For a \in \mathbb{C} let \overline{a} denote the complex conjugate of a. Recall that a matrix [a_{ij}] \in M_n(\mathbb{C})  is called Hermitian if a_{ij}=\overline{a_{ji}}, for all 1 \leq i,j \leq n. It is known that if A is Hermitian, then A is diagonalizable  and every eigenvalue of A is a real number. In this post, we will give a lower bound for the rank of a Hermitian matrix. In fact, the lower bound holds for any diagonalizable complex matrix whose eigenvalues are real numbers. To find the lower bound, we first need an easy inequality.

Problem 1. Prove that if a_1, \ldots , a_m \in \mathbb{R}, then (a_1 + \ldots + a_m)^2 \leq m(a_1^2 + \ldots + a_m^2).

Solution.  We have a^2+b^2 \geq 2ab for all a,b \in \mathbb{R} and so

(m-1)\sum_{i=1}^m a_i^2=\sum_{1 \leq i < j \leq m}(a_i^2+a_j^2) \geq \sum_{1 \leq i < j \leq m}2a_ia_j.

Adding the term \sum_{i=1}^m a_i^2 to both sides of the above inequality will finish the job. \Box

Problem 2. Prove that if 0 \neq A \in M_n(\mathbb{C}) is Hermitian, then {\rm{rank}}(A) \geq ({\rm{tr}}(A))^2/{\rm{tr}}(A^2).

Solution. Let \lambda_1, \ldots , \lambda_m be the nonzero eigenvalues of A. Since A is diagonalizable, we have {\rm{rank}}(A)=m. We also have {\rm{tr}}(A)=\lambda_1 + \ldots + \lambda_m and {\rm{tr}}(A^2)=\lambda_1^2 + \ldots + \lambda_m^2. Thus, by Problem 1,

({\rm{tr}}(A))^2 \leq {\rm{rank}}(A) {\rm{tr}}(A^2)

and the result follows. \Box

Throughout this post, U(R) and J(R) are the group of units and the Jacobson radical of a ring R. Assuming that U(R) is finite and |U(R)| is odd, we will show that |U(R)|=\prod_{i=1}^k (2^{n_i}-1) for some positive integers k, n_1, \ldots , n_k. Let’s start with a nice little problem.

Problem 1. Prove that if U(R) is finite, then J(R) is finite too and |U(R)|=|J(R)||U(R/J(R)|.

Solution. Let J:=J(R) and define the map f: U(R) \to U(R/J)) by f(x) = x + J, \ x \in U(R). This map is clearly a well-defined group homomorphism. To prove that f is surjective, suppose that x + J \in U(R/J). Then 1-xy \in J, for some y \in R, and hence xy = 1-(1-xy) \in U(R) implying that x \in U(R). So f is surjective and thus U(R)/\ker f \cong U(R/J). Now, \ker f = \{1-x : \ \ x \in J \} is a subgroup of U(R) and |\ker f|=|J|. Thus J is finite and |U(R)|=|\ker f||U(R/J)|=|J||U(R/J)|. \Box

Problem 2. Let p be a prime number and suppose that U(R) is finite and pR=(0). Prove that if p \nmid |U(R)|, then J(R)=(0).

Solution. Suppose that J(R) \neq (0) and 0 \neq x \in J(R). Then, considering J(R) as an additive group, H:=\{ix: \ 0 \leq i \leq p-1 \} is a subgroup of J(R) and so p=|H| \mid |J(R)|. But then p \mid |U(R)|, by Problem 1, and that’s a contradiction! \Box

There is also a direct, and maybe easier, way to solve Problem 2: suppose that there exists 0 \neq x \in J(R). On U(R), define the relation \sim as follows: y \sim z if and only if y-z = nx for some integer n. Then \sim is an equivalence relation and the equivalence class of y \in U(R) is [y]=\{y+ix: \ 0 \leq i \leq p-1 \}. Note that [y] \subseteq U(R) because x \in J(R) and y \in U(R). So if k is the number of equivalence classes, then |U(R)|=k|[y]|=kp, contradiction!

Problem 3. Prove that if F is a finite field, then |U(M_n(F))|=\prod_{i=1}^n(|F|^n - |F|^{i-1}). In particular, if |U(M_n(F))| is odd,  then n=1 and |F| is a power of 2.

Solution. The group U(M_n(F))= \text{GL}(n,F) is isomorphic to the group of invertible linear maps F^n \to F^n. Also, there is a one-to-one correspondence between the set of invertible linear maps F^n \to F^n and the set of (ordered) bases of F^n. So |U(M_n(F))| is equal to the number of bases of F^n. Now, to construct a basis for F^n, we choose any non-zero element v_1 \in F^n. There are |F|^n-1 different ways to choose v_1. Now, to choose v_2, we need to make sure that v_1,v_2 are not linearly dependent, i.e. v_2 \notin Fv_1 \cong F. So there are |F|^n-|F| possible ways to choose v_2. Again, we need to choose v_3 somehow that v_1,v_2,v_3 are not linearly dependent, i.e. v_3 \notin Fv_1+Fv_2 \cong F^2. So there are |F|^n-|F|^2 possible ways to choose v_3. If we continue this process, we will get the formula given in the problem. \Box

Problem 4. Suppose that U(R) is finite and |U(R)| is odd. Prove that |U(R)|=\prod_{i=1}^k (2^{n_i}-1) for some positive integers k, n_1, \ldots , n_k.

