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 $$ g(u)=0 \iff g(x)=0$$"

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 $$ g(u)=0 \iff g(x)=0$$"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.