$K[X^2,X^3]\subset K[X]$ is a Noetherian domain and all its prime ideals are maximal

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!

Solution 1:

Suppose $\mathfrak{p}$ is a nonzero prime ideal of $A := K[X^{2}, X^{3}]$; we want to show that $\mathfrak{p}$ is maximal. Note that $K[X^{2}]$ is a subring of $A$, and $K[X^{2}]$ is a principal ideal domain, since it is isomorphic to $K[X]$ via the morphism of $K$-algebras $K[X] \to K[X^{2}], X \mapsto X^{2}$. Note that $\mathfrak{m} := \mathfrak{p} \cap K[X^{2}]$ is a prime ideal of $K[X^{2}]$, since it is the preimage of $\mathfrak{p}$ under the inclusion morphism $K[X^{2}] \hookrightarrow A$. If $\mathfrak{m}$ is nonzero, then $\mathfrak{m}$ is maximal, since $K[X^{2}]$ is a PID. Moreover, since the inclusion $K[X^{2}] \hookrightarrow A$ is integral, so too is the induced (injective) morphism $K[X^{2}]/\mathfrak{m} \hookrightarrow A/\mathfrak{p}$. Since $K[X^{2}]/\mathfrak{m}$ is a field and $A$ is a domain, $A$ must be a field as well (this is, e.g., Proposition 5.7 in Atiyah Macdonald).

It therefore suffices to show that $\mathfrak{p} \cap K[X^{2}]$ is nonzero for any nonzero prime ideal $\mathfrak{p}$ of $A$. This amounts to showing that any nonzero $\mathfrak{p}$ contains a polynomial whose monomial terms all have even degree. Take $f(X) \in \mathfrak{p}$ nonzero, and write $f(X) = g(X) + h(X)$, where $g$ has monomial terms of even degree only, and $h$ has monomial terms of odd degree only. Then $f(-X) = g(X) - h(X)$, and $f(-X) \in A$, so $f(-X)f(X) = g(X)^{2} - h(X)^{2} \in \mathfrak{p}$, which clearly has monomial terms of even degree only.

Solution 2:

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.