Field Theory - Element u transcendental of F

[Math Amateur]

Field Theory - Element u transcendental over 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, $$F \subseteq E$$ 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]================================================================================================

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]. ... ... "

=================================================================================================

***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

 

[caffeinemachine]

Re: Field Theory - Element u transcendental over 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:

-----------------------------------------------------------------------------------------------------------

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

--------------------------------------------------------------------------------------------------------------------------

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]

Try doing the proofs of (3) and (4) without relying on (1) and (2). (3) is infact trivial. The proof of (4) is also easy but requires a bit more work. To prove (4), let $K$ be a subfield of $E$ which contains $F$ and $\alpha$. Since a field is closed under addition and multiplication, $K$ contains $f(\alpha)$ for all $f(x)\in F[x]$. Also, any field is closed under inverses, so $K$ contains $(g(\alpha))^{-1}$ for all $g(x)\in F[x]$ whenever $g(\alpha)\neq 0$. Lastly, any field is closed under multiplication, thus $f(\alpha)(g(\alpha))^{-1}\in K$ for all $f(x),g(x)\in F[x]$ whenever $g(\alpha)\neq 0$. Thus $K$ contains $F(\alpha)$.

I think this would explain the word 'immediate' above. However, I would suggest that you shouldn't take it too seriously. Proving the four statements independently is good enough.

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]. ... ... "

=================================================================================================

***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

Papantonopoulu and Nicholson have not given the same definition for $F(u)$.
Papanto gives $F(\alpha)=\{f(\alpha)(g(\alpha))^{-1}:f(x),g(x)\in F[x],{g(\alpha)}\neq 0\}$ while Nicholson says that if $\alpha$ is transcendental over $F$ then $F(\alpha)=\{f(\alpha)(g(\alpha))^{-1}:f(x),g(x)\in F[x], {g(x)}\neq 0\}$

(Note that in Papanto's expression of $F(\alpha)$ we have $g(\alpha)\neq 0$ and not $g(x)\neq 0$).

Papanto's def is general; it applies regardless of whether $\alpha$ is algebraic or transcendental over $F$.

Now to see this,
Papanto says that $F(u)=\{f(u)(g(u))^{-1}:f(x),g(x)\in F[x], g(u)\neq 0\}$ regardless of whether $u$ is transcendental over $F$ or not.

When $u$ is transcendental over $F$ then the above definition conincides with that of Nicholson as then $g(u)=0\iff g(x)=0$.

Now assume $u$ is algebraic over $F$. We will show that if $g(u)\neq 0$ then $(g(u))^{-1}\in F$.
Let $m(x)$ be the minimal polynomial of $u$ over $F$. now since $g(u)$ is not zero, $m(x)$ and $g(x)$ must be relatively prime over $F[x]$. Thus there exists $a(x),b(x)$ in $F[x]$ such that $a(x)m(x)+b(x)g(x)=1$. "Substitute" $u$ in place of $x$ in the LHS of the last equation (use the evaluation map to be rigorous) to get $b(u)g(u)=1$. Thus $(g(u))^{-1}=b(u)$ and hence we have shown that $(g(u))^{-1}\in F$.

 

[Math Amateur]

Re: Field Theory - Element u transcendental over F

Try doing the proofs of (3) and (4) without relying on (1) and (2). (3) is infact trivial. The proof of (4) is also easy but requires a bit more work. To prove (4), let $K$ be a subfield of $E$ which contains $F$ and $\alpha$. Since a field is closed under addition and multiplication, $K$ contains $f(\alpha)$ for all $f(x)\in F[x]$. Also, any field is closed under inverses, so $K$ contains $(g(\alpha))^{-1}$ for all $g(x)\in F[x]$ whenever $g(\alpha)\neq 0$. Lastly, any field is closed under multiplication, thus $f(\alpha)(g(\alpha))^{-1}\in K$ for all $f(x),g(x)\in F[x]$ whenever $g(\alpha)\neq 0$. Thus $K$ contains $F(\alpha)$.

