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Unformatted text preview: Math 302 Modern Algebra Spring 2009 Homework Solutions 4.3 IRREDUCIBLES AND UNIQUE FACTORIZATION 6. Show that x 2 + 1 is irreducible in Q [ x ]. Proof: Use a proof by contradiction. Suppose x 2 + 1 is reducible in Q [ x ]. We then need to obtain a contradiction. Since x 2 + 1 is reducible, then Theorem 4.10 implies that x 2 + 1 can be written as the product of two polynomials of lower degree. Since deg( x 2 + 1) = 2, each of the factors in the product must have degree 1. Therefore x 2 + 1 = ( ax + b )( cx + d ) for some a, b, c, d Q . Multiplying the two factors together, we obtain x 2 + 1 = ( ac ) x 2 + ( ad + bc ) x + bd. Equating coecients, we obtain ac = 1 , ad + bc = 0 , bd = 1 . The first and third equations imply that all of a , b , c , and d are nonzero. Then a = 1 /c and b = 1 /d . Substituting into the second equation, we obtain (1 /c )( d ) + (1 /d )( c ) = 0 d c + c d = 0 d 2 + c 2 cd = 0 d 2 + c 2 = 0 Since both c and d are rational and nonzero, this is a contradiction. Thus x 2 + 1 is irreducible. 9. Find all irreducible polynomials of (b) degree 3 in Z 2 [ x ]. Be sure to indicate why each of these is in fact irreducible. There are eight polynomials of degree 3 in Z 2 [ x ]: x 3 x 3 + x 2 x 3 + x x 3 + 1 x 3 + x 2 + x x 3 + x 2 + 1 x 3 + x + 1 x 3 + x 2 + x + 1 It follows from Theorem 4.10 that a polynomial of degree 3 in Z 2 [ x ] is reducible if and only if it can be written as the product of a polynomial of degree 1 and a polynomial of degree 2 in Z 2 [ x ]. The polynomials of degree 1 or degree 2 in Z 2 [ x ] are x x + 1 x 2 x 2 + x x 2 + 1 x 2 + x + 1 Taking all possible products of a polynomial of degree 1 and a polynomial of degree 2 in Z 2 [ x ], we obtain x ( x 2 ) = x 3 x ( x 2 + 1) = x 3 + x x ( x 2 + x ) = x 3 + x 2...
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This note was uploaded on 10/20/2009 for the course MATH 302 taught by Professor Edwards during the Spring '03 term at CSU Fullerton.
 Spring '03
 Edwards
 Algebra

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