Archive for the ‘Elementary Algebra; Problems & Solutions’ 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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Definition 1. Let n be an integer. A group G is called nabelian if (xy)^n=x^ny^n for all x,y \in G.
In other words, G is n-abelian if the map f : G \to G defined by f(x)=x^n is a group homomorphism.

Definition 2. If G is both n-abelian and m-abelian, for some integers m,n, then G is also mn-abelian because then for x,y \in G we will have

(xy)^{mn}=((xy)^m)^n=(x^my^m)^n=x^{mn}y^{mn}.

So the set

\text{E}(G)=\{n \in \mathbb{Z}: \ \ (xy)^n=x^ny^n, \ \ \forall x,y \in G\},

i.e., the set of those integers n for which G is n-abelian, is a multiplicative subset of \mathbb{Z}. Clearly 0,1 \in \text{E}(G). The set \text{E}(G) is called the exponent semigroup of G.

Remark 1. Since (xy)^n=x(yx)^{n-1}y, we have (xy)^n=x^ny^n if and only if (yx)^{n-1}=x^{n-1}y^{n-1}. So a group G is n-abelian if and only if (yx)^{n-1}=x^{n-1}y^{n-1} or, equivalently, (xy)^{1-n}=x^{1-n}y^{1-n} for all x,y \in G. So a group G is n-abelian if and only if it is (1-n)-abelian. In other words, n \in \text{E}(G) if and only if 1-n \in \text{E}(G).

Example 1. Every abelian group is obviously n-abelian for all n. It is also clear that 2-abelian groups are abelian because xyxy=(xy)^2=x^2y^2=xxyy gives yx=xy. As the next two examples show, there exists a non-abelian n-abelian group for any n > 2.

Example 2. Let n \ge 3 be an odd integer and consider the Heisenberg group G:=H(\mathbb{Z}/n\mathbb{Z}), which is a non-abelian group (why?). We show that n, n+1 \in \text{E}(G), i.e. G is both n-abelian and (n+1)-abelian. To see that, let

g= \displaystyle \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} \in G.

An easy induction shows that

\displaystyle g^m= \begin{pmatrix}1 & ma & mb + \frac{m(m-1)}{2}ac \\ 0 & 1 & mc \\ 0 & 0 & 1 \end{pmatrix}

for all integers m \ge 1. Thus g^n=I, the identity matrix, and so g^{n+1}=g, for all g \in G.
Hence I=(xy)^n=x^ny^n and xy=(xy)^{n+1}=x^{n+1}y^{n+1} for all x,y \in G proving that G is both n-abelian and (n+1)-abelian.

Remark 2.  Notice that, in Example 2, if n is even, then \displaystyle \frac{n(n-1)}{2} \ne 0 (as elements of \mathbb{Z}/n\mathbb{Z}) and so g^n is not always the identity element in this case. However, there’s a way to fix this, as the next example shows. But first, notice that H(\mathbb{Z}/n\mathbb{Z}) can also be viewed as the group of  triples

G:=\{(a,b,c): \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z}\}

with multiplication defined by

(a,b,c)(a',b',c')=(a+a',b+b'+ac',c+c').

In the next example, we modify the above multiplication such that we get g^n=1 for all g \in G.

Example 3. Let n \ge 3 be any integer and consider the set G:=\{(a,b,c): \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z}\}. Define multiplication in G by

(a,b,c)*(a',b',c')=(a+a',b+b'+2ac',c+c').

It’s easy to see that (G,*) is a non-abelian group. We show that n, n+1 \in \text{E}(G). Let g=(a,b,c) \in G. A quick induction shows that

g^m=(ma, mb+m(m-1)ac, mc)

for all integers m \ge 1. So g^n=(0,0,0)=1_G for all g \in G and thus 1_G=(xy)^n=x^ny^n for all x,y \in G proving that G is n-abelian.
Also, since g^{n+1}=g for all g \in G, we have xy=(xy)^{n+1}=x^{n+1}y^{n+1} and so G is (n+1)-abelian.

