Topics in Algebra, Chapter 3.2

2012-11-11 math algebra topics-in-algebra

There are no exercises from 3.1, so this section covers both 3.1 (“Definition and Examples of Rings”) and 3.2 (“Some Special Classes of Rings”). Throughout, R is a ring and p is a prime integer.

Topics covered: 3.1

  • Definition: An associative ring is a non-empty set R with two operations, + and obeying the usual axioms: R is an abelian group under +, R is closed under the associative multiplication , and the distributive law holds in R.

  • If R contains a multiplicative identity element then it is called unital.

  • R=Z is a ring under the usual operations. R is commutative and unital.

  • R=2Z, the even integers, form a commutative ring under the usual operations but R is not unital.

  • R=Q is a commutative, unital ring, but also a field.

  • R=Z/7Z is a field under addition and multiplication modulo 7.

  • R=Z/6Z is a ring with zero divisors, e.g. 23=0.

  • R=Mat2×2(Q), the set of 2×2 matrices with elements taken from Q, under the usual matrix addition and multiplication, is a non-commutative ring. It is unital and has zero-divisors.

  • R=C is a field.

  • The real quaternions (α0+α1i+α2j+α3k with the usual relations on i,j,k and αnR) form a non-commutative ring, but also a division ring, where every element is invertible but its left and right inverses may be distinct.

Topics covered: 3.2

  • Definition: With R commutative, a non-zero element aR is a zero divisor if there is non-zero element bR such that ab=0.

  • Definition: An integral domain is a commutative ring with no zero divisors.

  • Z is an integral domain.

  • Definition: A field is a commutative, unital ring where every element has a multiplicative inverse.

  • Lemma 3.2.1: For all a,bR, a0=0a=0, a(b)=(a)b=(ab) and (a)(b)=ab. Furthermore, if R is unital, then (1)a=a and (1)(1)=1. In essence, we can treat negatives and zero the same way that we do in ordinary arithmetic.

  • Pigeonhole principle

  • Lemma 3.2.2: If an integral domain is finite, then it is a field.

  • Z/pZ is a field.

  • Definition: With R an integral domain, aR non-zero, and mZ, R is of characteristic zero if ma=a+a++a=0 implies m=0.

  • Z, Q, etc. are of characteristic zero

  • Definition: With R an integral domain, R is of finite characteristic if there exists a positive integer m such that ma=0 for all aR. In that case, the smallest such m is the characteristic of R.

The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book.

Herstein 3.2.1: Let a,b,c,dR. Evaluate (a+b)(c+d).

By the distributive property, (a+b)(c+d)=(a+b)c+(a+b)d=ac+bc+ad+bd. Notably, the multiplication does not have to be commutative, so the order of the four products cannot be ignored.

Herstein 3.2.2: Let a,bR. Show that (a+b)2=a2+ab+ba+b2 where x2=xx.

This is an application of 3.2.1, with c=a and d=b.

Herstein 3.2.3: How does the binomial theorem generalize to an arbitrary ring?

First recall the traditional case. With a,b real numbers and n a non-negative integer, we have (a+b)n=k=0n(nk)akbnk.

Because integer multiplication is commutative, we were able to collect terms that were equivalent modulo moving factors around. Doing that collection gave us the factors of (nk): every term in the expansion has n factors, there are (nk) ways of choosing k of those factors to be a’s (and the rest are automatically b’s). In the non-commutative case, we won’t be able to collect and add up like terms in that way, but there will still be 2n terms to be added up, each being one of the ways of lining up n copies of the a’s and b’s.

Now consider a general ring R and let a,bR. The statement of the above generalization is that (a+b)n=ϵ1=01ϵn=01aϵ1b1ϵ1aϵ2b1ϵ2aϵnb1ϵn.

In fact, this is an almost contentless rewriting of (a+b)n because the n sums decouple into a bunch of ϵ=01aϵb1ϵ=a+b.

