$K[X^2,X^3]\subset K[X]$ 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!

• Have you tried considering $R = K[X^2, X^3] \cong \frac{K[Y, Z]}{(Y^3 - Z^2)}$? – André 3000 Jan 22 at 16:25
• @André3000 so the polynomial is irreducible and the ideal generated by it is prime. Moreover, $K[Y,Z]$ should be Noetherian by Hilbert's Basissatz, so the quotient is Noetherian. Right? The dimension $1$ (i.e., all its prime ideal are maximal) is not so evident for me... – Lei Feima Jan 22 at 16:35
• @LeiFeima: $\langle Y^{3}-Z^{2} \rangle$ is a prime ideal of $K[Y, Z]$, and any proper chain of prime ideals of $K[Y, Z]/\langle Y^{3}-Z^{2} \rangle$ lifts to a chain of prime ideals of $K[Y, Z]$ which begins $0 \subset \langle Y^{3}-Z^{2} \rangle \subset \cdots$. Since $K[Y, Z]$ has Krull dimension $2$, it follows that $K[Y, Z]/\langle Y^{3}-Z^{2} \rangle$ has Krull dimension at most $1$. One can see that it has Krull dimension exactly $1$ by (e.g.) considering the projection of the maximal ideal $\langle Y, Z\rangle$ of $K[Y, Z]$. – Alex Wertheim Jan 22 at 18:37
• I also have an alternative approach for showing that $K[X^{2}, X^{3}]$ has dimension $1$ which only uses basic facts on integrality, if you're interested. – Alex Wertheim Jan 22 at 18:39
• @LeiFeima It depends on how elementary you'd like your proof to be. For instance, $R$ is the coordinate ring of a curve, so its dimension is $1$. Or since the ideal $(Y^3 - Z^2)$ is principal, it has codimension $1$ by Krull's Hauptidealsatz. Or as Alex Wertheim suggests, $R$ is an integral (quadratic) extension of $K[Y]$. Integral extensions preserve dimension by the Cohen-Seidenberg theorems, so $R$ has the same dimension as $K[Y]$. I suspect you may be able to reproduce this result for integral quadratic extensions by hand, if you don't want to appeal to these theorems. – André 3000 Jan 22 at 22:26

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.
• Thanks a lot! This is exactly the proof that I had in mind when I posed the question, even if I showed that $p\cap K[X^2]$ is nonempty in a slightly different way... – Lei Feima Jan 23 at 9:00
• Just for the sake of completeness and clarity: I considered the $p$ prime ideal of $R$ and $p':=p\cap K[X^3]$. Since $R$ is integral over $K[X^3]$, I consider a nonzero $c\in p$ such that, for $b_i \in K[X^3]$ with $b_0 \neq 0$, $c^n+c^{n-1}b_{n-1}+...+b_0 =0$. Then $c(c^{n-1}+b_{n-1}c^{n-2}+...+b_1)=-b_0 \in p'$. – Lei Feima Jan 23 at 14:05
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.