Problem 1. Let G be a group. Show that

i) if n, n+1 \in \text{E}(G) for some integer n, then x^n \in Z(G) for all x \in G

ii) if n, n+1,n+2 \in \text{E}(G) for some integer n, then G is abelian.

Solution. i) Let x,y \in G. We have

xyx^ny^n=xy(xy)^n=(xy)^{n+1}=x^{n+1}y^{n+1}

and so yx^n=x^ny.

ii) Let x,y \in G. By i), we have yx^n=x^ny and yx^{n+1}=x^{n+1}y. Thus

x^{n+1}y=yx^{n+1}=yx^nx=x^nyx

and so xy=yx. \ \Box

Problem 2. Let G be an n-abelian group and let x,y \in G. Show that

i) x^{n-1}y^n=y^nx^{n-1}

ii) (xyx^{-1}y^{-1})^{n(n-1)}=1

Solution. i) By Remark 1, (ab)^{n-1}=b^{n-1}a^{n-1} for all a,b \in G. Thus

\begin{aligned} x^{n-1}y^{n-1}=(yx)^{n-1}=((yxy^{-1})y)^{n-1}=y^{n-1}(yxy^{-1})^{n-1}=y^{n-1}yx^{n-1}y^{-1}=y^nx^{n-1}y^{-1} \end{aligned}

and the result follows.

ii) Again, using the Remark 1, we have

\begin{aligned} (xyx^{-1}y^{-1})^{n(n-1)}=((x(yx^{-1}y^{-1}))^{n-1})^n=((yx^{-1}y^{-1})^{n-1}x^{n-1})^n=(yx^{-(n-1)}y^{-1}x^{n-1})^n \\ =y^n(x^{-(n-1)}y^{-1}x^{n-1})^n=y^nx^{-(n-1)}y^{-n}x^{n-1}=1, \ \ \ \text{by i)}.  \ \Box \end{aligned}

Remark 3. Let G be an n-abelian group for some integer n \ge 2. An immediate result of Problem 2, ii), is that if either G is torsion-free or G is finite and \text{gcd}(n(n-1), |G|)=1, then G is abelian.

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

Problem.  Show that if \displaystyle F \subseteq \mathbb{R} is a field and x_1, \cdots , x_n are positive real numbers such that \sum_{i=1}^n x_i \in F and x_i^2 \in F for all i, then x_i \in F for all i.

Solution. The proof is by induction over n. There’s nothing to prove for n=1. For n\ge 2, since \sum_{i=1}^n x_i \in F, we have

\displaystyle \sum_{i=2}^n x_i \in F(x_1)=F + Fx_1

and thus, by the induction hypothesis, x_i \in F+ Fx_1 for all i \ge 2. So x_2=a+bx_1 for some a,b \in F.

If b=0, then x_2 \in F and so \sum_{i \ne 2}x_i \in F. Hence, by the induction hypothesis, x_i \in F for all i.

If a=0, then x_2=bx_1 (so b > 0 because x_1,x_2 > 0). Thus

\displaystyle \sum_{i=1}^n x_i = (b+1)x_1 + \sum_{i \ge 3} x_i \in F

and so, by the induction hypothesis, x_i \in F for all i.

If a, b \ne 0, then x_2^2=a^2+b^2x_1^2+2abx_1 gives x_1 \in F and so \sum_{i \ge 2} x_i \in F implying, again by the induction hypothesis,that x_i \in F for all i. \ \Box

Example. Show that \mathbb{Q}(\sqrt{2}+ \sqrt{5}+ \cdots +\sqrt{n^2+1})=\mathbb{Q}(\sqrt{2}, \sqrt{5}, \cdots , \sqrt{n^2+1}). for all integers n \ge 1.

Solution. Let F:=\mathbb{Q}(x_1 + \cdots + x_n), where x_i=\sqrt{i^2+1} for all i. Since \sum_{i=1}^n x_i \in F and x_i^2 \in F for all i, we have, by the above problem, x_i \in F for all i. Thus \mathbb{Q}(x_1, \cdots , x_n) \subseteq F and we are done because obviously F \subseteq \mathbb{Q}(x_1, \cdots , x_n).

