Thread: Field Theory - Element u transcendental of F

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

I don't see that implication. I see that Nicholson uses that definition specifically for when $\displaystyle u$ is transcendental, but I do not see any indication by him that it would not be true when $\displaystyle u$ is algebraic. The proof that it is true for when $\displaystyle u$ is algebraic should be identical to the one where it is transcendental, possibly even easier. Given any $\displaystyle f \in F[x]$, can you show that $\displaystyle f(u) \in F[u]$? If so, then you are done.