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Unformatted text preview: MAT 322: Algebra with Galois Theory Professor Claus Sorensen Problem Set 8 Rik Sengupta December 4, 2009 1. Let F be a field of characteristic p > 0, and fix an algebraic closure F . For each k > 0, show that the following subset is a subfield, F p k = { x F : x p k F } F. Let F per be the union of all these subfields. Show that F per is the smallest perfect subfield of F , containing F . That is, the perfect closure of F in F . Solution. Fix some k > 0. Then, suppose a,b F p k . Then, from definition, a p k F , and b p k F . Then, we have, by the Frobenius mapping, ( a b ) p k = a p k b p k F , and ( a + b ) p k = a p k + b p k F . Also, obviously, ( ab ) p k = a p k b p k F , and ( a 1 ) p k = ( a p k ) 1 F . So, applying the definition once more, it clearly follows that a + b , a b , ab and a 1 are all in F p k . Furthermore, 0 p k = 0, and 1 p k = 1, and so 0 and 1 are obviously in F p k . So, it follows that F p k is a field. Now, since for all a F p k , we must have x p k f = 0 for some f F , and so a is algebraic over F . So, F p k is a subfield of F . 2 Now, note that F = F p F p 1 F p 2 .... Thus, F per is the union of an ascending chain of fields. Therefore, it is still a field, and in particular, we know F per is a subfield of F . Suppose we choose E to be the smallest subfield of F containing F that is perfect. In fact, since E is perfect, we know char E = char F = p , and so E p = E . Hence, we know F E implies that all p th roots of the elements of F are in E . In other words, F p 1 E . By induction, then, F p k E for all k = 1 , 2 ,... . 1 So, it follows that F per E . We will show now that F per is perfect. But this is immediate, because of the following reasoning. We know that char F per = p , which is obvious. Now, given any a F per , we must have a F p k for some k , and then there exists some b F such that b p = a , and so, b p k +1 F , which implies that b F p k 1 F per , i.e. for any a F per , there is a p th root of a also in F per . So, this is equivalent to proving that F per is perfect. But then, by the minimality of E , we must have E F per . Hence, F per = E , i.e. F is the smallest perfect subfield of F containing F . 2 2. Let E/F be a finite extension, and assume E is a perfect field. Show that F is perfect. Is this true in general for algebraic extensions? Solution. Obviously, if char E = 0, then we must have char F = 0, and so F is perfect. So suppose char E = p > 0. Then, because E is perfect, we have E p = E . Also, char F = p > 0. Consider the ascending chain of fields F F p 1 F p 2 ......
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 '09
 CLAUSSORENSEN
 Algebra

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