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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.

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  • $\begingroup$ ....not all the subrings are UFD's perhaps? $\endgroup$ – Brent J Apr 2 '13 at 18:12
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    $\begingroup$ Were you asking for help with these questions? Could you say how far you’ve gotten so far in considering them? $\endgroup$ – Lubin Apr 2 '13 at 18:34
  • $\begingroup$ This has been asked before, by the way. Have you tried searching before asking? $\endgroup$ – Mariano Suárez-Álvarez Apr 2 '13 at 18:51
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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.

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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.

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    $\begingroup$ @YACP, yes, I know it is true for $n=1$; see the last sentence of the first paragraph above. Also, as you wrote one can find counterexamples to the second claim already for $n=1$ ( $k[x^2,x^3]$ being the smallest) but there's something I like better about the example I gave. $\endgroup$ – Stephen Apr 2 '13 at 21:11
  • $\begingroup$ It's amusing that someone has downvoted this for no apparent reason. This was posted before the original post was edited down to the one-variable case; maybe the down-voter thought that my answer was a non-sequitur, when in fact it gives a perfectly valid answer to the original question? $\endgroup$ – Stephen Sep 8 '16 at 18:54

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