Theorem 21.2 Uniqueness Property If a is algebraic over a field F, then there is a unique monic irreduci- ble polynomial p(x) in F[x] such that p(a) = 0. Theorem 21.3 Divisibility Property Let a be algebraic over F, and let p(xr) be the minimal polynomial for a over F. If f(x) E F[x] and f(a) = 0, then p(x) divides f(x) in F[x].
Prove Theorem 21.2 and Theorem 21.3.
![Theorem 21.2 Uniqueness Property
If a is algebraic over a field F, then there is a unique monic irreduci-
ble polynomial p(x) in F[x] such that p(a) = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b69fc01-f34d-405a-8d49-06630fb36b27%2Fdaa42115-6b63-460d-a375-4194cdb79b89%2F9sgxxpq.jpeg&w=3840&q=75)
![Theorem 21.3 Divisibility Property
Let a be algebraic over F, and let p(xr) be the minimal polynomial for
a over F. If f(x) E F[x] and f(a) = 0, then p(x) divides f(x) in F[x].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b69fc01-f34d-405a-8d49-06630fb36b27%2Fdaa42115-6b63-460d-a375-4194cdb79b89%2F9acf3y7.jpeg&w=3840&q=75)
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Since you have posted multiple questions, we will provide the solution only to the first question as per our Q&A guidelines. Please repost the remaining questions separately.
As per the question we have to prove that if a is algebraic over a field F then there is a unique monic irreducible polynomial p(x) in the field F[x] such that : p(a) = 0
To prove that we will first show that there exists a monic polynomial p(x) in the field F[x] such that p(a) = 0, then we will prove that the polynomial is irreducible.
As per the question we have : a is algebraic over the field F
Which means the set :
G = {g(x) ∈ F[x] | g(a) = 0 ∀ g(x)}
is an ideal of of F[x] , hence F[x] is a Euclidean domain.
This implies that the ideal G is also a principal domain.
Hence we can write :
G = ⟨p(x)⟩ for some unique monic polynomial p(x) ∈ F[x]
→ ∃p(x) ∈ F[x] : p(a) = 0
Thus there exists a monic polynomial p(x) in F[x] such that p(a) = 0
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