22 When we make a field Q(α), where α is an abstract root of a polynomial p(x), we would prefer to avoid messy polynomials like x³±³½³x²+³½³x+½. The point of this exercise is to clean up such polynomials by switching to a different choice of α. For starters, we can assume that p(x) is monic (meaning that it has a leading coefficient of 1, as in the example) just by dividing through by the leading coefficient. (a) Show that if p(x) has degree n and is monic, then we can replace α by some equivalent a' to eliminate the x-1 term, and keep the new polynomial p'(x) monic. (Hint: Let a' = a+r for a suitable rational number r E Q. Also, the prime punctuation here has nothing to do with taking derivatives.) (b) Show that if p(x) is monic, then we can replace a by some equivalent a' so that the new polynomial p'(x) is an integer polynomial, and still monic if p(x) is monic, and without changing which of the terms in the polynomial are non-zero. (Hint: This time rescale α. Try an example like the one given in the problem, or a simpler example such as x³±³½³x+³½³.) +3 (c) If we apply parts (a) and (b) to quadratic polynomials, then we learn that every quadratic field extension of Q can be described as Q(x)/(x² - d), in other words as Q(√d). Show finally that d can be made square-free, meaning that no prime divides d twice.
22 When we make a field Q(α), where α is an abstract root of a polynomial p(x), we would prefer to avoid messy polynomials like x³±³½³x²+³½³x+½. The point of this exercise is to clean up such polynomials by switching to a different choice of α. For starters, we can assume that p(x) is monic (meaning that it has a leading coefficient of 1, as in the example) just by dividing through by the leading coefficient. (a) Show that if p(x) has degree n and is monic, then we can replace α by some equivalent a' to eliminate the x-1 term, and keep the new polynomial p'(x) monic. (Hint: Let a' = a+r for a suitable rational number r E Q. Also, the prime punctuation here has nothing to do with taking derivatives.) (b) Show that if p(x) is monic, then we can replace a by some equivalent a' so that the new polynomial p'(x) is an integer polynomial, and still monic if p(x) is monic, and without changing which of the terms in the polynomial are non-zero. (Hint: This time rescale α. Try an example like the one given in the problem, or a simpler example such as x³±³½³x+³½³.) +3 (c) If we apply parts (a) and (b) to quadratic polynomials, then we learn that every quadratic field extension of Q can be described as Q(x)/(x² - d), in other words as Q(√d). Show finally that d can be made square-free, meaning that no prime divides d twice.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Transcribed Image Text:22
When we make a field Q(α), where α is an abstract root of a polynomial p(x), we would
prefer to avoid messy polynomials like x³±³½³x²+³½³x+½. The point of this exercise is to
clean up such polynomials by switching to a different choice of α. For starters, we can
assume that p(x) is monic (meaning that it has a leading coefficient of 1, as in the example)
just by dividing through by the leading coefficient.
(a) Show that if p(x) has degree n and is monic, then we can replace α by some equivalent
a' to eliminate the x-1 term, and keep the new polynomial p'(x) monic. (Hint: Let
a' = a+r for a suitable rational number r E Q. Also, the prime punctuation here has
nothing to do with taking derivatives.)
(b) Show that if p(x) is monic, then we can replace a by some equivalent a' so that the
new polynomial p'(x) is an integer polynomial, and still monic if p(x) is monic, and
without changing which of the terms in the polynomial are non-zero. (Hint: This time
rescale α. Try an example like the one given in the problem, or a simpler example such
as x³±³½³x+³½³.)
+3
(c) If we apply parts (a) and (b) to quadratic polynomials, then we learn that every quadratic
field extension of Q can be described as Q(x)/(x² - d), in other words as Q(√d). Show
finally that d can be made square-free, meaning that no prime divides d twice.
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