Topics in Algebra, Chapter 3.6

This page covers section 3.6 (“The Field of Quotients of an Integral Domain”). Throughout, R is an integral domain and F is its field of fractions, defined in “topics covered”.

Topics covered: 3.6

  • The section focuses on generalizing the relationship between Z and Q (field of fractions) to integral domains in general.

  • Definition: Ring S can be imbedded in ring S if there exists an injective ring homomorphism SS. If both S and S are unital then we also require the imbedding to map 1 to 1. If S imbeds in S then S is an extension of S.

  • Theorem 3.6.1: Every integral domain may be imbedded in a field.

  • The theorem is proven by constructing the field of fractions F=R×(R0)/ where [a,b][c,d] if and only if ad=bc; here a,b,c,dR and b,d0. This is motivated by equivalence of fractions ab=cd if and only if ad=bc. R then imbeds in F in the natural way, r[rx,x] for any non-zero xR. The notation [a,b] is shorthand for [(a,b)], the equivalence class under of the element (a,b)R×(R0).

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 3.6.1: Show that multiplication in F is well-defined. (See “topics covered” for definition of notation)

Let a,a,b,b,c,c,d,dR with b,b,d,d0 and such that [a,b]=[a,b] and [c,d]=[c,d] in F. In the rational numbers, this would look like a/b=1/2 and a/b=4/8.

We have [a,b][c,d]=ac/bd and [a,b][c,d]=ac/bd. The two products are the same if acbd=acbd. Of course, we have that ab=ba and cd=dc so that acbd=badc=acbd and the multiplication is well-defined (recall that R is commutative because it is an integral domain).

Herstein 3.6.2: Show that the distributive law holds in F.

Let a,b,c,d,e,fR. We have [a,b]([c,d]+[e,f])=[a,b][cf+de,df]=[acf+ade,bdf]

and [a,b]([c,d]+[e,f])=[ac,bd]+[ae,bf]=[acbf+aebd,b2df].

The two forms are equal if (acf+ade)b2df and bdf(acbf+aebd) are equal. Multiplying each out, we see that they are both acdb2f2+aefb2d2

so that the distributive law holds.

Herstein 3.6.3: Show that ϕ:RF given by ϕ(r)=[r,1] is an injective ring homomorphism.

We do not require that R be unital, but rather we define [r,1]=[rx,x] for any non-zero xR. This is well-defined because if y is another non-zero element of R, then [rx,x]=[ry,y] because (rx)y=rxy=(ry)x.

Let a,b,xR with x0. The map respects ring addition because ϕ(a)+ϕ(b)=[ax,x]+[bx,x]=[axx+bxx,xx]=[a+b,1]=ϕ(a+b)

where we see by the comments above that [rxx,xx]=[r,1]. It also respects ring multiplication because ϕ(a)ϕ(b)=[ax,x][bx,x]=[abxx,xx]=[ab,1]=ϕ(ab).

Finally, the map is injective: ϕ(a)=ϕ(b) means that [ax,x]=[bx,x], or axx=bxx. This can be rearranged as (ab)xx=0. Now, xx0 so we must conclude that a=b because R is an integral domain.

Herstein 3.6.4: Prove that if K is a field containing R, then K contains a sub-field isomorphic to F.

In K, every element rR has a multiplicative inverse r1K which may or may not exist in R. We consider the map ϕ:FK given by ϕ([r,s])=rs1.

We must show that this map is well-defined. If [r,s]=[t,u] then ru=st. Then we see that s(rs1tu1)=rstu1=rruu1=0

which has us conclude that rs1=tu1, which is the desired statement. The map is also a homomorphism, because ϕ([r,s]+[t,u])=(ru+ts)(su)1=rs1+tu1=ϕ([r,s])+ϕ([t,u])

and ϕ([r,s][t,u])=(rt)(su)1=rs1tu1=ϕ([r,s])ϕ([t,u]).

