Archive for the ‘Noncommutative Ring Theory Notes’ Category

Problem. Let R be a finite ring with 1 and let n be the number of elements of R. Show that if n is square-free, then R \cong \mathbb{Z}/n\mathbb{Z}.

Solution. Since (R,+) is an abelian group of order n and n is square-free, (R,+) is cyclic. Let a \in R be a generator of (R,+). Then

R=\{0,a,2a, \cdots , (n-1)a\}.


pa=1, \ \ \ qa=a^2,

for some integers p,q and so


Thus n \mid 1-pq because n is the order of a. Hence q,n are coprime. Now consider the map

f: R \to \mathbb{Z}/n\mathbb{Z}, \ \ \ f(ka)=kq+n\mathbb{Z}, \ \ \ \ \ k \in \mathbb{Z}.

We show that f is a ring isomorphism, which is what we need.

i) f is well-defined. Because if ka=0 for some integer k, then n \mid k and so f(ka)=kq+n\mathbb{Z}=0.

ii) f is additive. That’s clear.

iii) f is bijective. Suppose that f(ka)=kq+n\mathbb{Z}=0 for some integer k. Then n \mid kq and so n \mid k, because q,n are coprime, implying that ka=0. So f is injective and hence bijective because R and \mathbb{Z}/n\mathbb{Z} have the same number of elements.

iv) f is multiplicative. Because for integers k,k' we have

\begin{aligned} f(kak'a)=f(kk'a^2)=f(kk'qa)=kk'q^2+n\mathbb{Z}=(kq+n\mathbb{Z})(k'q+n\mathbb{Z})=f(ka)f(k'a).\end{aligned} \ \Box


Problem. Let R be a finite ring with 1 and suppose that R satisfies the following properties

i) \displaystyle \sum_{r \in R} r \ne 0

ii) \displaystyle \prod_{0 \ne r \in R} r \ne 0.

Show that either R \cong \mathbb{Z}_2 or R \cong \mathbb{Z}_4.

Solution. First notice that both \mathbb{Z}_2 and \mathbb{Z}_2 satisfy i), ii). Since \mathbb{Z}_2 is the only ring with two elements, we may assume that |R| > 2. Let n:=|R|. First we show that n is even. Suppose, to the contrary, that n is odd. Then the group (R,+) would have no element of order 2, i.e. r \ne -r for all 0 \ne r \in R, and thus

\displaystyle \sum_{r \in R} r = \sum (r+(-r))=0,

contradicting the property i).
Now let m:=\text{char}(A). We show that m=p^k for some prime p and k \in \{1,2\}. To see that, let m=\prod_{i=1}^{\ell}p_i^{k_i} be the prime factorization of m. If \ell > 1, then

\displaystyle \prod_{i=1}^{\ell} p_i^{k_i}1_R=m1_R=0,

contradicting the property ii), because p_i^{k_i}1_R are non-zero distinct elements of R. So \ell=1, i.e. m=p^k for some prime number p and positive integer k. Now if k > 2, then we can write


and again, by the minimal property of m, the elements p1_R and p^{k-1}1_R are non-zero and distinct and that contradicts the property ii). So k \le 2.
We now show that in fact m=4. We just proved that m=p^k for some prime p and k \in \{1,2\}. Suppose that q \ne p is a prime divisor of n. Then there exists r \in (R,+) that has order q. But then qa=ma=0 which gives the false result r=0, because q,m are coprime. So n is a power of p. But we have already showed that n is even. So p=2 and thus either m=2 or m=4. If m=2, then R would be a vector space over \mathbb{Z}_2, i.e. R=\sum_{i=1}^s \mathbb{Z}_2r_i for some integer s \ge 2, because n > 2, and r_i \in R. But then

\displaystyle \sum_{r \in R}r =2^{s-1}\sum_{i=1}^s r_i=0,

contradicting the property ii). So m=4 and hence R contains a subring R_0 \cong \mathbb{Z}_4.
Finally, we show that R=R_0. Suppose, to the contrary, that R \neq R_0 and let r \in R \setminus R_0. We have (21_R)(2r)=0, because \text{char}(R)=4, and so, since 21_R \ne 0, we must have either 21_R=2r or 2r=0, by property ii). If 21_R=2r, then (21_R)(r-1_R)=0, which gives r=1_R \in R_0, by property ii), and that’s a contradiction. So 2r=0, i.e. we have shown that 2r=0 for any element r \in R \setminus R_0. But if r \in R \setminus R_0, then r+1_R \in R \setminus R_0 too, because 1_R \in R_0, and so 2r=2(r+1_R)=0, which gives 21_R=0, contradiction. So R=R_0 \cong \mathbb{Z}_4. \ \Box

All rings in this post are assumed to have the multiplicative identity 1.

