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Unformatted text preview: Math 461 F Spring 2011 Fundamental Theorem of Algebra Drew Armstrong When we proved the impossibility of the classical construction problems, we were in terested in the existence of certain roots of polynomials. The flavor of what we did is contained in the following example: Suppose that a cubic polynomial f ( x ) with rational coefficients has 1 + √ 2 as a root. Applying conjugation in the field extension Q ⊆ Q [ √ 2] we conclude that 1 √ 2 is also a root, and then we can use Descartes’ Factor Theorem to conclude that f ( x ) = x (1 + √ 2) x (1 √ 2) g ( x ) = ( x 2 2 x 3) g ( x ) , for some polynomial g ( x ) of degree 1 with coefficients in Q [ √ 2]. However, if we use long division to divide f ( x ) by the polynomial x 2 2 x 3 we find that g ( x ) in fact has rational coefficients. That is, g ( x ) = ax + b for some a,b ∈ Q , in which case a/b is a root of g ( x ), and hence f ( x ). We conclude that f ( x ) has a rational root. This argument depends vitally on the fact that f ( x ) is cubic; for higher degrees the proof falls apart. In general it is quite hard to tell whether a given polynomial has a root in a given field. (The exception is the field Q in which we can use the Rational Root Test.) Over time people began to suspect that “every” polynomial has a root in the com plex numbers C . The precise statement of this is one of the most famous theorems in mathematics....
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This note was uploaded on 01/08/2012 for the course MATH 561 taught by Professor Armstrong during the Spring '11 term at University of Miami.
 Spring '11
 Armstrong
 Algebra, Factor Theorem, Fundamental Theorem Of Algebra, Polynomials

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