By now we have exhibited that the field of fractions F is homomorphic to a subset of K. Last of all, the map ϕ is injective because ϕ([r,s])=ϕ([t,u]) means rs1=tu1. Rewriting this as ru=st, we see that it’s the condition that [r,s] and [t,u] are the same equivalence class.

Therefore any field containing R also contains the field of fractions of R. In that sense, F is the smallest field containing R.

Herstein 3.6.5*: Let R be a commutative, unital ring and let SR be non-empty. S is a multiplicative system if 0S and S is closed under multiplication. Define a relation on R×S such that (r,s)(r,s) if there exists sS such that

s(rssr)=0.

(a) Show that is an equivalence relation on R×S.

(b) Denote the equivalence class of (r,s) by [r,s] and let RS=R×S/. Define addition and multiplication in RS as was done for the field of fractions. Prove that RS is a ring.

(c) Can R be imbedded in RS?

(d) Prove that ϕ:RRS given by ϕ(r)=[rs,s] is a homomorphism and find its kernel.

(e) Prove that kerϕ has no intersection with S.

(f) Prove that every element in RS of the form [s1,s2] with s1,s2S is invertible in RS.

(a) Let r1,r2,r3R and s1,s2,s3S.

  1. is reflexive: We have (r1,s1)(r1,s1) because any element sS satisfies s(r1s1s1r1)=0.

  2. is symmetric: Suppose (r1,s1)(r2,s2) so there exists sS with s(r1s2s1r2)=0. As R is commutative, we also have s(r2s1s2r1)=0, which shows that (r2,s2)(r1,s1).

  3. is transitive: Suppose (r1,s1)(r2,s2) and (r2,s2)(r3,s3). Then there exist s,sS with s(r1s2s1r2)=0 and s(r2s3s2r3)=0. This is slightly tricky. We want to combine the two equations in such a way that we get terms r1s3 and s1r3, and the r2 dependence cancels out. One way to do that is to multiply the first equation by s3s and the second equation by s1s and then add the two equations together. This gives sss2(r1s3s1r3)=0

which shows that (r1,s1)(r3,s3).

(b) Let r,tR and s,uS. We define multiplication on RS by [r,s][t,u]=[rt,su] and see that RS is closed under it because suS. Suppose [r,s]=[r,s] and [t,u]=[t,u]. Then [r,s][t,u]=[rt,su]

but we can show that (rt,su)(rt,su). If s1,s2S are such that s1(rssr)=0 and s2(tuut)=0, then multiplying the first equation by s2tu and the second equation by s1sr and adding gives the desired result. Therefore multiplication is well-defined.

We define addition on RS by [r,s]+[t,u]=[ru+st,su] and again see that RS is closed under it because suS. Suppose again that [r,s]=[r,s] and [t,u]=[t,u]. Then [r,s]+[t,u]=[ru+st,su]

and again we can show that (ru+st,su)(ru+st,su) with the same technique. Therefore addition is well-defined.

The only other ring axiom to verify is the distributive property, and the proof is identical to that in problem 3.6.2. Thus RS is a ring as defined.

(c) This part deserves some commentary. First, we note that this problem (3.6.5) is concerned with generalizing the field of fractions. Before, we took S=R0 and had an equivalence relation (r,s)(r,s) if rssr=0. Now we let S be more general and employ a new equivalence relation which makes sense in the context. The other big change is that R may have zero divisors. In fact, if we choose S to have no zero divisors, then the new, more complicated-looking equivalence relation immediately reduces to the old equivalence relation. This problem is foundational for ring localizations.

The question is now whether R imbeds in RS. It can but need not. If S contains no zero divisors, then the map f(r)=[rs,s] for fixed sS is an imbedding (i.e. is one-to-one). This is true because f(r)f(r) means there exists s such that 0=s(rssr)=ss(rr). As ssS and S has no zero divisors, this implies r=r so the map f is injective.

