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August 31st, 2013, 15:01 Finitely Generated k-algebra

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Peter

Finitely Generated k-algebra

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

Re: Finitely Generated k-algebra

Hint: Any finite set of generators of only produce a finite number of irreducible factors in the denominators.

Peter

Re: Finitely Generated k-algebra Quote:

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

Re: Finitely Generated k-algebra

This is the idea: suppose for example that is generated by only one This means

that But the elements of have the form

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

k(x)

Originally Posted by Fernando Revilla

Hint: Any finite set of generators of only produce a finite number of irreducible factors in the

denominators.

k(x)

k(x) p

1

(x)/ (x) ∈ k(x). q

1

k(x) = k[ (x)/ (x)]. p

1

q

1

k[ (x)/ (x)] p

1

q

1

+ … + + = ( ∈ k, p(x) ∈ k[x])

a

n

( p

1

(x) ) (x) q

1

n

a

1

p

1

(x)

(x)

q

1

a

0

p(x)

( (x)) q

1 n

a

i

k(x) k(x)

Finitely Generated k-algebra http://mathhelpboards.com/linear-abstract-algebra-14/finitely...

1 de 2 18/02/14 18:34

(2)

August 31st, 2013, 22:00

Peter

Re: Finitely Generated k-algebra Quote:

Thanks Fernando, I appreciate your help.

Peter

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Originally Posted by Fernando Revilla

This is the idea: suppose for example that is generated by only one

This means that But the elements of have the form

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

k(x) p

1

(x)/ (x) ∈ k(x). q

1

k(x) = k[ (x)/ (x)]. p

1

q

1

k[ (x)/ (x)] p

1

q

1

+ … + + = ( ∈ k, p(x) ∈ k[x])

a

n

( p

1

(x) ) (x) q

1

n

a

1

p

1

(x)

(x)

q

1

a

0

p(x)

( (x)) q

1 n

a

i

k(x) k(x)

Finitely Generated k-algebra http://mathhelpboards.com/linear-abstract-algebra-14/finitely...

2 de 2 18/02/14 18:34

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