Archive for the ‘Rings and Modules’ Category

Suppose that R \ne (0) is a ring with no proper left ideals. If R has 1, then  R is a division ring. To see this, let 0 \ne x \in R. Then Rx=R and so yx=1 for some y \in R. Since y \ne 0, we have Ry=R and hence zy=1 for some z \in R. Then x=zyx=z and so yx=xy=1 proving that R is a division ring.

But what if R doesn’t have 1 ? The following problem answers this question.

Problem. Let R \ne (0) be a ring, which may or may not have 1. Show that if R has no proper left ideals, then either R is a division ring or R^2=(0) and |R|=p for some prime number p.

Solution. Let

I:=\{r \in R: \ \ Rr=(0)\}.

Then I is a left ideal of R because it’s clearly a subgroup of (R,+) and, for s \in R and r \in I, we have Rsr \subseteq Rr =(0) and so Rsr=(0), i.e. sr \in I. So either I=(0) or I=R.

Case 1: I=R. That means sr=0 for all r,s \in R or, equivalently, R^2=(0). Thus every subgroup of (R,+) is a left (in fact, two-sided) ideal of R. Hence (R,+) has no proper subgroup (because R has no proper left ideals) and thus |R|=p for some prime p.

Case 2: I=(0). Choose 0 \ne r \in R. So r \notin I and hence Rr=R because Rr is clearly a left ideal of R. Thus there exists e \in R such that er=r. Now

\text{ann}(r):=\{s \in R: \ sr=0\},

the left-annihilator of r in R, is obviously a left ideal of R and we can’t have \text{ann}(r)=R because then Rr=(0). So \text{ann}(r)=(0). Since

(re-r)r=rer-r^2=r^2-r^2=0,

we have re-r \in \text{ann}(r)=(0). Thus re=er=r. Let

J=\{x \in R: \ \ xe=x\}.

Clearly J is a left ideal of R and 0 \ne r \in J. Thus J=R. So xe=x for all x \in R. Now let 0 \ne r' be any element of R. Then, by what we just proved, r'e=r'. On the other hand, by the same argument we used for r, we find e' \in R such that r'e'=e'r'=r'. Thus r'(e-e')=0, i.e. r' \in \text{ann}(e-e').
So \text{ann}(e-e') \ne (0) and hence \text{ann}(e-e')=R, i.e. R(e-e')=(0) and thus e-e' \in I=(0).
So e=e' and hence r'e=er'=r' for all r' \in R. Thus e=1_R proving that R is a division ring. \Box

Remark. The same result given in the above problem holds if R has no proper right ideals.

Example. Let p be a prime number. The ring

\displaystyle R:= \left \{\begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}: \ \ \ a \in \mathbb{Z}/p\mathbb{Z}\right\} \subset M_2(\mathbb{Z}/p\mathbb{Z})

is not a division ring and it has no proper left (or right) ideals.

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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, 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

Example. It’s clear from Problem 4 that there’s no ring with only 5 units. We can prove that directly without using the result given in Problem 4, as follows. Suppose, to the contrary, that there exists a ring R such that |U(R)|=5. Then U(R) is cyclic and so it has a generator g. If 2 \ne 0 in R, then \{1,-1\} would be a subgroup of order 2 in U(R) and that is not possible because |U(R)| is odd. So 2=0. Now see that

(1+g^2+g^3)(1+g+g^4)=1

and so 1+g^2+g^3 \in U(R). Thus 1+g^2+g^3=g^n for some integer 0 \le n \le 4. Clearly we can’t have n=0,2,3 because the order of g is 5. If n=1, then 1+g^2+g^3=g gives g^2+g^4+1=g^3 and hence g^4=g^3+g^2+1=g implying that  g^3=1, which is false. Similarly, the case n=4 is impossible. So |U(R)| \ne 5.

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 \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!

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. Thus (x^2y+yx^2)x^2=x^2(x^2y+yx^2), which gives xy=yx. \ \Box