In this post, we showed that the ring R:= \mathbb{Z}/n\mathbb{Z} has 2^m idempotents, where m is the number of prime divisors of n. Clearly m is also the number of prime (= maximal) ideals of R (recall that, in general, in commutative Artinian rings, prime ideals are maximal). Let \varphi be the Euler’s totient function. The number of units of R is \varphi(n). If n=\prod_{i=1}^m p_i^{n_i} is the prime factorization of n, then \varphi(n)=\prod_{i=}^m(p_i^{n_i}-p_i^{n_i-1}). If n is odd, then p_i^{n_i}-p_i^{n_i-1} is even for all i and hence 2^m divides \varphi(n). So if n is odd, then the number of idempotents of R divides the number of units of R. As we are going to show now, this is a property of any finite commutative ring of odd order.

Recall that a commutative ring R is called semilocal if the number of maximal ideals of R is finite and it is called local if it has only one maximal ideal.

Example 1. If p is a prime and k \ge 1 is an integer, then \mathbb{Z}/p^k\mathbb{Z} is a local ring with the unique maximal ideal p\mathbb{Z}/p^k\mathbb{Z}. If n \ge 2 is an integer, then \mathbb{Z}/n\mathbb{Z} is a semilocal ring (what are its maximal ideals?).

Example 2. Generalizing the above example, Artinian rings are semilocal. This is easy to see; let R be an Artinian ring and let S be the set of all finite intersections of maximal ideals of R. Since R is Artinian, S has a minimal element I:=\bigcap_{i=1}^m M_i. Let M be any maximal ideal of R. Since I \bigcap M \in S and I is a minimal element of S, we must have M_1M_2 \cdots M_m \subseteq I \subseteq M. So M_i \subseteq M for some i and hence M_i=M. Therefore M_1, \cdots , M_m are all the maximal ideals of R.

Problem 1. Show that if R is a local ring, then 0,1 are the only idempotents of R.

Solution. Let M be the maximal ideal of R and let e be an idempotent of R. Then e(1-e)=0 \in M. Thus either e \in M or 1-e \in M. If e \in M, then 1-e \notin M because otherwise 1=e+(1-e) \in M, which is false. So 1-e is a unit because there’s no other maximal ideal to contain 1-e. So (1-e)r=1 for some r \in R and thus e=e(1-e)r=0. Similarly, if 1-e \in M, then e \notin M and thus e is a unit. So er=1 for some r \in R implying that 1-e=(1-e)er=0 and hence e=1. \ \Box

Problem 2 Show that the number of idempotents of a commutative Arinian ring R is 2^m, where m is the number of maximal ideals of R.

Solution. Since R is Artinian, it has only finitely many maximal ideals, say M_1, \cdots , M_m (see Example 2). The Jacobson radical of R is nilpotent, hence there exists an integer k \ge 1 such that

(0)=\left(\bigcap_{i=1}^m M_i \right)^k=\prod_{i=1}^m M_i^k=\bigcap_{i=1}^m M_i^k.

Thus, by the Chinese remainder theorem for commutative rings, R \cong \prod_{i=1}^m R/M_i^k. Since each R/M_i^k is a local ring, with the unique maximal ideal M_i/M_i^k, it has only two idempotents, by Problem 1, and so R has exactly 2^m idempotents. \Box

Problem 3. Let R be a finite commutative ring. Show that if |R|, the number of elements of R, is odd, then the number of idempotents of R divides the number of units of R.

