Request explain where has (m(x)) < Kero. Now, let g(x) = Ker , then g(a) = 0 and so p(x)|g(x) in k[x]. Consequently, g(x)= (m²(x)), and hence Ker(m(x)), proving the claim. Now, by the Fundamental Theorem of Ring Homomorphisms, k[x]=Im = k[x]. o por come from
Request explain where has (m(x)) < Kero. Now, let g(x) = Ker , then g(a) = 0 and so p(x)|g(x) in k[x]. Consequently, g(x)= (m²(x)), and hence Ker(m(x)), proving the claim. Now, by the Fundamental Theorem of Ring Homomorphisms, k[x]=Im = k[x]. o por come from
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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![Theorem 7: Let F/k be an extension field and ae F be algebraic over k.
Then the following holds:
i)
k[x]/(m₂(x)) = k[a];
ii)
k[a] =k(α);
iii) [k[a]:k] = deg α.
Proof: Let :k[x]→F:o(f(x))=f(a), be the evaluation map at a. You
should check that is a well-defined ring homomorphism. Let deg α = n.
i)
Since a is algebraic over k, Kero is a non-trivial ideal of k[x]. We
claim that Ker = (m₁(x)).
Since m(a) = 0, and Ker o consists of all the non-zero polynomials
in k[x] satisfied by a, we see that m(x) = Ker . Hence
(m₁ (x)) ≤ Ker o.
C
Request explain where has
Now, let g(x) = Ker , then g(a) = 0 and so p(x)|g(x) in k[x].
Consequently, g(x) = (m(x)), and hence Kero ≤(m(x)), proving
the claim. Now, by the Fundamental Theorem of Ring Homomorphisms,
k[x]/
K[X]{m, (x)) ≈ Im ✨ = k[c].
o
ii) By (1) above, k[x]/(m²(x)) = k[c]. Since m. (x) is reducible,
k[x)/(m₂(x))
coincide with its quotient field, i.e., k[a] = k(α).
is a field. Consequently, k[a] is a field. Therefore, it must
pers
come from](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2Fcaa32bb0-7175-4a24-998d-e1c33d01906a%2Fc9huedp_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem 7: Let F/k be an extension field and ae F be algebraic over k.
Then the following holds:
i)
k[x]/(m₂(x)) = k[a];
ii)
k[a] =k(α);
iii) [k[a]:k] = deg α.
Proof: Let :k[x]→F:o(f(x))=f(a), be the evaluation map at a. You
should check that is a well-defined ring homomorphism. Let deg α = n.
i)
Since a is algebraic over k, Kero is a non-trivial ideal of k[x]. We
claim that Ker = (m₁(x)).
Since m(a) = 0, and Ker o consists of all the non-zero polynomials
in k[x] satisfied by a, we see that m(x) = Ker . Hence
(m₁ (x)) ≤ Ker o.
C
Request explain where has
Now, let g(x) = Ker , then g(a) = 0 and so p(x)|g(x) in k[x].
Consequently, g(x) = (m(x)), and hence Kero ≤(m(x)), proving
the claim. Now, by the Fundamental Theorem of Ring Homomorphisms,
k[x]/
K[X]{m, (x)) ≈ Im ✨ = k[c].
o
ii) By (1) above, k[x]/(m²(x)) = k[c]. Since m. (x) is reducible,
k[x)/(m₂(x))
coincide with its quotient field, i.e., k[a] = k(α).
is a field. Consequently, k[a] is a field. Therefore, it must
pers
come from
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