89Now so that the product lies inF4. It also follows from this for-mula that multiplication inF4is commutative. All the other ax-ioms of a commutative ring hold inF4because they hold in the fullmatrix ring(for the cogniscenti, every subring of a (not necessarilycommutative) ring is a ring).(ii)Prove thatF4is afeld having exactly 4 elements.Solution.If a matrixA=£abba+b±inI4is nonzero, thena±=0orb±=0. Thus,det(A)=a±=0ifb=0;b±=fa=0;1ifa=1=b.Thus, ifA±=0, then det(A)=1, and so the matrixA−1exists.As usual,A−1=£a+bb±, and so it is only a question of whetherthis matrix lies inI4; that is, is the 2,2 entry the sum of the entriesin row 1? The answer is yes, for(a+b)+b=a.(iii)Show thatI4is not afeld.Solution.The commutative ringI4is not even a domain, for ±=and=.3.20Prove that every domainRwith afnite number of elements must be afeld.Solution.LetR×denote the set of nonzero elements ofR. The cancella-tion law can be restated: for eachr∈R×, the functionµr:R×→R×,defned byµr:x7→rx, is an injectionR×→R×. SinceR×isfnite,Exercise 2.13 shows that every
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