For any set X, we denote by {\rm{Sym}}(X) the group of bijective maps from X to X.

Problem. Let \alpha : \mathbb{N} \longrightarrow [1, \infty) be any map which satisfies the following two conditions: \alpha(p) \geq p and \alpha(p+q) \geq \alpha(p) \alpha(q) for all p,q \in \mathbb{N}. Let X be a set with |X|=n. Prove that if G is an abelian subgroup of {\rm{Sym}}(X), then |G| \leq \alpha(n).

Solution.  The proof is by induction on n. If n=1, then |G|=1 \leq \alpha(1). Let X be a set with |X|= n \geq 2 and suppose that the claim is true for any set of size < n. Let G be an abelian subgroup of {\rm{Sym}}(X). Clearly gx=g(x), \ g \in G, x \in X, defines an action of G on X. Let X_1, \ldots , X_k be the orbits corresponding to this action and consider two cases.

Case 1. k=1: Fix an element x_1 \in X. Then X=X_1=Gx_1. Suppose that g_1x_1=x_1 for some g_1 \in G and let x \in X. Then x=gx_1 for some g \in G. Thus, since G is abelian, we have


Hence g_1x=x for all x \in X and thus g_1=1. So the stabilizer of x_1 is trivial and therefore, by the orbit-stabilizer theorem, |G|=|X|=n \leq \alpha(n).

Case 2k \geq 2: Let |X_i|=n_i, \ i=1,2, \ldots, k. Clearly \sum_{i=1}^k n_i=n and, since k \geq 2, we have n_i < n for all i. For every g \in G and 1 \leq i \leq k let g_i=g|_{X_i}, the restriction of g to X_i, and put

G_i=\{g_i: \ g \in G\}.

Then g_i \in {\rm{Sym}}(X_i) and G_i is an abelian subgroup of {\rm{Sym}}(X_i). Thus, by the induction hypothesis

|G_i| \leq \alpha(n_i),

for all i. Now, define \varphi : G \longrightarrow \bigoplus_{i=1}^k G_i by \varphi(g)=(g_1, g_2, \ldots , g_k) for all g \in G. It is obvious that \varphi is one-to-one and so

|G| \leq |\bigoplus_{i=1}^k G_i|=\prod_{i=1}^k |G_i| \leq \prod_{i=1}^k \alpha(n_i) \leq \alpha(\sum_{i=1}^k n_i)=\alpha(n). \ \Box

Remark. The map \alpha: \mathbb{N} \longrightarrow [1, \infty) defined by \alpha(p)=3^{p/3}, for all p \in \mathbb{N}, satisfies both conditions in the above Problem. So if |X|=n and if G is an abelian subgroup of {\rm{Sym}}(X), then |G| \leq 3^{n/3}.


Let R be a ring, which may or may not have 1. We proved in here that if x^3=x for all x \in R, then R is commutative.  A similar approach shows that if x^4=x for all x \in R, then R is commutative.

Problem. Prove that if x^4=x for all x \in R, then R is commutative.

Solution. Clearly R is reduced, i.e. R has no nonzero nilpotent element. Note that 2x=0 for all x \in R because x=x^4=(-x)^4=-x. Hence x^2+x is an idempotent for every x \in R because


Thus x^2+x is central for all x \in R, by Remark 3 in this post.  Therefore (x^2+y)^2+x^2+y is central for all x,y \in R. But

(x^2+y)^2+x^2+y=x^2+x+y^2+y+ x^2y+yx^2

and hence x^2y+yx^2 is central. Therefore (x^2y+yx^2)x^2=x^2(x^2y+yx^2) which gives us xy=yx. \ \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


Throughout R is a ring with 1 and all modules are left R-modules. In Definition 2 in this post, we defined Z(M), the singular submodule of a module M.

Problem 1. Let M be an R-module and suppose that N_1, \cdots, N_k are submodules of M. Prove that \bigcap_{i=1}^k N_i \subseteq_e M if and only if N_i \subseteq_e M for all i.

Solution. We only need to solve the problem for k = 2. If N_1 \cap N_2 \subseteq_e M, then N_1 \subseteq_e M and N_2 \subseteq_e M because both N_1 and N_2 contain N_1 \cap N_2. Conversely, let P be a nonzero submodule of M. Then N_1 \cap P \neq \{0\} because N_1 \subseteq_e M and therefore (N_1 \cap N_2) \cap P = N_2 \cap (N_1 \cap P) \neq \{0\} because N_2 \subseteq_e M. \ \Box

Problem 2. Prove that if M is an R-module, then Z(M) is a submodule of M and Z(R) is a proper two-sided ideal of R. In particular, if R is a simple ring, then Z(R)=\{0\}.