On the other hand, if S contains a zero divisor then it may not imbed. We can build some intuition by studying a particular example, such as R=Z/6Z with S={2,4}. In that case, RS={[0,2],[1,4],[1,2]}; all other elements fall into these equivalence classes. We see that RS has 3 elements – it is smaller than R, so R clearly does not imbed in RS! The zero divisors in S cause some of the fractions to “reduce” in ways that we would not expect.

Judging by the content of the subsequent parts of the problem, it doesn’t seem that a proof of the general criterion for imbedding is called for. It appears to suffice to show that there are situations where R imbeds in RS but also situations where it does not.

(d) Let r,rR and fix sS. We have ϕ(r)ϕ(r)=[rs,s][rs,s]=[rrss,ss]. Now, ϕ(rr)=[rrs,s][rrss,ss]=ϕ(r)ϕ(r) because (rrs)(ss)s(rrss)=0. Also, ϕ(r)+ϕ(r)=[rs,s]+[rs,s]=[(r+r)s,s]=ϕ(r+r).

Hence ϕ is a homomorphism from R to RS. The kernel is the set of elements r with ϕ(r)0. That is, those elements for which there exists sS with ssr=0. Recalling the notation of previous sections, we see that kerϕ=sSAnn(s).

(e) If rkerϕ, then there exists sS with sr=0 by the comments of part (d). On the other hand, we know that S is closed under multiplication and 0S, so if r were in S then sr0. Hence kerϕS is empty.

(f) The multiplicative identity element in RS is [s,s] for any sS. We see this is true because [r,s][s,s]=[rs,ss][r,s] since (rs)s(ss)r=0. As [r,s] is some sort of generalization of the rational number r/s, the natural thing to consider is [s1,s2][s2,s1]=[s1s2,s1s2].

But s1s2S and, as we have just seen, this is the identity element! Hence any element [s1,s2] with s1,s2S is invertible in RS.

In this way, we have created something like the field of fractions. However, with the looser constraints on R and S, we end up having only some subset of the construction being invertible.

Herstein 3.6.6: Let R be an integral domain and let a,bR be such that an=bn and am=bm for m,nZ positive and coprime. Show that a=b.

The subject of the section is the field of fractions of an integral domain, which is a big hint. We know that m,n coprime implies that there exist α,βZ such that αm+βn=1. The natural thing to do is say that a=aαm+βn=bαm+βn=b. If α and β are both positive then this is fine, but they cannot be as we will see momentarily. Negative powers are tricky in R because we do not assume elements to have multiplicative inverses.

If m or n is 1 then the result is trivially true, so we will consider m,n>1 which implies that (1) neither α nor β is zero, and (2) one of α or β is positive and the other is negative. We may assume without loss of generality that α>0 and β<0. Then we prefer to write αmβn=1.

In the field of fractions F of R, we may state the fact that [aαm,aβn]=[bαm,bβn]

which follows directly from the hypotheses of the problem. This implies that aαmbβn=aβnbαm.

Now use that βn=αm1 and subtract to find aαm1bαm1(ab)=0.

Because R is an integral domain, we have either a=b or aαm1bαm1=0. The latter case is easily shown to imply that a=b=0. Therefore, a=b.

Herstein 3.6.7: Let R be a ring (not necessarily commutative) in which xy=0 implies x=0 or y=0. Let a,bR be such that an=bn and am=bm for m,nZ positive and coprime. Show that a=b.

We have no notion of field of fractions or even ring localization (problem 3.6.5) for an arbitrary ring. However, the argument presented above was only suggested by the field of fractions and didn’t actually rely on the construction. Therefore we again take α,βZ as before, with α>0 and β<0 and αmβn=1. We have aαmbβn=aαmaβn=aβnaαm=aβnbαm.

As before, we use that βn=αm1 and subtract to find aαm1(ab)bβn1=0

which again forces us to conclude that a=b.