Suppose $k$ is a field, $A$ is a $k$-subalgebra of the polynomial ring $k[x]$. Must $A$ be a finitely generated $k$-algebra?



2 Answers 2


If $A=k$, nothing need to prove. We assume $f\in A$ monic non-constant, then $k[x]$ is integral over $k[f],$ thus $k[x]$ is a noetherian $k[f]$-module. It follows $A$ is also a noetherian $k[f]$-module, thus $A$ is a finitely generated $k$-algebra.

Remark. If one uses the Artin-Tate lemma we directly get $A$ is a finitely generated $k$-algebra.

  • 3
    $\begingroup$ Great answer! But there's no need to use Artin-Tate's lemma. In fact, since $A$ is finite over $k[f]$ and $k[f]$ is finitely generated over $k$, you have that $A$ is finitely generated over $k$. $\endgroup$
    – Andrea
    Aug 24, 2011 at 21:39

I remembered that I answered this question in my Master's thesis, but unfortunately I do not remember any details. So I just copy from my thesis:

In [1], Robbiano and Sweedler prove that every subalgebra of the polynomial ring in one variable has a finite SAGBI basis (SAGBI = Subalgebra Analog of Gröbner Bases for Ideals). Since a SAGBI basis, as defined in [1], always is a generating set of the subalgebra, it follows that every subalgebra of $k[x]$ is finitely generated.

[1] L. Robbiano and M. Sweedler, Subalgebra bases. In: Commutative algebra (Salvador 1988), pages 61-87, Springer, 1990.

The very obvious question is whether there is a proof that does not use these fancy SAGBI bases. I'll write again if I find an answer to that. The important thing to know about them is, however, that a SAGBI basis for a subalgebra is a generating set with certain properties well-suited for computations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.