TITLE: $K[X^2,X^3]\subset K[X]$ is a Noetherian domain and all its prime ideals are maximal
QUESTION [6 upvotes]: Consider $K$ field and consider the ring $R=K[X^2,X^3]\subset K[X]$. It is clear that $R$ is not a Dedekind domain, since with the element $X$ one immediately see that it is not integrally closed. But $R$ is a Noetherian domain and every non trivial prime ideal is maximal. I have a proof for the last two part, but it is a bit heavy and maybe someone can help me finding a better proof!
Thanks in advance!

REPLY [3 votes]: Merely because $K[X]$ is an integral extension of $K[X^2,X^3]$. 
More precisely, the going up theorem ensures any prime of $K[X^2,X^3]$ is a contraction of $K[X]$, and you must know that an ideal $\mathfrak{m}\subseteq K[X]$ is maximal if and only if $\mathfrak{m}\cap K[X^2,X^3]\subseteq K[X^2,X^3]$ is maximal.