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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.

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter8: Polynomials
Section8.4: Zeros Of A Polynomial
Problem 19E
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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.
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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