Homework11 - x 4-2 x 2-4 which is irreducible by the Mod-p...

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Homework 11 Solutions 1. Since 3 5 is the root of x 3 - 5 which is irreducible over Q we know Q ( 3 5) = Q [ x ] / ( x 3 - 5) and the elements in the latter have the form ax 2 + bx + c + ( x 3 - 5) and so the corresponding elements in the former have the form a ( 3 5) 2 + b ( 3 5) + c with a, b, c Q . 2. Q ( 2 + 3) Q ( 2 , 3) clearly. note that ( 2+ 3) 3 = 11 2+9 3 thus (11 2+9 3)+( - 9 2 - 9 3) = 2 2 Q ( 2+ 3) and hence 1 2 · 2 2 = 2 Q ( 2 + 3). And therefore 2 + 3 - 2 = 3 Q ( 2 + 3). Hence Q ( 2 , 3) Q ( 2 + 3). Thus Q ( 2 , 3) = Q ( 2 + 3). 3. x 3 - 1 = ( x - 1)( x 2 + x + 1) Q [ x ] and ( x 2 + x + 1) is irreducible over Q . The roots (in C ) of x 2 + x + 1 are - 1 2 ± 3 2 i . Q ( - 1 2 + 3 2 i ) = Q [ x ] / ( x 2 + x + 1) and since x 2 + x + 1 is quadratic, it splits completely in this field extension. Q ( - 1 2 + 3 2 i ) = Q ( 3 i ) ( Q absorbs its own members). So x 3 - 1 splits completely in Q ( 3 i ). 4. Note: x 4 + 1 = ( x 2 + 2 x + 1)( x 2 - 2 x + 1) and its roots are ± 2 ± 2 i 2 . If we extend to Q ( i, 2) we get all these roots. In fact, we can write this as Q ( i + 2). Note: Q ( i, 2) = { a + bi + c 2 + d 2 i | a, b, c, d Q } . 5. x = p 1 + 5 x 2 - 1 = 5 x 4 - 2 x 2 + 1 = 5 x 4 - 2 x 2 - 4 = 0. Thus p 1 + 5 is a root of
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Unformatted text preview: x 4-2 x 2-4 which is irreducible by the Mod-p test for p = 3. Therefore Q ( p 1 + 5) = Q [ x ] / ( x 4-2 x 2-4). 6. F ( a ) = F [ x ] / ( x 3 + x + 1) since x 3 + x + 1 is irreducible in F [ x ]. F [ x ] / ( x 3 + x + 1) = { ax 2 + bx + c + ( x 3 + x + 1) | a, b, c F } thus it has 8 elements. Hence F ( a ) has 8 elements and F ( a ) = { fa 2 + ga + h | f, g, h F } . 1 a a + 1 a 2 a 2 + 1 a 2 + a a 2 + a + 1 1 1 a a + 1 a 2 a 2 + 1 a 2 + a a 2 + a + 1 a a a 2 a 2 + a a + 1 1 a 2 + a + 1 a 2 + 1 a + 1 a + 1 a 2 + a a 2 + 1 a 2 + a + 1 a 2 1 a a 2 a 2 a + 1 a 2 + a + 1 a 2 + a a a 2 + 1 1 a 2 + 1 a 2 + 1 1 a 2 a a 2 + a + 1 a + 1 a 2 + a a 2 + a a 2 + a a 2 + a + 1 1 a 2 + 1 a + 1 a a 2 a 2 + a + 1 a 2 + a + 1 a 2 + 1 a 1 a 2 + a a 2 a + 1...
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