Nevertheless, the insight is that we have to enumerate all 2n of the length-n strings made up of a’s and b’s. It is unlikely that there is much more that we can do in a general non-commutative setting.

Herstein 3.2.4: Show that R is commutative if x2=x for every xR. (“Boolean ring”)

We must show that xy=yx for any x,yR. The natural thing to consider, in light of what is given, is (x+y)2=x2+xy+yx+y2=x+y+xy+yx.

Of course, we are also given that (x+y)2=x+y, so xy+yx=0.

In fact, (r)2=r2 by Lemma 3.2.1, so the Boolean condition also gives us that x=x2=(x)2=x, i.e. that every element is its own additive inverse. Therefore xy=yx=yx,

and the claim is proven.

Herstein 3.2.5: With a,bR and m,nN, we define ma as the sum of m copies of a. Prove that (ma)(nb)=(mn)ab.

By the distributive property, (ma)(nb)=(a++a)(b++b)=a(b++b)++a(b++b)=n(ab)++n(ab)=(mn)(ab).\begin{aligned} (ma)(nb)&=(a+\cdots+a)(b+\cdots+b)=a(b+\cdots+b)+\cdots+a(b+\cdots+b)\
=&n(ab)+\cdots+n(ab)=(mn)(ab). \end{aligned}

This illustrates the idea, and the inductive proof is trivial.

Herstein 3.2.6: Show that if R is an integral domain of finite characteristic, then its characteristic is prime.

Let the characteristic of R be k so that ka=0 for all aR. Furthermore, suppose that k is composite and write k=mn with m,n integers larger than 1. Let bR be an element such that nb0. Such an element exists because otherwise we would say that n<k was the characteristic. Also, let aR be arbitrary. By problem 3.2.5, we have that (ma)(nb)=(mn)ab=0 because mn=k is the characteristic. Now, because R is an integral domain, this forces either ma=0 or nb=0. The latter is explicitly ruled out by assumption, and therefore ma=0 for every aR, which is the contradiction we seek (the characteristic is defined as the least such positive integer). Thus, k must not admit any non-trivial factorizations, i.e. it must be prime.

Herstein 3.2.7: Give an example of an integral domain with infinitely many elements but finite characteristic.

Let R be the set of polynomials, in a variable x, with coefficients taken from Z/2Z. This set is infinite (we can enumerate a countably infinite subset by considering {xkk=0,1,2,}) and forms a ring under the familiar addition and multiplication (i.e. powers of x are increased by multiplication). Furthermore, it is an integral domain because multiplying non-zero polynomials produces a polynomial of equal or larger degree. More concretely, if p,qR have leading terms xn and xm, respectively (the coefficients must be one to be relevant as leading terms), then pqR has leading term xn+m, and hence pq0.

Finally, R has characteristic 2 because twice any coefficient is zero in Z/2Z.

Herstein 3.2.8: Let R be an integral domain and suppose there exists nN and non-zero aR such that na=0. Prove that R has finite characteristic.

Let bR and consider 0=(na)b=a(nb). The last manipulation is justified by problem 3.2.5. As R is an integral domain and a0, we have that nb=0 for any bR. This implies that the characteristic of R is finite, less than or equal to n.

Herstein 3.2.9: If R is a set satisfying all the ring axioms except possibly commutativity of addition, and R has a multiplicative identity 1, then prove that addition must be commutative in R, and so R is a ring.

Following the hint in Herstein, we let a,bR and compute (a+b)(1+1) in two different ways. First, (a+b)(1+1)=(a+b)+(a+b)=a+b+a+b.

Second, (a+b)(1+1)=a(1+1)+b(1+1)=a+a+b+b.

These must be equal, a+b+a+b=a+a+b+b, and subtracting a from the left and b from the right on both sides, we find b+a=a+b, the desired result.

Herstein 3.2.10: Let R be a commutative ring and let a,b,cR. Show that R is an integral domain if and only if ab=ac with a0 implies that b=c. In other words, R is an integral domain if and only if one can cancel out a common non-zero factor in an equation.

