Topics in Algebra, Chapter 3.5

2012-12-07 math algebra topics-in-algebra

This page covers section 3.5 (“More Ideals and Quotient Rings”). Throughout, R is a ring.

Topics covered: 3.5

  • Lemma 3.5.1: Let R be commutative and unital and only have the ideals (0) and R. Then R is a field.

  • Definition: An ideal IR is a maximal ideal if any other ideal J such that IJR is either J=I or J=R.

  • The prime ideals of Z are exactly the ideals (p)={kpkZ} for p a prime integer. See also: problem 3.4.8.

  • Theorem 3.5.1: Let R be commutative and unital and let I be an ideal of R. Then R/M is maximal if and only if R/M is a field.

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 3.5.1: Let R be unital but not necessarily commutative, and such that the only right ideals of R are (0) and R. Show that R is a division ring.

Let a0 be an element of R and consider aR={arrR} which was shown to be a right ideal in problem 3.4.14. Because 1R, we have that a10 is a non-zero element of aR. By the conditions of the problem, we conclude that aR=R. In particular, there exists bR such that ab=1. We would like to show also that ba=1. If that is the case, then we have exhibited a multiplicative inverse in R for an arbitrary non-zero ring element.

Considering bR and arguing as above, we find an element cR such that bc=1. However, we see that abc=a(bc)=a1=a while abc=(ab)c=1c=c so that a=c. Therefore, ab=ba=1 and we have proved that every non-zero element of R has a multiplicative inverse, so that R is a division ring.

Herstein 3.5.2*: Let R be such that the only right ideals of R are (0) and R. Prove that either R is a division ring or that R has a prime number of elements and ab=0 for every a,bR.

This is related to problem 3.5.1 in that we only relax the condition on R being unital. My solution was heavily influenced by the nice solution by AllTheCheese in this forum thread.

Case 1: there exists some aR such that aR=R. In that case, there exists an element eR such that ae=a. We would like to show that this e is the multiplicative unit for the ring, which will put us in the realm of problem 3.5.1. First recall the annihilator Ann(a)={xRax=0} shown in problem 3.4.17 to be a right ideal of R. As eAnn(a), we must have that Ann(a)=(0), so a annihilates nothing but the zero element. However, right-multiplying the equation ae=a by a, we find 0=aeaaa=a(eaa).

In light of the comment about Ann(a), we have that ea=a.

Now take any other element bR; we would like to show that eb=be=e so that e is truly the unit element. Recalling that aR=R, there exists xR such that ax=b, so eb=eax=ax=b. It is more difficult to show that be=b.

Let be=c and multiply on the right by e to see that be=ce (e2=e, see below for the proof*). Then (bc)e=0. Recall problem 3.4.18 which tells us that λ(Re)={xRxre=0 for all rR} is a right ideal. Putting x=a and r=e, we see that aee=ae=a0, so there exists at least one element of R which doesn’t belong to λ(Re). With our strict condition on ideals, this means that λ(Re)=(0). In other words, there does not exist a non-zero element xR for which xre=0 for all rR. Letting x=bc, and assuming x0, we see that there is at least some rR such that (bc)re0, which forces us to conclude that Ann(bc)=(0). However, we have already seen that (bc)e=0, a contradiction. The only way out is to have that c=b, which shows that be=b, after all. Now we have shown that case 1 leads to the ring having a unit, and therefore being a division ring by problem 3.5.1.

Case 2: there is no aR such that aR=R; this is the complement of case 1. Then aR=(0) always so that ab=0 for any a,bR. Now, let aR be a non-zero element. If we consider the set I={0,a,2a,3a,}, we see that it is a right ideal because multiplication from the outside always yields 0 which belongs to I, while closure under addition is clear. Also, I contains the non-zero element a, so I=R. Now, if R is finite, the sequence 0,a,2a will yield R distinct elements and then loop around to elements it has already visited. In particular, if R is composite, say R=mn with m,n>1 integers, then ma is non-zero, and J={0,ma,2ma,} is a right ideal with a non-zero element, but it is a proper subset of I. By the condition of the problem, such a J may not exist. Therefore if R is in case 2 and is finite, then R must be prime.

Finally, we must also consider case 2 with R infinite. In fact, this is not possible. Simply consider the ideal J={0,2a,4a,} which is a proper subset of I with a non-zero element. As before, this is a contradiction of the terms of the problem.

Here is a brief digression from the main course of the proof to show that e2=e for the alleged identity element e of R. Recall that the non-zero element a satisfied ae=ea=a and ax=0 implied x=0. Now, ea0 implies that eR(0), so eR=R. In particular, there exists dR with ed=e. Multiplying on the left by a gives 0=aedae=adae=a(de). As a only annihilates the zero element, we must have that d=e, so e2=e as claimed.