Definition. For an integer n \ge 1, the n\times n Hilbert matrix is defined by H_n=[a_{ij}], where

\displaystyle a_{ij}=\frac{1}{i+j-1}, \ \  1 \le i,j \le n.

It is known that H_n is invertible and if H_n^{-1}=[b_{ij}], then \displaystyle \sum_{i,j}b_{ij}=n^2. We are going to use these two properties of Hilbert matrices to solve the following calculus problem.

Problem. Let n \ge 1 be an integer and let f : [0,1] \longrightarrow \mathbb{R} be a continuous function. Suppose that \displaystyle \int_0^1 x^kf(x) \ dx = 1 for all  0 \le k \le n-1. Show that \displaystyle \int_0^1 (f(x))^2 dx \ge n^2.

Solution. Since H_n, the n\times n Hilbert matrix, is invertible, there exist real numbers p_0, p_1, \cdots , p_{n-1} such that

\displaystyle \sum_{i=1}^n\frac{p_{i-1}}{i+j-1}=1, \ \ \ 1 \le j \le n.

So the polynomial \displaystyle p(x)=\sum_{k=0}^{n-1}p_kx^k satisfies the conditions

\displaystyle \int_0^1x^k p(x) \ dx =1, \ \ \ 0 \le k \le n-1.

Clearly \displaystyle \sum_{k=0}^{n-1}p_k is the sum of all the entries of H_n^{-1} and so \displaystyle \sum_{k=0}^{n-1}p_k=n^2. Now let f be a real-valued continuous function on [0,1] such that

\displaystyle \int_0^1x^kf(x) \ dx  = 1, \ \ \ 0 \le k \le n-1.

Let p(x) be the above polynomial.Then since

\displaystyle (f(x))^2-2f(x)p(x)+(p(x))^2 =(f(x)-p(x))^2 \ge 0,

integrating gives

\displaystyle \begin{aligned} \int_0^1 (f(x))^2dx \ge 2\int_0^1f(x)p(x) \ dx -\int_0^1(p(x))^2dx=2\sum_{k=0}^{n-1}p_k \int_0^1 x^kf(x) \ dx - \\ \sum_{k=0}^{n-1}p_k\int_0^1x^kp(x) \ dx = 2\sum_{k=0}^{n-1}p_k-\sum_{k=0}^{n-1}p_k=\sum_{k=0}^{n-1}p_k =n^2. \ \Box \end{aligned}

Problem. Let A \in M_3(\mathbb{R}) be orthogonal and suppose that \det(A)=-1. Find \det(A-I).

Solution. Since A is orthogonal, its eigenvalues have absolute value 1 and it can be be diagonalized. Let D be a diagonal matrix such that PDP^{-1}=A for some invertible matrix P \in M_3(\mathbb{C}). Then

\det(D)=\det(A)=-1, \ \ \det(D-I)=\det(A-I).

We claim that the eigenvalues of A are \{-1,e^{i\theta},e^{-i\theta}\} for some \theta. Well, the characteristic polynomial of A has degree three and so it has either three real roots or only one real root. Also, the complex conjugate of a root of a polynomial with real coefficients is also a root. So, since \det(A)=-1, the eigenvalues of A are either all -1, which is the case \theta=\pi, or two of them are 1 and one is -1, which is the case \theta = 0, or one is -1 and the other two are in the form \{e^{i\theta}, e^{-i\theta}\} for some \theta. So

\displaystyle \begin{aligned} \det(D-I)=\det(A-I)=(-2)(e^{i\theta}-1)(e^{-i\theta}-1)=-4(1-\cos \theta). \ \Box \end{aligned}

Note that given \theta, the matrix

\displaystyle A=\begin{pmatrix} -1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{pmatrix}

is orthogonal, \det(A)=-1 and \det(A-I)=-4(1-\cos \theta). \ \Box