Field Theory - Element u transcendental of F

**Bernhard** · Oct 8th 2013, 09:59 PM

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, $F \subseteq E$ a subfield of E, and $\alpha \in E$ an element of E.

In E let

$F[ \alpha ] = \{ f( \alpha ) \ | \ f(x) \in F[x] \}$

$F ( \alpha ) = \{ f ( \alpha ) / g ( \alpha ) \ | \ f(x), g(x) \in F[x] \ , \ g( \alpha ) \ne 0 \}$

Then

(1) $F[ \alpha ]$ is a subring of E containing F and $\alpha$

(2) $F[ \alpha ]$ is the smallest such subring of E

(3) $F( \alpha )$ is a subfield of E containing F and $\alpha$

(4) $F( \alpha )$ 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 $F[ \alpha ] \subseteq E$ and E is a field, $F[ \alpha ]$ is an integral domain, and $F( \alpha )$ is simply the field of quotients of $F[ \alpha ]$ . "

[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

$F(u) = \{ f(u){g(u)}^{-1} \ | \ f(x), g(x) in F[x] \ ; \ g(x) \ne 0$

Hence $F(u) \cong F(x)$ 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 $F( \alpha )$ but Nicholson implies that the relation $F(u) = \{ f(u){g(u)}^{-1} \ | \ f(x), g(x) in F[x] \ ; \ g(x) \ne 0 \}$ is only the case if u is transcendental???

Can someone please clarify this issue for me.

Peter

**SlipEternal** · Oct 9th 2013, 05:54 AM

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

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