If then , which would imply that the minimal polynomial would be . So, implies , which implies the minimal polynomial must be at least degree 2. Since is a degree 2 polynomial with a root, it is the minimal polynomial.
I am reading Nicholson: Introduction to Abstract Algebra, Section 6.2 - Algebraic Extensions.
Example 15 on page 282 (see attachment) reads as follows:
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Example 15.
Let .
Find , exhibit a -basis of E, and show that . Then find the minimum polynomial of over .
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In the solution we read:
Solution: We write for convenience so that ... ... etc
... ... ... We claim that is the minimal polynomial of over L. Because and are the only roots of in , we merely need to show that . ... ... etc
My problem is the following:
How does showing imply that is the minimal polynomial of over L?
Can someone please help with this issue?
Peter
If then , which would imply that the minimal polynomial would be . So, implies , which implies the minimal polynomial must be at least degree 2. Since is a degree 2 polynomial with a root, it is the minimal polynomial.