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An extension field
of field is an algebraic extension of if every element in is algebraic over . -
If an extension field
of a field is of finite dimension as a vector space over , then is a finite extension of degree over . The degree of over is denoted - Note that
.
- Note that
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(Fraleigh 31.3) A finite extension field
of a field is an algebraic extension of . - Proof: Take
. The set forms a linearly dependent set. Their linear combination gives a nonzero polynomial that evaluates to . Therefore is algebraic over .
- Proof: Take
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(Fraleigh 31.4) If
is a finite extension field of and is a finite extension field of , then is a finite extension of and (Fraleigh 31.6) By induction if we have fields
such that is a finite extension of , then is a finite extension of and -
Intuition: If
is a basis for over is a basis for over . Then, is a basis for over . -
(Fraleigh 31.7) If
is an extension field of , is algebraic over and , then divides
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(Fraleigh 31.11) Let
be an algebraic extension of a field . Then there exists a finite number of elements such that if and only if is a finite-dimensional vector space over . That is, if and only if is a finite extension of -
Proof: In the forward case,
is algebraic over . is algebraic over and so on. Now get Which is finite.
Conversely, if
is a finite algebraic extension of , we can keep adding elements where . We do it such that This construction gives us
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(Fraleigh 31.23) If
is a finite extension of field and is a prime number, then is a simple extension of . Thus - Proof: We have for any Note that the factors of a prime
are and . Clearly since , Therefore since . Consequently, and thus
- Proof: We have for any
Links
Footnotes
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Note the analogy with Lagrange’s Theorem and Fraleigh 10.14 in cosets. ↩