Corollary 5.31. If p(x) is irreducible in F[x] and p[x] divides the product r¡(x)· · · rn(x) for r:(x) E F[x], then p(x) divides r;(x) for at least one i.
Corollary 5.31. If p(x) is irreducible in F[x] and p[x] divides the product r¡(x)· · · rn(x) for r:(x) E F[x], then p(x) divides r;(x) for at least one i.
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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Is it possible to give a proof for this? It is about Unique Factorization. It says that is follows a Mathematical Induction from the above Theorem.
![Theorem 5.30. Let p(x) be an irreducible polynomial in F[x]. If p(x) divides r(x)s(x)
for r(x), s(x) E F[x], then either p(x) divides r(x) or p(x) divides s(x).
Proof. Suppose p(x) divides r(x)s(x). Then r(x)s(x) e< p(x) > which is maximal by
previous theorem. Thus, < p(x) > is a prime ideal which means that either
r(x) E< p(x) > or s(x) E< p(x) >, that is, r(x) = q1 (x)p(x) or s(x) = 92(x)p(x), q1(x), q2(x) E
F[a].
Therefore, p(x) divides srx) or p(x) divides s(x).
Corollary 5.31. If p(x) is irreducible in F[x] and p[æ] divides the product r1(x)· · rn(x)
for r;(x) E F[x], then p(x) divides r;(x) for at least one i.
Proof. Follows from the above Theorem by Mathematical Induction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F564e48ba-306d-4dea-8d90-383ba19026e8%2F81445bb7-7c0b-4876-b903-33d01986a32c%2F4wp2gh_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem 5.30. Let p(x) be an irreducible polynomial in F[x]. If p(x) divides r(x)s(x)
for r(x), s(x) E F[x], then either p(x) divides r(x) or p(x) divides s(x).
Proof. Suppose p(x) divides r(x)s(x). Then r(x)s(x) e< p(x) > which is maximal by
previous theorem. Thus, < p(x) > is a prime ideal which means that either
r(x) E< p(x) > or s(x) E< p(x) >, that is, r(x) = q1 (x)p(x) or s(x) = 92(x)p(x), q1(x), q2(x) E
F[a].
Therefore, p(x) divides srx) or p(x) divides s(x).
Corollary 5.31. If p(x) is irreducible in F[x] and p[æ] divides the product r1(x)· · rn(x)
for r;(x) E F[x], then p(x) divides r;(x) for at least one i.
Proof. Follows from the above Theorem by Mathematical Induction.
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