# HW 05 - MAT 322 Algebra with Galois Theory Professor Claus...

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MAT 322: Algebra with Galois Theory Professor Claus Sorensen Problem Set 5 Rik Sengupta October 23, 2009 1. Let F 4 be a field with four elements { 0 , 1 , x, y } . Write down the tables for addition and multiplication, hence showing F 4 is unique. Solution. The tables are as follows: + 0 1 x y 0 0 1 x y 1 1 0 y x x x y 0 1 y y x 1 0 and × 0 1 x y 0 0 0 0 0 1 0 1 x y x 0 x y 1 y 0 y 1 x Those are the addition and multiplication tables, and they serve an im- portant purpose. They show that the operations on the field are fixed if the field axioms are assumed, and so the field is essentially unique. Hence, they prove well-definedness and uniqueness. 2. Let D n be the dihedral group of order 2 n , where n > 2. Pick generators r and s such that r n = e , s 2 = e , and rs = sr - 1 . Write n as 2 m k , where k is odd. For each i m , show that Z i ( D n ) is generated by r n/ 2 i . Moreover, verify that Z i ( D n ) = Z m ( D n ) for every i m . Solution. When i = 0, Z 0 ( D n ) = { e } by definition, and r n = e , hence it trivially follows. We consider the case i = 1. Then, for x D n , write x = r a s b . So now, xs = r a s b +1 and sx = sr a s b = r - a s b +1 , so that sx = xs iff r - a = r a , which is true iff n | 2 a . Now, rx = r a +1 s b , and xr = r a s b r , which equals r a +1 if b = 0 and r a - 1 s if b = 1. Hence, rx = xr iff b = 0 or b = 1 and n | 2. Since n 3, we have that x Z ( D n ) iff x = r a where n | 2 a . 1

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Now, n | 2 a = n 2 | a . But 0 a < n , so that a = n 2 or a = 0, and it follows immediately that Z ( D n ) = { e, r n 2 } = h r n 2 i . Now, suppose we have shown that Z i ( D n ) = h r n 2 i i for some 0 i m - 1. Once again, let x = r a s b . Then, we have [ s, x ] = sr a sr - a = r - 2 a . On the other hand, [ x, s ] = r a s b ss b r - a s = r 2 a . Note that [ x, r ] = r a s b rs b r - a r - 1 , which equals e when b = 0 and r - 2 when b = 1. On the other hand, [ r, x ] = rr a s b r - 1 s b r - a , which equals e when b = 0 and r 2 when b = 1. Hence, we easily get [ x, y ] Z i ( D n ) for all y D n iff n 2 i | 2 a and b = 0, or when b = 1 and n 2 i | 2. It follows immediately that x Z i +1 ( D n ) iff x = r a where n 2 i +1 | a . Thus, we get Z i +1 ( D n ) = h r n 2 i +1 i , and hence, by an inductive argument, the problem is solved. For the second part, note that Z m ( D n ) = h r n 2 m i = h r k i . Once again, setting x = r a s b and using the same commutators from our calculations above, we see that x Z m +1 ( D n ) iff k | 2 a and b = 0, or when b = 1 and k | 2. Since k > 1 is odd, we have x = r a where k | a . Since 0 a < 2 m k , it follows immediately that Z m +1 ( D n ) = h r k i = Z m ( D n ). As for the trivial final step, since Z m +2 ( D n ) is defined recursively from Z m +1 ( D n ) = Z m ( D n ), it follows that Z m +2 ( D n ) = h r k i = Z m ( D n ). Therefore, by another inductive argument, we see that Z i ( D n ) = Z m ( D n ) for all i m . 3. Let R be an integral domain: A commutative ring such that R 6 = 0 , a, b R : ab = 0 = a = 0 b = 0 . Suppose R is finite, and show that R is necessarily a field. Solution. Pick any nonzero element a R . Define ϕ : R R by x 7→ ax .
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