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- #1

- Jun 22, 2012

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"Prove that the field k(x) of rational functions over k in the variable x is not a finitely generated k-algebra." (Dummit and Foote Chapter 15, page 668)

Peter

[This has also been posted on MHF]

- Thread starter Peter
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- Thread starter
- #1

- Jun 22, 2012

- 2,918

"Prove that the field k(x) of rational functions over k in the variable x is not a finitely generated k-algebra." (Dummit and Foote Chapter 15, page 668)

Peter

[This has also been posted on MHF]

- Jan 29, 2012

- 661

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- #3

- Jun 22, 2012

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Thanks Fernando.

My apologies ... I need a little more help ... Can you start me on the formal proof ..

I also need some help as to why irreducible elements enter the picture ...

Peter

- Jan 29, 2012

- 661

$$a_n\left(\dfrac{p_1(x)}{q_1(x)}\right)^n+\ldots+a_1\dfrac{p_1(x)}{q_1(x)}+a_0=\dfrac{p(x)}{\left(q_1(x)\right)^n}\quad (a_i\in k,\;p(x)\in k[x])$$

The denominator has a finite number of irreducible factors. This means that we can't generate $k(x)$ with only one rational fraction (the number of irreducible plolynomials in $k(x)$ is infinite). Try to generalize to any finite number of generators.

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- #5

- Jun 22, 2012

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Thanks Fernando, I appreciate your help.

$$a_n\left(\dfrac{p_1(x)}{q_1(x)}\right)^n+\ldots+a_1\dfrac{p_1(x)}{q_1(x)}+a_0=\dfrac{p(x)}{\left(q_1(x)\right)^n}\quad (a_i\in k,\;p(x)\in k[x])$$

The denominator has a finite number of irreducible factors. This means that we can't generate $k(x)$ with only one rational fraction (the number of irreducible plolynomials in $k(x)$ is infinite). Try to generalize to any finite number of generators.

Peter