Solution. First note that 0 \in Z(M) because \text{ann}(0)=R \subseteq_e R. Now suppose that x_1,x_2 \in Z(M). Then \text{ann}(x_1+x_2) \supseteq \text{ann}(x_1) \cap \text{ann}(x_2) \subseteq_e M, by Problem 1. Therefore \text{ann}(x_1+x_2) \subseteq_e M and hence x_1+x_2 \in Z(M). Now let r \in R and x \in Z(M). We need to show that rx \in Z(M). Let J be a nonzero left ideal of R. Then Jr is also a left ideal of R. If Jr = \{0\}, then J \subseteq \text{ann}(rx) and thus \text{ann}(rx) \cap J = J \neq \{0 \}. If Jr \neq \{0\}, then \text{ann}(x) \cap Jr \neq \{0\} because x \in Z(M). So there exists s \in J such that sr \neq 0 and srx = 0. Hence 0 \neq s \in \text{ann}(rx) \cap J. So rx \in Z(M) and thus Z(M) is a submodule of M. Now, considering R as a left R-module, Z(R) is a left ideal of R, by what we have just proved. To see why Z(M) is a right ideal, let r \in R and x \in Z(R). Then \text{ann}(xr) \supseteq \text{ann}(x) \subseteq_e R and so \text{ann}(xr) \subseteq_e R, i.e. xr \in Z(R). Finally, Z(R) is proper because \text{ann}(1)=\{0\} and so 1 \notin Z(R). \ \Box

Problem 3. Prove that if M_i, \ i \in I, are R-modules, then Z(\bigoplus_{i \in I} M_i) = \bigoplus_{i \in I} Z(M_i). Conclude that if R is a semisimple ring, then Z(R)=\{0\}.

Solution. The first part is a trivial result of Problem 1 and this fact that if x = x_1 + \cdots + x_n, where the sum is direct, then \text{ann}(x) = \bigcap_{i=1}^n \text{ann}(x_i). The second now follows trivially from the first part, Problem 2 and the Wedderburn-Artin theorem. \Box

Problem 4. Suppose that R is commutative and let N(R) be the nilradical of R. Prove that

1) N(R) \subseteq Z(R);

2) it is possible to have N(R) \neq Z(R);

3) if Z(R) \neq \{0\}, then N(R) \subseteq_e Z(R), as R-modules or Z(R)-modules.

Solution. 1) Let a \in N(R). Then a^n = 0 for some integer n \geq 1. Now suppose that 0 \neq r \in R. Then ra^n=0. Let m \geq 1 be the smallest integer such that ra^m = 0. Then 0 \neq ra^{m-1} \in \text{ann}(a) \cap Rr and hence a \in Z(R).

2) Let R_i = \mathbb{Z}/2^i \mathbb{Z}, \ i \geq 1 and put R=\prod_{i=1}^{\infty}R_i. For every i, let a_i = 2 + 2^i \mathbb{Z} and consider a = (a_1,a_2, \cdots ) \in R. It is easy to see that a \in Z(R) \setminus N(R).

3) Let a \in Z(R) \setminus N(R). Then \text{ann}(a) \cap Ra \neq \{0\} and thus there exists r \in R such that ra \neq 0 and ra^2=0. Hence (ra)^2 = 0 and so ra \in N(R). Thus 0 \neq ra \in N(R) \cap Ra implying that N(R) is an essential R-submodule of Z(R). Now, we view Z(R) as a ring and we want to prove that N(R) as an essential ideal of Z(R). Again,  let a \in Z(R) \setminus N(R). Then \text{ann}(a) \cap Ra^2 \neq \{0\} and thus there exists r \in R such that ra^2 \neq 0 and ra^3 = 0. Let s = ra \in Z(R). Then (sa)^2=0 and thus 0 \neq sa \in N(R) \cap Z(R)a implying that N(R) is an essential ideal of Z(R). \ \Box


We will assume that R is a ring (not necessarily commutative) with 1 and all modules are left R-modules.

Definition 1. Let M be an R-module and N a nonzero submodule of M. We say that N is an essential submodule of M, and we will write N \subseteq_e M, if N \cap X \neq (0) for any nonzero submodule X of M. Clearly, that is equivalent to saying N \cap Rx \neq (0) for any nonzero element x \in M. So, in particular, a nonzero left ideal I of R is an essential left ideal of R if I \cap J \neq (0) for any nonzero left ideal J of R, which is equivalent to the condition I \cap Rr \neq (0) for any nonzero element r \in R.

