Finally, let f(x) = S. Then, by the division algorithm in k[x], there exist polynomials q(x) and r(x) in k[x] such that f(x) = p(x)q(x) +r(x), where either r(x)=0 or deg r(x) < deg p(x)., Request explain Now if r(x) #0, then f(α) = 0 implies that r(a) = 0. But deg r(x) < deg p(x), which again contradicts the choice of p(x). So r(x) = 0, and hence f(x) = p(x)q(x), i.e., p(x)|f(x). Thus, p(x) f(x)=f(x) = S.
Finally, let f(x) = S. Then, by the division algorithm in k[x], there exist polynomials q(x) and r(x) in k[x] such that f(x) = p(x)q(x) +r(x), where either r(x)=0 or deg r(x) < deg p(x)., Request explain Now if r(x) #0, then f(α) = 0 implies that r(a) = 0. But deg r(x) < deg p(x), which again contradicts the choice of p(x). So r(x) = 0, and hence f(x) = p(x)q(x), i.e., p(x)|f(x). Thus, p(x) f(x)=f(x) = S.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
100%
![Theorem 6: Let F/k be an extension field, and let a F be algebraic over k.
Then there exists a unique least degree non-zero irreducible monic polynomial
p(x) = K[x] s.t. p(x) = 0. Further, if f(a) = 0 for some polynomial
f(x)= k[x], then p(x)|f(x).
Proof: Since a is algebraic over k, there exists a non-zero polynomial g(x)
in k[x] such that g(a) = 0. This shows that the set S = {f(x)= k[x]|f(x)=0}
is non-empty. So we can find p(x) = S of smallest degree, say n. Multiplying
p(x) by the inverse of its leading coefficient (if necessary), we can assume
that p(x) is a monic polynomial. Thus, there exists a monic polynomial p(x)
over k, of least degree, satisfied by a.
Irreducibility: Suppose, to the contrary, that p(x) is reducible over k. Then
there exist non-constant polynomials a(x), b(x) = k[x] with
deg a(x), deg b(x)<n such that p(x)= a(x)b(x). Then
p(a) = 0 ⇒ a(α)b(a) = 0⇒ a(α) = 0 or b(a) = 0.
⇒ a(x) = S or b(x) = S.
In either case, this contradicts the minimality of n. Hence, p(x) must be
irreducible.
Uniqueness: Suppose q(x) = S is another such irreducible monic polynomial
of least degree n. Then h(x) = p(x)-q(x) is a non-zero polynomial
(: p(x) = q(x)) of degree less than n such that h(a)=p(a)-q(a) = 0. This
contradicts the minimality of n. Hence our assumption must be wrong. That
is, p(x) is the unique monic polynomial of least degree satisfied by o..
Finally, let f(x) = S. Then, by the division algorithm in k[x], there exist
polynomials q(x) and r(x) in k[x] such that f(x) = p(x)q(x) +r(x), where
either r(x)= 0 or deg r(x) < deg p(x).
Request
explain
Now if r(x) = 0, then f(x) = 0 implies that r(α) = 0. But
deg r(x) < deg p(x), which again contradicts the choice of p(x). So r(x) = 0,
and hence f(x) = p(x)q(x), i.e., p(x)|f(x).
Thus, p(x)|f(x)=f(x) = S.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2Fbdb4a625-d4c4-4b81-be7f-08fe3b176847%2Ft500fkx_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem 6: Let F/k be an extension field, and let a F be algebraic over k.
Then there exists a unique least degree non-zero irreducible monic polynomial
p(x) = K[x] s.t. p(x) = 0. Further, if f(a) = 0 for some polynomial
f(x)= k[x], then p(x)|f(x).
Proof: Since a is algebraic over k, there exists a non-zero polynomial g(x)
in k[x] such that g(a) = 0. This shows that the set S = {f(x)= k[x]|f(x)=0}
is non-empty. So we can find p(x) = S of smallest degree, say n. Multiplying
p(x) by the inverse of its leading coefficient (if necessary), we can assume
that p(x) is a monic polynomial. Thus, there exists a monic polynomial p(x)
over k, of least degree, satisfied by a.
Irreducibility: Suppose, to the contrary, that p(x) is reducible over k. Then
there exist non-constant polynomials a(x), b(x) = k[x] with
deg a(x), deg b(x)<n such that p(x)= a(x)b(x). Then
p(a) = 0 ⇒ a(α)b(a) = 0⇒ a(α) = 0 or b(a) = 0.
⇒ a(x) = S or b(x) = S.
In either case, this contradicts the minimality of n. Hence, p(x) must be
irreducible.
Uniqueness: Suppose q(x) = S is another such irreducible monic polynomial
of least degree n. Then h(x) = p(x)-q(x) is a non-zero polynomial
(: p(x) = q(x)) of degree less than n such that h(a)=p(a)-q(a) = 0. This
contradicts the minimality of n. Hence our assumption must be wrong. That
is, p(x) is the unique monic polynomial of least degree satisfied by o..
Finally, let f(x) = S. Then, by the division algorithm in k[x], there exist
polynomials q(x) and r(x) in k[x] such that f(x) = p(x)q(x) +r(x), where
either r(x)= 0 or deg r(x) < deg p(x).
Request
explain
Now if r(x) = 0, then f(x) = 0 implies that r(α) = 0. But
deg r(x) < deg p(x), which again contradicts the choice of p(x). So r(x) = 0,
and hence f(x) = p(x)q(x), i.e., p(x)|f(x).
Thus, p(x)|f(x)=f(x) = S.
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