(2) Let K | F be a field extension and A € M₁ (F). Denote its minimal polynomial by A,F, and denote it by A,K if we consider A as an element of Mn(K). From the definition of minimal polynomials it's clear that μA,K divides A,F in K[x]. Explain why here (as opposed to the situation for mini- mal polynomials of elements of field extensions) we always have μA,K = μA,F. Remark: There are multiple ways to approach this question.
(2) Let K | F be a field extension and A € M₁ (F). Denote its minimal polynomial by A,F, and denote it by A,K if we consider A as an element of Mn(K). From the definition of minimal polynomials it's clear that μA,K divides A,F in K[x]. Explain why here (as opposed to the situation for mini- mal polynomials of elements of field extensions) we always have μA,K = μA,F. Remark: There are multiple ways to approach this question.
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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This is an advanced problem in abstract algebra. Please provide as many details as you can. Thanks!
![(2) Let K| F be a field extension and A € M₁ (F). Denote its minimal
polynomial by A,F, and denote it by MA,K if we consider A as an element
of Mn(K). From the definition of minimal polynomials it's clear that μA,K
divides μA,F in K[r]. Explain why here (as opposed to the situation for mini-
mal polynomials of elements of field extensions) we always have μA,K = μA,F.
Remark: There are multiple ways to approach this question.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2dbf05e8-4609-48ee-b8a3-5dc8c1c2617c%2Fe754f6c7-3872-4ef2-9ecf-9dbff488b4e4%2Fqlyxhq9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(2) Let K| F be a field extension and A € M₁ (F). Denote its minimal
polynomial by A,F, and denote it by MA,K if we consider A as an element
of Mn(K). From the definition of minimal polynomials it's clear that μA,K
divides μA,F in K[r]. Explain why here (as opposed to the situation for mini-
mal polynomials of elements of field extensions) we always have μA,K = μA,F.
Remark: There are multiple ways to approach this question.
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