6. If D is a division ring, then the ring of Laurent series D((x)) is a division ring too.

Proof. Recall the definition of the ring of (formal) power series D[[x]] and the ring of Laurent series D((x))

\displaystyle D[[x]]=\left \{\sum_{k=0}^{\infty} d_kx^k: \ \ \ \ d_k \in D\right\},

\displaystyle D((x))=\left \{\sum_{k=n}^{\infty} d_kx^k: \ \ \ n \in \mathbb{Z}, \ d_k \in D\right\},

where we assume that x is in the center of D((x)). Clearly D[[x]] \subset D((x)), and it’s easy to see that \sum_{k=0}^{\infty} d_kx^k \in D[[x]] is invertible in D[[x]] if and only if d_0 \ne 0. Now, let 0 \ne z \in D((x)). Then z=\sum_{k=n}^{\infty}d_kx^k for some integer n and d_k \in D, where we choose n to be as large as possible, i.e. an integer such that d_n \ne 0. So we can write z=x^nu, where u=d_n+d_{n+1}x + \cdots \in D[[x]]. So u is invertible and thus z^{-1}=x^{-n}u^{-1} \in D((x)). \ \Box

The next example generalizes Example 6.

Example 7. Let D be a division ring, and let \sigma \in \text{Aut}(D), the group of automorphisms of D. Then the ring of twisted Laurent series D((x; \sigma)) is a division ring too.

Proof. The definitions of the ring of twisted power series D[[x;\sigma]] and twisted Laurent series D((x;\sigma)) are the same as the definitions of D[[x]] and D((x)) given in Example 6 with one difference: unlike in D((x)), the element x \in D((x;\sigma)) is not central unless \sigma is the identity map, in which case D((x;\sigma))=D((x)). Let’s make that precise. We have, by definition,

\displaystyle D[[x; \sigma]]=\left \{\sum_{k=0}^{\infty} d_kx^k: \ \ \ \ d_k \in D\right\},

\displaystyle D((x; \sigma))=\left \{\sum_{k=n}^{\infty} d_kx^k: \ \ \ n \in \mathbb{Z}, \ d_k \in D\right\},

with this rule that xd=\sigma(d)x, for all d \in D. It now follows that x^nd=\sigma^n(d)x^n for all n \in \mathbb{Z}, d \in D. When multiplying two elements of D((x;\sigma)), we will need the rule x^nd=\sigma^n(d)x^n. For example, if d, d_1,d_2 \in D, then

(d_1x^{-2}+d_2x)dx=d_1(x^{-2}d)x+d_2(xd)x=d_1\sigma^{-2} (d)x^{-2}x+d_2\sigma(d)xx

= d_1\sigma^{-2}(d)x^{-1}+d_2\sigma(d)x^2 \in D((x; \sigma).

Note that D[[x;\text{id}]]=D[[x]] and D((x;\text{id}))=D((x)), where \text{id}: D \to D is the identity map. Again, as in Example 6, an element \sum_{k=0}^{\infty} d_kx^k \in D[[x;\sigma]] is invertible in D[[x;\sigma]] if and only if d_0 \ne 0. Now, let 0 \ne z \in D((x;\sigma)). Then z=\sum_{k=n}^{\infty}d_kx^k for some integer n and d_k \in D, \ d_n \ne 0. So we can write z=ux^n, where u=d_n+d_{n+1}x + \cdots \in D[[x;\sigma]] is invertible and we get z^{-1}=x^{-n}u^{-1}. \ \Box

Example 8. Let D_1, D_2 be finite dimensional central division k-algebras. If \gcd(\dim_k D_1, \dim_k D_2)=1, then D_1 \otimes_k D_2 is a division ring.

Proof. See the Theorem in this post. \ \Box

Exercise. Let D be a division ring, and let \sigma: D \to D be an automorphism. Let R:=D[[x;\sigma]], as defined in Example 7. Show that

i) R is a domain,

ii) an element \sum_{k=0}^{\infty}d_kx^k \in R is invertible in R if and only if d_0 \ne 0,

iii) if I \ne (0) is a left ideal of R, then I=Rx^n for some integer n \ge 0, and so R is a Noetherian domain.
Hint. For i), since x^md=\sigma^m(d)x^m, for all d \in D, we have

(c_mx^m+c_{m+1}x^{m+1} + \cdots )(d_nx^n+d_{n+1}x^{n+1}+\cdots )=c_m\sigma^m(d_n)x^{n+m} + \text{higher degree terms}.

So if c_m, d_n \ne 0, then c_m\sigma^m(d_n) \ne 0. For iii), choose n to be the smallest integer n \ge 0 for which there exists \sum_{k=n}^{\infty}d_kx^k, \ d_n \ne 0.

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

Quotient rings; some facts (1)

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