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Finitely Generated k-algebra

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I would be grateful if someone could get me started on the following problem:

"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]
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hint: Any finite set of generators of $k(x)$ only produce a finite number of irreducible factors in the denominators.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Hint: Any finite set of generators of $k(x)$ only produce a finite number of irreducible factors in the denominators.
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
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
This is the idea: suppose for example that $k(x)$ is generated by only one $p_1(x)/q_1(x)\in k(x).$ This means that $k(x)=k\left[p_1(x)/q_1(x)\right].$ But the elements of $k\left[p_1(x)/q_1(x)\right]$ have the form
$$a_n\left(\dfrac{p_1(x)}{q_1(x)}\right)^n+\ldots+a_1\dfrac{p_1(x)}{q_1(x)}+a_0=\dfrac{p(x)}{\left(q_1(x)\right)^n}\quad (a_i\in k,\;p(x)\in k[x])$$
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

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
This is the idea: suppose for example that $k(x)$ is generated by only one $p_1(x)/q_1(x)\in k(x).$ This means that $k(x)=k\left[p_1(x)/q_1(x)\right].$ But the elements of $k\left[p_1(x)/q_1(x)\right]$ have the form
$$a_n\left(\dfrac{p_1(x)}{q_1(x)}\right)^n+\ldots+a_1\dfrac{p_1(x)}{q_1(x)}+a_0=\dfrac{p(x)}{\left(q_1(x)\right)^n}\quad (a_i\in k,\;p(x)\in k[x])$$
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.
Thanks Fernando, I appreciate your help.

Peter