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.


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