# Any subring of $K[X]$ that contains $K$ is noetherian; not all of such subrings are UFDs [closed]

Let $K$ be a field. Show that any subring of $K[X]$ that contains $K$ is noetherian. Give an example that demonstrates not all of these subrings are UFDs.

• Were you asking for help with these questions? Could you say how far you’ve gotten so far in considering them? – Lubin Apr 2 '13 at 18:34
• This has been asked before, by the way. Have you tried searching before asking? – Mariano Suárez-Álvarez Apr 2 '13 at 18:51

Let $K\subset R\subset K[X]$. There exists $f\in R$ such that $\deg f\ge 1$. Since $R[X]=K[X]$ and $X$ is integral over $R$ the extension $R\subset K[X]$ is integral and finitely generated, so it is finite. But $K[X]$ is noetherian, and the Eakin-Nagata theorem implies that $R$ is noetherian.
Now take $R=K[X^2,X^3]$. This is not an UFD since $X^2$ is irreducible and not prime ($X^2\mid X^6$, but $X^2\not\mid X^3$.)
Remark. An elementary argument for showing that $R$ is noetherian can be found here.
Let $R$ be the $k$-subalgebra of $k[x_1,x_2]$ generated by all the monomials of the form $x_1 x_2^i$ for $i \in \mathbb{Z}_{\geq 0}$ (so a $k$-linear basis of $R$ is the set of monomials $x_1^i x_2^j$ such that $i \neq 0$ if $j \neq 0$). The ideal of $R$ generated by these monomials is not finitely generated, so $R$ is not Noetherian and the first claim is false. This kind of thing can't happen in just one variable (details upon request).
Here is an example that shows that even for a finitely generated subring, the UFD property isn't inherited: Let $R$ be the subring generated by $K$ and $x_1 x_2$, $x_3 x_4$, $x_1 x_4$, and $x_2 x_3$. Then $x_1 x_2 x_3 x_4=(x_1 x_2)(x_3 x_4)=(x_1 x_4)(x_2 x_3)$ shows that $R$ is not a UFD.