If R is an integral domain and a0, then 0=abac=a(bc) forces b=c.

On the other hand, suppose that we can cancel out common non-zero factors in an equation. Let a,bR be such that ab=0 and assume that a0. Then ab=a0 implies b=0. Alternately, suppose b0 and we have no information about a. Then ab=0b implies a=0. Therefore, for the product to be zero, one of a or b must be zero, which is the statement that R is an integral domain.

Herstein 3.2.11: Prove that an infinite integral domain need not be a field (compare with Lemma 3.2.2).

Z is an infinite integral domain which is not a field.

Herstein 3.2.12: Prove that a field is an integral domain.

A field is a commutative ring where the non-zero elements form a group under multiplication, so a product of non-zero elements will never give zero. This is the definition of an integral domain.

Herstein 3.2.13: Let a,b,m,nZ with (m,n)=1. Use the pigeonhole principle to show that there exists an integer x such that xamodm and xbmodm.

Note that this problem is identical to problem 1.3.15, except for the new suggestion to use the pigeonhole principle. Also note that this is a special case of the chinese remainder theorem.

As suggested by Herstein, consider the set S={a,a+m,a+2m,,a+(n1)m}

which contains n integers. Clearly every element of S is congruent to a modulo m. If we could also show that one element of S was congruent to b modulo n, then that would be the desired x. If it were the case that no element of S is congruent to b modulo n, then the pigeonhole principle says that, of the n elements of S computed modulo n, some output occurs at least twice, say a+ima+jmmodn for some 0i,j<n and ij. Then we have (a+im)(a+jm)=(ij)m0modn.

Because m is coprime to n, it can be canceled, leaving ij0modn which is impossible under the conditions imposed on i and j. Therefore, there is an element of S congruent to b modulo n, and it is the desired x.

Herstein 3.2.14: Use the pigeonhole principle to prove that the decimal expansion of a rational number must eventually become repeating.

Consider a rational number m/n, with m and n coprime integers and n0. We can compute the decimal expansion by the Euclidean algorithm (i.e. long division). We proceed as m=a0n+r0,10r0=a1n+r1,10r1=a2n+r2,,

where each remainder rk{0,,n1}, and then we find the decimal expansion to be the string of digits a0a1a2 with a decimal point in the appropriate position. Now, once we find the kth remainder, rk, we find the next digit in a straightforward way, ak+1=10rkn.

The subsequent remainder rk+1 is similarly found, and the sequence proceeds from there entirely deterministically. Therefore, if a remainder is ever repeated, it signals that the decimal has begun to repeat and will continue indefinitely. There are n values allowed for the remainder, and infinitely many digits to compute, so the pigeonhole principle says that this repetition must happen, but also that the first period will terminate before the (n+1)-st digit.

As an example, consider the fraction 5/7, m=5 and n=7. The process described above manifests as 5=07+5a0=0,r0=550=77+1a1=7,r1=110=17+3a2=1,r2=330=47+2a3=4,r3=220=27+6a4=2,r4=660=87+4a5=8,r5=440=57+5a6=5,r6=550=77+1a7=7,r7=1\begin{gathered} 5=0\cdot7+5\qquad\longrightarrow&\qquad a_0=0,\qquad r_0=5\
50=7\cdot7+1\qquad\longrightarrow&\qquad a_1=7,\qquad r_1=1\
10=1\cdot7+3\qquad\longrightarrow&\qquad a_2=1,\qquad r_2=3\
30=4\cdot7+2\qquad\longrightarrow&\qquad a_3=4,\qquad r_3=2\
20=2\cdot7+6\qquad\longrightarrow&\qquad a_4=2,\qquad r_4=6\
60=8\cdot7+4\qquad\longrightarrow&\qquad a_5=8,\qquad r_5=4\
40=5\cdot7+5\qquad\longrightarrow&\qquad a_6=5,\qquad r_6=5\
50=7\cdot7+1\qquad\longrightarrow&\qquad a_7=7,\qquad r_7=1\

comments powered by Disqus