Solution. If 1 \neq -1 in R, then \{1,-1\} would be a subgroup of order 2 in U(R) and this is not possible because |U(R)| is odd. So 1=-1. Hence 2R=(0) and \mathbb{Z}/2\mathbb{Z} \cong \{0,1\} \subseteq R. Let S be the ring generated by \{0,1\} and U(R). Obviously S is finite, 2S=(0) and U(S)=U(R). We also have J(S)=(0), by Problem 2. So S is a finite semisimple ring and hence S \cong \prod_{i=1}^k M_{m_i}(F_i) for some positive integers k, m_1, \ldots , m_k and some finite fields F_1, \ldots , F_k, by the Artin-Wedderburn theorem and Wedderburn’s little theorem. Therefore |U(R)|=|U(S)|=\prod_{i=1}^k |U(M_{m_i}(F_i))|. The result now follows from the second part of Problem 3. \Box

See part (1) here! Again, we will assume that R is a PID and x is a varibale over x. In this post, we will take a look at the maximal ideals of R[x]. Let I be a maximal ideal of R[x]. By Problem 2, if I \cap R \neq (0), then I=\langle p, f(x) \rangle for some prime p \in R and some f(x) \in R[x] which is irreducible modulo p. If I \cap R =(0), then I=\langle f(x) \rangle for some irreducible element f(x) \in R[x]. Before investigating maximal ideals of R[x] in more details, let’s give an example of a PID R which is not a field but R[x] has a maximal ideal I which is principal. We will see in Problem 3 that this situation may happen only when the number of prime elements of R is finite.

Example 1. Let F be a filed and put R=F[[t]], the formal power series in the variable t over F. Let x be a variable over R. Then I:=\langle xt - 1 \rangle is a maximal ideal of R[x].

Proof. See that R[x]/I \cong F[[t,t^{-1}]] and that F[[t,t^{-1}]] is the field of fractions of R. Thus R[x]/I is a field and so I is a maximal ideal of R[x]. \ \Box

Problem 3. Prove that if R has infinitely many prime elements, then an ideal I of R[x] is maximal if and only if I=\langle p, f(x) \rangle for some prime p \in R and some f(x) \in R[x] which is irreducible modulo p.

Solution. We have already proved one direction of the problem in Problem 1. For the other direction, let I be a maximal ideal of R[x]. By the first case in the solution of Problem 2 and the second part of Problem 1, we  only need to show that I \cap R \neq (0). So suppose to the contrary that I \cap R=(0). Then, by the second case in the solution of Problem 2, I=\langle f(x) \rangle for some f(x) \in R[x]. We also know that R[x]/I is a field because I is a maximal ideal of R[x]. Since R has infinitely many prime elements, we can choose a prime p \in R such that p does not divide the leading coefficient of f(x). Now, consider the natural ring homomorphism \psi : R[x] \to R[x]/I. Since I \cap R=(0), \psi(p) \neq 0 and so \psi(p) is invertible in R[x]/I. Therefore pg(x)-1 \in \ker \psi = I for some g(x) \in R[x]. Hence pg(x)-1=h(x)f(x) for some h(x) \in R[x]. If p \mid h(x), then we will have p \mid 1 which is non-sense. So h(x)=pu(x) + v(x) for some u(x),v(x) \in R[x] where p does not divide the leading coefficient of v(x). Now pg(x) - 1 =h(x)f(x) gives us p(g(x)-u(x)f(x)) - 1 =v(x)f(x) and so the leading coefficient of v(x)f(x) is divisible by p. Hence the leading coefficient of f(x) must be divisible by p, contradiction! \Box

Example 2. The ring of integers \mathbb{Z} is a PID and it has infinitely many prime elements. So, by Problem 3, an ideal I of \mathbb{Z}[x] is maximal if and only if I=\langle p, f(x) \rangle for some prime p \in \mathbb{Z} and some f(x) which is irreducible modulo p. By Problem 2, the prime ideals of \mathbb{Z}[x] are the union of the following sets:
1) all maximal ideals
2) all ideals of the form \langle p \rangle, where p \in \mathbb{Z} is a prime
3) all ideals of the form \langle f(x) \rangle, where f(x) is irreducible in \mathbb{Z}[x].

