Unformatted text preview: x 42 x 24 which is irreducible by the Modp test for p = 3. Therefore Q ( p 1 + √ 5) ∼ = Q [ x ] / ( x 42 x 24). 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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This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.
 Spring '11
 Johnson
 Algebra

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