Adv Alegbra HW Solutions 89

Adv Alegbra HW Solutions 89 - 89 Now so that the product...

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89 Now so that the product lies in F 4 . It also follows from this for- mula that multiplication in F 4 is commutative. All the other ax- ioms of a commutative ring hold in F 4 because they hold in the full matrix ring(for the cogniscenti, every subring of a (not necessarily commutative) ring is a ring). (ii) Prove that F 4 is a f eld having exactly 4 elements. Solution. If a matrix A = £ ab ba + b ± in I 4 is nonzero, then a ±= 0or b ±= 0. Thus, det ( A ) = a ±= 0i f b = 0 ; b ±= f a = 0 ; 1i f a = 1 = b . Thus, if A ±= 0, then det ( A ) = 1, and so the matrix A 1 exists. As usual, A 1 = £ a + bb ± , and so it is only a question of whether this matrix lies in I 4 ; that is, is the 2 , 2 entry the sum of the entries in row 1? The answer is yes, for ( a + b ) + b = a . (iii) Show that I 4 is not a f eld. Solution. The commutative ring I 4 is not even a domain, for [ 2 ] ±= [ 0 ] and [ 2 ][ 2 ]=[ 4 0 ] . 3.20 Prove that every domain R with a f nite number of elements must be a f eld. Solution. Let R × denote the set of nonzero elements of R . The cancella- tion law can be restated: for each r R × , the function µ r : R × R × , de f ned by µ r : x 7→ rx , is an injection R × R × . Since R × is f nite, Exercise 2.13 shows that every
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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