We know that if R is a field and if x is a variable over R, then R[x] is a PID and a non-zero ideal I of R[x] is maximal if and only if I is prime if and only if I is generated by an irreducible element of R[x]. If R is a PID which is not a field, then R[x] could have prime ideals which are not maximal. For example, in \mathbb{Z}[x] the ideal I:=\langle 2 \rangle is prime but not maximal. In this two-part post, we will find prime and maximal ideals of R[x] when R is a PID.

Notation. Throughout this post, R is a PID and R[x] is the polynomial ring in the variable x over R. Given a prime element p \in R, we will denote by \phi_p the natural ring homomorphism R[x] \to R[x]/pR[x].

Definition Let p be a prime element of R. An element f(x) \in R[x] is called irreducible modulo p if \phi_p(f(x)) is irreducible in R[x]/pR[x]. Let \gamma : R \to R/pR be the natural ring homomorphism. Then, since R[x]/pR[x] \cong (R/pR)[x], an element f(x)=\sum_{i=0}^n a_ix^i \in R[x] is irreducible modulo p if and only if \sum_{i=0}^n \gamma(a_i)x^i is irreducible in (R/pR)[x]. Note that R/pR is a field because R is a PID.

Problem 1. Prove that if p \in R is prime and if f(x) \in R[x] is irreducible modulo p, then I:=\langle p, f(x) \rangle is a maximal ideal of R[x]. If f =0, then I is a prime but not a maximal ideal of R[x].

Solution. Clearly I/pR[x]=\phi_p(I)=\langle \phi_p(f(x)) \rangle. So \phi_p(I) is a maximal ideal of R[x]/pR[x] because \phi_p(f(x)) is irreducible in R[x]/pR[x] \cong (R/pR)[x] and R/pR is a field. So I is a maximal ideal of R[x]. If f =0, then I=\langle p \rangle=pR[x] and so R[x]/I \cong (R/pR)[x] is a domain which implies that I is prime. Finally, I= \langle p \rangle is not maximal because, for example, I \subset \langle p,x \rangle \ \Box

Problem 2. Prove that a non-zero ideal I of R[x] is prime if and only if either I= \langle f(x) \rangle for some irreducible element f(x) \in R[x] or I=\langle p, f(x) \rangle for some prime p \in R and some f(x) \in R[x] which is either zero or irreducible modulo p.

Solution. If f(x) \in R[x] is irreducible, then \langle f(x) \rangle is a prime ideal of R[x] because R[x] is a UFD. If f(x)=0 or f(x) is irreducible modulo a prime p \in R, then I=\langle p, f(x) \rangle is a prime ideal of R[x] by Problem 1.
Conversely, suppose that I is a non-zero prime ideal of R[x]. We consider two cases.
Case 1. I \cap R \neq (0) : Let 0 \neq r \in I \cap R. Then r is clearly not a unit because then I wouldn’t be a proper ideal of R[x]. So, since r \in I and I is a prime ideal of R[x], there exists a prime divisor p of r such that p \in I.  So pR[x] \subseteq I and hence \phi_p(I)=I/pR[x] is a prime ideal of R[x]/pR[x] \cong (R/pR)[x]. Thus we have two possibilities. The first possibility is that \phi_p(I)=(0), which gives us I \subseteq \ker \phi_p = pR[x] and therefore I=pR[x]=\langle p \rangle. The second possibility is that \phi_p(I)=\langle \phi_p(f(x)) \rangle= \phi_p(\langle f(x) \rangle) for some irreducible element \phi_p(f(x)) \in R[x]/pR[x], which gives us I=\langle p, f(x) \rangle because \ker \phi_p =pR[x].
Case 2. I \cap R = (0) : Let Q be the field of fractions of R and put J:=IQ[x]. Then J is a non-zero prime ideal of Q[x] because I is a prime ideal of R[x]. Note that J=\{g(x)/r : \ g(x) \in I, \ 0 \neq r \in R \}. So, since Q[x] is a PID, J=q(x)Q[x] for some irreducible element q(x) \in Q[x]. Obviously, we can write q(x)=\alpha f(x), where \alpha \in Q and f(x) \in R[x] is irreducible and the gcd of the coefficients of f(x) is one. Thus J = f(x)Q[x] and, since f(x) \in J, we have f(x) = g(x)/r for some 0 \neq r \in R and g(x) \in I. But then rf(x)=g(x) \in I and so f(x) \in I because I is prime and I \cap R = (0). Hence \langle f(x) \rangle \subseteq I. We will be done if we prove that I \subseteq \langle f(x) \rangle. To prove this, let h(x) \in I \subseteq J=f(x)Q[x]. So h(x)=f(x)q_0(x) for some q_0(x) \in Q[x]. Therefore, since the gcd of the coefficients of f(x) is one, we must have q_0(x) \in R[x] by Gauss’s lemma. Hence h(x) \in \langle f(x) \rangle and the solution is complete. \Box

See the next part here!

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

x=gx_1=gg_1x_1=g_1gx_1=g_1x.

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

(x^2+x)^2=x^4+2x^3+x^2=x^2+x.

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).

Thus

{\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