Proposition 6: Let F/k be an extension field. Let E = {α = F α is algebraic over k}. Then E is a subfield of F.
Proposition 6: Let F/k be an extension field. Let E = {α = F α is algebraic over k}. Then E is a subfield of F.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![Requent explain this proposition 4 proof in
simple teims. Unable to understand the proof
Proposition 6: Let F/k be an extension field. Let
E = {α = F α is algebraic over k}. Then E is a subfield of F.
Proof: Firstly, k≤E, since k/k is algebraic.
Next, let α,ße E and let n,m denote, respectively, their degrees over k.
Then, by Theorem 7, [k(α): k]=n and [k(ß): k] = m. Since ẞ is algebraic
over k and kk(a), ß is also algebraic over k(a). Hence,
[k(a)(B): k(a)] ≤ m. Now, by Theorem 3, it follows that
[k(a)(B): k] = [k(a)(ß): k(a)][k(a): k] ≤mn.
So k(a)(B)/k is a finite extension, and hence an algebraic extension. Since
k(a)(B) is a field containing a and B, it contains a ±ß,aß and a/ß (for
B = 0). Hence all these elements are algebraic over k, and thus, lie in E.
Thus, E is a subfield of F.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2F26baab44-eb0f-48a0-8065-e8b16348ba30%2Foy63yck_processed.png&w=3840&q=75)
Transcribed Image Text:Requent explain this proposition 4 proof in
simple teims. Unable to understand the proof
Proposition 6: Let F/k be an extension field. Let
E = {α = F α is algebraic over k}. Then E is a subfield of F.
Proof: Firstly, k≤E, since k/k is algebraic.
Next, let α,ße E and let n,m denote, respectively, their degrees over k.
Then, by Theorem 7, [k(α): k]=n and [k(ß): k] = m. Since ẞ is algebraic
over k and kk(a), ß is also algebraic over k(a). Hence,
[k(a)(B): k(a)] ≤ m. Now, by Theorem 3, it follows that
[k(a)(B): k] = [k(a)(ß): k(a)][k(a): k] ≤mn.
So k(a)(B)/k is a finite extension, and hence an algebraic extension. Since
k(a)(B) is a field containing a and B, it contains a ±ß,aß and a/ß (for
B = 0). Hence all these elements are algebraic over k, and thus, lie in E.
Thus, E is a subfield of F.
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