I think this would explain the word 'immediate' above. However, I would suggest that you shouldn't take it too seriously. Proving the four statements independently is good enough.Papantonopoulu and Nicholson have not given the same definition for $F(u)$.
Papanto gives $F(\alpha)=\{f(\alpha)(g(\alpha))^{-1}:f(x),g(x)\in F[x],{g(\alpha)}\neq 0\}$ while Nicholson says that if $\alpha$ is transcendental over $F$ then $F(\alpha)=\{f(\alpha)(g(\alpha))^{-1}:f(x),g(x)\in F[x], {g(x)}\neq 0\}$

(Note that in Papanto's expression of $F(\alpha)$ we have $g(\alpha)\neq 0$ and not $g(x)\neq 0$).

Papanto's def is general; it applies regardless of whether $\alpha$ is algebraic or transcendental over $F$.

Now to see this,
Papanto says that $F(u)=\{f(u)(g(u))^{-1}:f(x),g(x)\in F[x], g(u)\neq 0\}$ regardless of whether $u$ is transcendental over $F$ or not.

When $u$ is transcendental over $F$ then the above definition conincides with that of Nicholson as then $g(u)=0\iff g(x)=0$.

Now assume $u$ is algebraic over $F$. We will show that if $g(u)\neq 0$ then $(g(u))^{-1}\in F$.
Let $m(x)$ be the minimal polynomial of $u$ over $F$. now since $g(u)$ is not zero, $m(x)$ and $g(x)$ must be relatively prime over $F[x]$. Thus there exists $a(x),b(x)$ in $F[x]$ such that $a(x)m(x)+b(x)g(x)=1$. "Substitute" $u$ in place of $x$ in the LHS of the last equation (use the evaluation map to be rigorous) to get $b(u)g(u)=1$. Thus $(g(u))^{-1}=b(u)$ and hence we have shown that $(g(u))^{-1}\in F$.

Thanks for the help Caffeinemachine.

Just one issue.

You write:

"When u is transcendental over F then the above definition coincides with \(\displaystyle g(u)=0 \iff g(x)=0\)"

But ... when u is transcendental over F then there is no polynomial in F such that g(u) = 0?

Can you help further?

Peter

 

[caffeinemachine]

Re: Field Theory - Element u transcendental over F

Thanks for the help Caffeinemachine.Just one issue. You write:"When u is transcendental over F then the above definition coincides with \(\displaystyle g(u)=0 \iff g(x)=0\)"But ... when u is transcendental over F then there is no polynomial in F such that g(u) = 0?

No. When $u$ is transcendental over $F$ then there is exactly one polynomial $g(x)$ in $F[x]$ for which $g(u)=0$. The zero polynomial!

Try proving this: Let $g(x)\in\mathbb Q(x)$ be such that $g(\pi)=0$. Show that $g(x)=0$. The converse is obvious.

Tell me if you have any further doubts.

 

[blue_raver22]

The statement that F(u) = { f(u){g(u)}^{-1} | f(x), g(x) in F[x] ; g(x) ≠ 0 } is true regardless of whether u is transcendental or algebraic over F. However, the significance of this statement is different for transcendental and algebraic elements.

For an algebraic element, this statement simply defines F(u) as the field of quotients of the integral domain F[x], which is the smallest subfield of E containing both F and u. This is the same as the statement (4) in Proposition 10.2.2.

For a transcendental element, this statement shows that F(u) is isomorphic to the field of rational functions F(x) over F. This is because any rational function in F(x) can be written as a quotient of two polynomials in F[x], and this is the same structure as the elements in F(u) according to the statement above. Therefore, for a transcendental element, F(u) is not just the field of quotients of F[x], but it is isomorphic to F(x). This is the significance of the statement in Nicholson's book.

In summary, the statement F(u) = { f(u){g(u)}^{-1} | f(x), g(x) in F[x] ; g(x) ≠ 0 } is true for both transcendental and algebraic elements, but it has different implications for each case.