In Section 10.2 Algebraic Extensions in Papantonopoulou: Algebra - Pure and Applied, Proposition 10.2.2 on page 309 (see attachment) reads as follows:
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10.2.2 Proposition
Let E be a field, $\displaystyle F \subseteq E $ a subfield of E, and $\displaystyle \alpha \in E $ an element of E.
In E let
$\displaystyle F[ \alpha ] = \{ f( \alpha ) \ | \ f(x) \in F[x] \} $
$\displaystyle F ( \alpha ) = \{ f ( \alpha ) / g ( \alpha ) \ | \ f(x), g(x) \in F[x] \ , \ g( \alpha ) \ne 0 \} $
Then
(1) $\displaystyle F[ \alpha ] $ is a subring of E containing F and $\displaystyle \alpha $
(2) $\displaystyle F[ \alpha ] $ is the smallest such subring of E
(3) $\displaystyle F( \alpha ) $ is a subfield of E containing F and $\displaystyle \alpha $
(4) $\displaystyle F( \alpha ) $ is the smallest such subfield of E
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Papantonopoulou proves (1) and (2) (see attachment) and then writes:
" ... ... (3) and (4) are immediate from (1) and (2) since $\displaystyle F[ \alpha ] \subseteq E $ and E is a field, $\displaystyle F[ \alpha ] $ is an integral domain, and $\displaystyle F( \alpha ) $ is simply the field of quotients of $\displaystyle F[ \alpha ] $. "
[Note: I do not actually follow this statement - can someone help clarify this "immediate" proof]
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However ...
... in Nicholson: Introduction to Abstract Algebra, Section 6.2 Algebraic Extensions, page 279 (see attachment) we read:
" ... ... If u is transcendental over , it is routine to verify that
$\displaystyle F(u) = \{ f(u){g(u)}^{-1} \ | \ f(x), g(x) in F[x] \ ; \ g(x) \ne 0 $
Hence $\displaystyle F(u) \cong F(x) $ where F(x) is the field of quotients of the integral domain F[x]. ... ... "
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***My problem with the above is that Papantonopoulou and Nicholson both give the same expression for $\displaystyle F( \alpha ) $ but Nicholson implies that the relation $\displaystyle F(u) = \{ f(u){g(u)}^{-1} \ | \ f(x), g(x) in F[x] \ ; \ g(x) \ne 0 \} $ is only the case if u is transcendental???
Can someone please clarify this issue for me.
Peter