Solution. Since R is finite, it is Artinian. Let \{M_1, \cdots , M_m\} be the set of maximal ideals of R. By problem 2, the umber of idempotents of R is 2^m and R \cong \prod_{i=1}^m R/M_i^k for some integer k \ge 1.
Since |R| is odd, each |M_i| is odd too because (M_i,+) is a subgroup of (R,+) and so |M_i| divides |R|. Also, units in a local ring are exactly those elements of the ring which are not in the maximal ideal. So the number of units of each R/M_i^k is |R/M_i^k|-|M_i/M_i^k|, which is an even number because both |R| and |M_i| are odd. So the number of units of R, which is the product of the number of units of R/M_i^k, \ 1 \le i \le m, is divisible by 2^m, which is the number of idempotents of R. \ \Box

Remark 1. The result given in Problem 3 is not always true if the number of elements of the ring is even. For example, \mathbb{Z}/2\mathbb{Z} has one unit and two idempotents. However, the result is true in \mathbb{Z}/2^n\mathbb{Z}, \ n \ge 2, which has \varphi(2^n)=2^n-2^{n-1} units and two idempotents.
Can we find all even integers n for which the result given in Problem 3 is true in \mathbb{Z}/n\mathbb{Z}? Probably not because this question is equivalent to finding all integers n such that 2^m \mid \varphi(n), where m is the number of prime divisors of n, and that is not an easy thing to do.

Remark 2. The result given in Problem 3 is not necessarily true in noncommutative rings with an odd number of elements. For example, consider R:=M_2(\mathbb{F}_3), the ring of 2\times 2 matrices with entries from the field of order three. Then R has (3^2-3)(3^2-1)=48 units (see Problem 3 in this post!) but, according to my calculations, R has 14 idempotents and 14 does not divide 48.

Problem. Let R be a finite ring with 1 and let R^*=R \setminus \{0\}. Show that for every integer n \ge 1 there exist x,y,z \in R^* such that x^n+y^n=z^n if and only if R is not a division ring.

Solution. Suppose first that R is a division ring. Then, since R is a finite ring, R is a finite field, by the Wedderburn’s little theorem. Let |R|=q. Then x^{q-1}=1 for all x \in R^* and so x^{q-1}+y^{q-1}=z^{q-1} has no solution in R^*.
Conversely, suppose that R is not a division ring (equivalently, a field because R is finite). So |R| > 2 and hence the equation x+y=z has solutions in R^* (just choose y=1, \ z \ne 0,1 and x=z-1).
Let J(R) be the Jacobson radical of R. Since R is finite, it is Artinian and so J(R) is nilpotent.
So if J(R) \neq (0), then there exists a \in R^* such that a^2=0 and so a^n=0 for all n \ge 2. Therefore the equation x^n+y^n=z^n, \ n \ge 2, has a solution x=a, y=z=1 in R^*.
If J(R)=(0), then by the Artin-Wedderburn’s theorem,

\displaystyle R=\prod_{i=1}^k M_{n_i}(F_i),

for some finite fields F_i. Since R is not a field, we have either n_i >1 for some i or n_i=1 for all i and k \ge 2. If n_i > 1 for some i, then M_{n_i}(F_i), and hence R, will have a non-zero nilpotent element and we are done. If n_i=1 for all i and k \ge 2, then

x=(1,0,0, \cdots ,0), \ y = (0,1,0, \cdots , 0), \ z = (1,1,0, \cdots , 0)

will satisfy x^n+y^n=z^n. \ \Box

Throughout this post, R is a ring with 1.

Theorem (Jacobson). If x^n=x for some integer n > 1 and all x \in R, then R is commutative.

In fact n, in Jacobson’s theorem, doesn’t have to be fixed and could depend on x, i.e. Jacobson’s theorem states that if for every x \in R there exists an integer n > 1 such that x^n=x, then R is commutative. But I’m not going to prove that here.
In this post, we’re going to prove Jacobson’s theorem. Note that we have already proved the theorem for n=3, 4 (see here and here) and we didn’t need R to have 1, we didn’t need that much ring theory either. But to prove the theorem for any n > 1, we need a little bit more ring theory.

Lemma. If Jacobson’s theorem holds for division rings, then it holds for all rings with 1.

Proof. Let R be a ring with 1 such that x^n=x for some integer n > 1 and all x \in R. Then clearly R is reduced, i.e. R has no non-zero nilpotent element. Let \{P_i: \ i \in I\} be the set of minimal prime ideals of R.
By the structure theorem for reduced rings, R is a subring of the ring \prod_{i\in I}D_i, where D_i=R/P_i is a domain. Clearly x^n=x for all x \in D_i and all i \in I. But then, since each D_i is a domain, we get x=0 or x^{n-1}=1, i.e. each D_i is a division ring. Therefore, by our hypothesis, each D_i is commutative and hence R, which is a subring of \prod_{i\in I}D_i, is commutative too. \Box

Example. Show that if x^5=x for all x \in R, then R is commutative.