Herstein 3.5.3: Let p be a prime integer and let (p) be the ideal of Z consisting of all multiples of p.

(a) Show that Z/(p)Zp, where Zp is the ring with elements {0,1,,p1} under addition and multiplication modulo p.

(b) Show that Z/(p) is a field.

(a) There is a natural homomorphism to consider, namely ϕ:Z/(p)Zp

given by ϕ(a+(p))=amodp, where on the right hand side a is reduced modulo p so as to live in Zp. To show that this is well-defined, let a,aR be such that a+(p)=a+(p). Then aa(p) so they differ by a multiple of p and the map is well-defined. It is also easily seen to be a homomorphism. Because the kernel is trivial, ϕ is an isomorphism.

(b) By problem 3.4.8, we know that (p) is a maximal ideal of Z, and by theorem 3.5.1, we therefore have that Z/(p) is a field, because it is a commutative unital ring (Z) modded out by a maximal ideal.

Herstein 3.5.4**: Let R=C[0,1], the ring of continuous functions [0,1]R. If M is a maximal ideal of R, prove that there exists γ[0,1] such that M={fRf(γ)=0}.

First we note that an invertible element in this ring is one which has no zeroes. The multiplicative identity element is the constant function x1, and if f has no roots then it has a multiplicative inverse given by x1/f(x). If an ideal contains an invertible element, then it is all of R. Therefore every element of a proper ideal must have a root.

The sketch of the proof is as follows:

  1. Let I be a proper ideal of R. Any two elements f,gI have a root in common.
  2. Every element of I has at least one root in common. In other words, there is a non-empty set ZI such that every element f of I satisfies f(ZI)=0.
  3. A maximal ideal M must have ZM=1 and thus be of the stated form.

(1) Let f,gI. We have that f2, g2 and therefore f2+g2 are also in I because it is an ideal. Suppose that f and g share no root. Then at any root x of f, we have f(x)2+g(x)2=g(x)2>0. Similarly f(x)2+g(x)2=f(x)2>0 at any root x of g. Of course, f2+g2>0 elsewhere. Therefore if f and g share no root, then I contains the invertible element f2+g2, and hence I=R. We are interested in proper ideals of R, in which, as we have just seen, any pair of elements has a root in common.

We can say something more general than the pairwise statement, though. Given any finite subset of a proper ideal I, say {f1,,fn}I, the same argument shows that all members of that subset share at least one zero. Otherwise ifi2 would be an invertible element which forces I=R.

With fI and writing Zf for the (non-empty) set of zeroes of f, we have just shown that the collection {ZffI} has the finite intersection property. Any finite subcollection of that collection has non-empty intersection.

(2) We should now like to extend the statement above to cover the existence of a root shared by everything in the ideal. That is, we would like to prove that ZI=fIZf.

It is not true in general that the finite intersection property extends to a non-empty intersection on the whole collection. However, we have additional conditions at our disposal.

Let fR. Because f is continuous, we know that the inverse image under f of an open set is open. In particular, f1(R0) is open. Its complement, f1(0)=Zf, is therefore closed in [0,1]. That is, every zero set Zf that we consider is topologically closed.

Additionally, [0,1] is a compact set. In a compact metric space such as [0,1], any collection of closed sets with the finite intersection property has a non-empty intersection (proof). This follows simply by taking the definition that “every open cover of a compact set has a finite subcover” and turning it into a statement about the closed sets given by the complements of the sets in the cover. Therefore we have that, for any proper ideal I of R, there is a nonempty set ZI on which every member of I takes the value zero.

(3) Now we want to determine what a maximal ideal M looks like. If ZM>1, then M may not be maximal. If ZM>1, then we could take a proper, non-empty subset ZJZM and construct the proper ideal J={fRf(ZJ)=0}. If fM, then f(ZJ)f(ZM)=0, so we also have fJ and therefore MJ. In other words, reducing the size of the zero set makes the ideal bigger — the bigger ideal contains all the functions that were zero on the expanded zero set, but it also contains new functions that take other values on the roots that were excluded. On a side note, it may be true that zero sets are in one-to-one correspondences with ideals, but I do not know at this time. For the purposes of this exercise, it suffices to construct one ideal from a given zero set, as we constructed J earlier in this paragraph.

Thus it must be the case that a maximal ideal M has exactly one shared root among its elements, ZM=1. Calling that root γ, we must have M as a subset of M={fRf(γ)=0}. However, we easily verify that M is itself an ideal, so any proper subset of it would clearly not be maximal. Therefore, our maximal ideal is M and the claim is proven.

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