Definition 2. Let M be an R-module and x \in M. Recall that the (left) annihilator of x in R is defined by \text{ann}(x)=\{r \in R: \ rx = 0 \}, which is obviously a left ideal of R. Now, consider the set Z(M):=\{x \in M: \ \text{ann}(x) \subseteq_e R \}. It is easy to see that Z(M) is a submodule of M (see Problem 2 in this post for the proof!) and we will call it the singular submodule of M. If Z(M)=M, then M is called singular. If Z(M)=(0), then M is called nonsingular. We will not discuss nonsingular modules in this post.

Problem 1. Prove that if F is a free R-module, then Z(F) \neq F, i.e. a free module is never singular.

Solution. Let x \in F be any element of an R-basis of F. Let r \in \text{ann}(x). Then rx = 0 and so r = 0. Thus \text{ann}(x)=(0) and so x \notin Z(F). \ \Box

Next problem characterizes singular modules.

Problem 2. Prove that an R-module M is singular if and only if M = A/B for some R-module A and some submodule B \subseteq_e A.

Solution. Suppose first that M=A/B where A is an R-module and B \subseteq_e A. Let x = a + B \in M and let J be a nonzero left ideal of R. If Ja = (0), then Ja \subseteq B and so \text{ann}(x) \cap J = J \neq (0). If Ja \neq (0), then B \cap Ja \neq (0) because B \subseteq_e A. So there exists r \in J such that 0 \neq ra \in B. That means 0 \neq r \in \text{ann}(x) \cap J. So we have proved that x \in Z(M) and hence Z(M)=M, i.e. M is singular. Conversely, suppose that M is singular. We know that every R-module is the homomorphic image of some free R-module. So there exists a free R-module F and a submodule K of F such that M \cong F/K. So we only need to show that K \subseteq_e F. Note that K \neq (0), by Problem 1. Let \{x_i \} be an R-basis for F and suppose that 0 \neq x \in F. We need to show that there exists s \in R such that 0 \neq sx \in K. We can write, after renaming the indices if necessarily,

x = \sum_{i=1}^n r_ix_i, \ \ \ \ \ \ \ \ \ \ (1)

where r_1 \neq 0. For any y \in F, let \overline{y}=y+K \in F/K. Now, since F/K is singular, \text{ann}(\overline{x_1}) \subseteq_e R and so \text{ann}(\overline{x_1}) \cap Rr_1 \neq (0). So there exists s_1 \in R such that s_1r_1 \neq 0 and s_1r_1x_1 \in K. Hence (1) gives us

s_1x= s_1r_1x_1 +\sum_{i=2}^n s_1r_ix_i. \ \ \ \ \ \ \ \ \ (2)

Note that s_1x \neq 0 because s_1r_1 \neq 0. Now, if s_1r_i = 0 for all 2 \leq i \leq n, then s_1x =s_1r_1x_1 \in K and we are done. Otherwise, after renaming the indices in the sum on the right hand side of (2) if necessary, we may assume that s_1r_2 \neq 0. Repeating the above process gives us some s_2 \in R such that s_2s_1r_2 \neq 0 and s_2s_1r_2x_2 \in K. Then (2) implies

s_2s_1x = s_2s_1r_1x_1 + s_2s_1r_2x_2 + \sum_{i=3}^n s_2s_1r_ix_i. \ \ \ \ \ \ \ (3)

The first two terms on the right hand side of (3) are in K and s_2s_1x \neq0 because s_2s_1r_2 \neq 0. If we continue this process, we will eventually have a positive integer 1 \leq m \leq n and s = s_ms_{m-1} \cdots s_1 \in R such that 0 \neq sx \in K. \ \Box


Problem. Let F be a field and suppose that f(x),g(x), p(x) are three polynomials in F[x]. Prove that if both f(x) and p(x) are irreducible and p(x) \mid f(g(x)), then \deg f(x) \mid \deg p(x).

Solution. Let \mathfrak{m} and \mathfrak{n} be the ideals of F[x] generated by f(x) and p(x), respectively. Let E=F[x]/\mathfrak{m} and L = F[x]/\mathfrak{n}. Since both f(x) and p(x) are irreducible, E and L are field extensions of F. Now, define the map \varphi : E \longrightarrow L by \varphi(h(x) + \mathfrak{m})=h(g(x)) + \mathfrak{n}, for all h(x) \in F[x]. We first show that \varphi is well-defined. To see this, suppose that h(x) \in \mathfrak{m}. Then h(x)=f(x)u(x) for some u(x) \in F[x] and hence h(g(x)) = f(g(x))u(g(x)) \in \mathfrak{n}, because p(x) \mid f(g(x)). So \varphi is well-defined. Now \varphi is clearly a ring homomorphism and, since E is a field and \ker \varphi is an ideal of E, we must have \ker \varphi = \{0\}. Therefore we may assume that F \subseteq E \subseteq L and hence \deg p(x) = [L : F] = [L: E][E:F]=(\deg f(x))[L : E]. \ \Box