Solution. By the lemma, we may assume that R is a division ring.
Then 0=x^5-x=x(x-1)(x+1)(x^2+1) gives x=0,1,-1 or x^2=-1. Suppose that R is not commutative and choose a non-central element x \in R. Then x+1,x-1 are also non-central and so x^2=(x+1)^2=(x-1)^2=-1 which gives 1=0, contradiction! \Box

Remark 1. Let D be a division ring with the center F. If there exist an integer n \ge 1 and a_i \in F such that x^n+a_{n-1}x^{n-1}+ \cdots + a_1x+a_0=0 for all x \in D, then F is a finite field. This is obvious because the polynomial x^n+a_{n-1}x^{n-1}+ \cdots + a_1x+a_0 \in F[x] has only a finite number of roots in F and we have assumed that every element of F is a root of that polynomial.

Remark 2. Let D be a domain and suppose that D is algebraic over some central subfield F. Then D is a division ring and if 0 \ne d \in D, then F[d] is a finite dimensional division F-algebra.

Proof. Let 0 \ne d \in D. So d^m +a_{m-1}d^{m-1}+ \cdots + a_1d+a_0=0 for some integer m \ge 1 and a_i \in F. We may assume that a_0 \ne 0. Then d(d^{m-1} + a_{m-1}d^{m-2}+ \cdots + a_1)(-a_0^{-1})=1 and so d is invertible, i.e. D is a division ring.
Since F[d] is a subring of D, it is a domain and algebraic over F and so it is a division ring by what we just proved. Also, since d^m \in \sum_{i=0}^{m-1} Fd^i for some integer m \ge 1, we have F[d]=\sum_{i=0}^{m-1} Fd^i and so \dim_F F[d] \le m. \ \Box

Proof of the Theorem. By the above lemma, we may assume that R is a division ring.
Let F be the center of R. By Remark 1, F is finite. Since R is a division ring, it is left primitive. Since every element of R is a root of the non-zero polynomial x^n-x \in F[x], \ R is a polynomial identity ring.
Hence, by the Kaplansky-Amtsur theorem, \dim_F R < \infty and so R is finite because F is finite. Thus, by the Wedderburn’s little theorem, R is a field. \Box

We defined the n-th Weyl algebra A_n(R) over a ring R in here.  In this post we will find the GK dimension of A_n(R) in terms of the GK dimension of R. The result is similar to what we have already seen in commutative polynomial rings (see corollary 1 in here). We will assume that k is a field and R is a k-algebra.

Theorem. {\rm{GKdim}}(A_1(R))=2 + {\rm{GKdim}}(R).

Proof. Suppose first that R is finitely generated and let V be a frame of R. Let U=k+kx+ky. Since yx = xy +1, we have

\dim_k U^n = \frac{(n+1)(n+2)}{2}. \ \ \ \ \ \ \ \ \ (*)

Let W=U+V. Clearly W is a frame of A_1(R) and

W^n = \sum_{i+j=n} U^i V^j,

for all n, because every element of V commutes with every element of U. Therefore, since V^j \subseteq V^n and U^i \subseteq U^n for all i,j \leq n, we have W^n \subseteq U^nV^n and W^{2n} \supseteq U^nV^n. Thus W^n \subseteq U^nV^n \subseteq W^{2n} and hence

\log_n \dim_k W^n \leq \log_n \dim_k U^n + \log_n \dim_k V^n \leq \log_n \dim_k W^{2n}.

Therefore {\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)), by (*), and we are done.

For the general case, let R_0 be any finitely generated k– subalgebra of R. Then, by what we just proved,

2 + {\rm{GKdim}}(R_0)={\rm{GKdim}}(A_1(R_0)) \leq {\rm{GKdim}}(A_1(R))

and hence 2+{\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)). Now, let A_0 be a k-subalgebra of A_1(R) generated by a finite set \{f_1, \ldots , f_m\}. Let R_0 be the k-subalgebra of R generated by all the coefficients of f_1, \ldots , f_m. Then A_0 \subseteq A_1(R_0) and so

{\rm{GKdim}}(A_0) \leq {\rm{GKdim}}(A_1(R_0))=2 + {\rm{GKdim}}(R_0) \leq 2 + {\rm{GKdim}}(R).


