# Must $k$-subalgebra of $k[x]$ be finitely generated?

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

Thanks.

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.

• 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 $k$. 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.