Archive for the ‘Elementary Algebra; Problems & Solutions’ Category

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

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

We will assume that R is a commutative ring with 1. We will allow the case R=(0), i.e. where 1_R=0_R. The goal is to describe  the Krull dimension of R in terms of elements of R and not prime ideals of R. The reference is this paper: A Short Proof for the Krull Dimension of a Polynomial Ring. Recall that the Krull dimension of R is the largest integer n \geq 0 for which there exist prime ideals P_i, \ 0 \leq i \leq n, of R such that P_0 \subset P_1 \subset \ldots \subset P_n. Then we write \text{K.dim}(R)=n. If there is no such integer, then we define \text{K.dim}(R)=\infty and if R=(0), we define \text{K.dim}(R)=-1.

Notation. Let x \in R and set S_x = \{x^n(1+ax): \ n \geq 0, \ a \in R \}. Clearly S_x is multiplicatively closed. Let R_x be the localization of R at S_x. If 0 \in S_x, then we define R_x =(0).

Problem 1. Let x \in R and suppose that \mathfrak{m} is a maximal ideal of R. Then \mathfrak{m} \cap S_x \neq \emptyset. Moreover, if P \subset \mathfrak{m} is a prime ideal and x \in \mathfrak{m} \setminus P, then P \cap S_x = \emptyset.

Solution. if x \in \mathfrak{m}, then x \in \mathfrak{m} \cap S_x and we are done. Otherwise, \mathfrak{m} + Rx = R and thus 1 + ax \in \mathfrak{m} for some a \in R. Then 1+ax \in \mathfrak{m} \cap S_x. For the second part, suppose to the contrary that x^n(1+ax) \in P for some integer n \geq 0 and a \in R. Then, since x \notin P, we have 1+ax \in P \subset \mathfrak{m} and thus 1 \in \mathfrak{m} because x \in \mathfrak{m}. \Box

Problem 2. \text{K.dim}(R)=0 if and only if R_x=\{0\} for all x \in R.

Solution. As we already mentioned, R_x = \{0\} means 0 \in S_x. Suppose that 0 \notin S_x for some x \in R. Then there exists a prime ideal P of R such that P \cap S_x = \emptyset, because S_x is multiplicatively closed. By Problem 1, P is not a maximal ideal and thus \text{K.dim}(R) > 0. Conversely, suppose that 0 \in S_x for all x \in R and P is a prime ideal of R. Let \mathfrak{m} be  a maximal ideal of R which contains P. Choose x \in \mathfrak{m} \setminus P. By Problem 1, P \cap S_x =\emptyset, contradicting 0 \in P \cap S_x. \ \Box

Problem 3. Let n \geq 0 be an integer. Then \text{K.dim}(R) \leq n if and only if \text{K.dim}(R_x) \leq n-1 for all x \in R.

Solution. The case n=0 was done in Problem 2. So we’ll assume that n \geq 1. Suppose that \text{K.dim}(R) \leq n and let x \in R. A prime ideal of R_x is in the form PR_x, where P is a prime ideal of R and P \cap S_x = \emptyset. Also, if PR_x and QR_x are two prime ideals of R_x with PR_x \subset QR_x, then P \subset Q. Now, suppose to the contrary that \text{K.dim}(R_x) \geq n and consider the chain P_0R_x \subset P_1R_x \subset \ldots \subset P_n R_x of prime ideals of R_x. By Problem 1, P_n is not a maximal ideal and so there exists a prime ideal P_{n+1} such that P_n \subset P_{n+1} and that will give us the chain P_0 \subset P_1 \subset \ldots \subset P_{n+1} of prime ideals of R, contradicting \text{K.dim}(R) \leq n. Conversely, suppose that \text{K.dim}(R_x) \leq n-1 for all x \in R. Suppose also, to the contrary, that there exists a chain P_0 \subset P_1 \subset \ldots \subset P_{n+1} of prime ideals of R. Let x \in P_{n+1} \setminus P_n. By the second part of Problem 1, P_n \cap S_x = \emptyset and thus P_0R_x \subset P_1R_x \subset \ldots \subset P_nR_x is a chain of prime ideals of R_x, contradicting \text{K.dim}(R_x) \leq n-1. \ \Box

Problem 4. Let n \geq 0 be an integer. Then \text{K.dim}(R) \leq n if and only if for every x_0, x_1, \ldots , x_n \in R there exist integers k_0, k_1, \ldots , k_n \geq 0 and a_0, a_1, \ldots ,a_n \in R such that

x_n^{k_n}( \ldots (x_2^{k_2}(x_1^{k_1}(x_0^{k_0}(1+a_0x_0)+a_1x_1) + a_2x_2) + \ldots) + a_nx_n)=0.

Solution. By Problem 3, \text{K.dim}(R) \leq n if and only if \text{K.dim}((\ldots ((R_{x_0})_{x_1})_{x_2} \ldots)_{x_n})=-1. Therefore \text{K.dim}(R) \leq n if and only if (\ldots ((R_{x_0})_{x_1})_{x_2} \ldots)_{x_n}=\{0\}. \ \Box