{\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R)

and the proof is complete. \Box

Corollary. {\rm{GKdim}}(A_n(R))=2n + {\rm{GKdim}}(R) for all n. In particular, {\rm{GKdim}}(A_n(k))=2n.

Proof. It follows from the theorem and the fact that A_n(R)=A_1(A_{n-1}(R)). \Box

As usual, I’ll assume that k is a field. Recall that if a k-algebra A is an Ore domain, then we can localize A at S:=A \setminus \{0\} and get the division algebra Q(A):=S^{-1}A. The algebra Q(A) is called the quotient division algebra of A.

Theorem (Borho and Kraft, 1976) Let A be a finitely generated k-algebra which is a domain of finite GK dimension. Let B be a k-subalgebra of A and suppose that {\rm{GKdim}}(A) < {\rm{GKdim}}(B) + 1. Let S:=B \setminus \{0\}. Then S is an Ore subset of A and S^{-1}A=Q(A). Also, Q(A) is finite dimensional as a (left or right) vector space over Q(B).

Proof. First note that, by the corollary in this post, A is an Ore domain and hence both Q(A) and Q(B) exist and they are division algebras. Now, suppose, to the contrary, that S is not (left) Ore. Then there exist x \in S and y \in A such that Sy \cap Ax = \emptyset. This implies that the sum By + Byx + \ldots + Byx^m is direct for any integer m. Let W be a frame of a finitely generated subalgebra B' of B. Let V=W+kx+ky and suppose that A' is the subalgebra of A which is generated by V. For any positive integer n we have

V^{2n} \supseteq W^n(kx+ky)^n \supseteq W^ny + W^nyx + \ldots + W^nyx^{n-1}

and thus \dim_k V^{2n} \geq n \dim_k W^n because the sum is direct. So \log_n \dim_k V^{2n} \geq 1 + \log_n \dim_k W^n and hence {\rm{GKdim}}(A) \geq {\rm{GKdim}}(A') \geq 1 + {\rm{GKdim}}(B'). Taking supremum of both sides over all finitely generated subalgebras B' of B will give us the contradiction {\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(B). A similar argument shows that S is right Ore. So we have proved that S is an Ore subset of A. Before we show that S^{-1}A=Q(A), we will prove that Q(B)A=S^{-1}A is finite dimensional as a (left) vector space over Q(B). So let V be a frame of A. For any positive ineteger n, let r(n) = \dim_{Q(B)} Q(B)V^n. Clearly Q(B)V^n \subseteq Q(B)V^{n+1} for all n and

\bigcup_{n=0}^{\infty}Q(B)V^n =Q(B)A

because \bigcup_{n=0}^{\infty}V^n=A. So we have two possibilities: either Q(B)V^n=Q(B)A for some n or the sequence \{r(n)\} is strictly increasing. If Q(B)V^n = Q(B)A, then we are done because V^n is finite dimensional over k and hence Q(B)V^n is finite dimensional over Q(B). Now suppose that the sequence \{r(n)\} is strictly increasing. Then r(n) > n because r(0)=\dim_{Q(B)}Q(B)=1. Fix an integer n and let e_1, \ldots , e_{r(n)} be a Q(B)-basis for Q(B)V^n. Clearly we may assume that e_i \in V^n for all i. Let W be a frame of a finitely generated subalgebra of B. Then

(V+W)^{2n} \supseteq W^nV^n \supseteq W^ne_1 + \ldots + W^ne_{r(n)},

which gives us

\dim_k(V+W)^{2n} \geq r(n) \dim_k W^n > n \dim_k W^n,

because the sum W^ne_1 + \ldots + W^ne_{r(n)} is direct. Therefore {\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(B), which is a contradiction. So we have proved that the second possibility is in fact impossible and hence Q(B)A is finite dimensional over Q(B). Finally, since, as we just proved, \dim_{Q(B)}Q(B)A < \infty, the domain Q(B)A is algebraic over Q(B) and thus it is a division algebra. Hence Q(B)A=Q(A) because A \subseteq Q(B)A \subseteq Q(A) and Q(A) is the smallest division algebra containing A. \Box